certain new grÜss type inequalities involving saigo fractional q-integral operator
TRANSCRIPT
Research ArticleSome New Saigo Type Fractional Integral Inequalities andTheir 119902-Analogues
Junesang Choi1 and Praveen Agarwal2
1 Department of Mathematics Dongguk University Gyeongju 780-714 Republic of Korea2Department of Mathematics Anand International College of Engineering Jaipur-303012 India
Correspondence should be addressed to Junesang Choi junesangmaildonggukackr
Received 6 January 2014 Accepted 21 February 2014 Published 2 April 2014
Academic Editor G Wang
Copyright copy 2014 J Choi and P Agarwal This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
The main objective of this paper is to establish some presumably new Saigo type fractional integral inequalities whose specialcases are shown to yield corresponding inequalities associated with Riemann-Liouville and Erdelyi-Kober type fractional integraloperators respectively We also investigate and present 119902-extensions of the above results and some other presumably new onesRelevant connections of the results presented here with those earlier ones are also indicated
1 Introduction and Preliminaries
Throughout this paper N R C and Zminus0
denote the setsof positive integers real numbers complex numbers andnonpositive integers respectively andN
0
= Ncup0 Considerthe following functional
119879 (119891 119892 119901 119902) = int
119887
119886
119902 (119909) 119889119909int
119887
119886
119901 (119909) 119891 (119909) 119892 (119909) 119889119909
+ int
119887
119886
119901 (119909) 119889119909int
119887
119886
119902 (119909) 119891 (119909) 119892 (119909) 119889119909
minus (int
119887
119886
119902 (119909) 119891 (119909) 119889119909)(int
119887
119886
119901 (119909) 119892 (119909) 119889119909)
minus (int
119887
119886
119901 (119909) 119891 (119909) 119889119909)(int
119887
119886
119902 (119909) 119892 (119909) 119889119909)
(1)
where119891 119892 [119886 119887] rarr R are two integrable functions on [119886 119887]and 119901(119909) and 119902(119909) are positive integrable functions on [119886 119887]If 119891 and 119892 are synchronous on [119886 119887] that is
(119891 (119909) minus 119891 (119910)) (119892 (119909) minus 119892 (119910)) ge 0 (2)
for any 119909 119910 isin [119886 119887] then we have (see eg [1 2])119879 (119891 119892 119901 119902) ge 0 (3)
The inequality in (2) is reversed if 119891 and 119892 are asynchronouson [119886 119887] that is
(119891 (119909) minus 119891 (119910)) (119892 (119909) minus 119892 (119910)) le 0 (4)for any 119909 119910 isin [119886 119887] If 119901(119909) = 119902(119909) for any 119909 119910 isin [119886 119887]we get the Chebyshev inequality (see [3]) Ostrowski [4]established the following generalization of the Chebyshevinequality
If 119891 and 119892 are two differentiable and synchronous func-tions on [119886 119887] and 119901 is a positive integrable function on [119886 119887]with |119891
1015840
(119909)| ge 119898 ge 0 and |1198921015840
(119909)| ge 119903 ge 0 for 119909 isin [119886 119887] thenwe have119879 (119891 119892 119901) = 119879 (119891 119892 119901 119901) ge 119898119903119879 (119909 minus 119886 119909 minus 119886 119901) ge 0
(5)
If 119891 and 119892 are asynchronous on [119886 119887] then we have119879 (119891 119892 119901) le 119898119903119879 (119909 minus 119886 119909 minus 119886 119901) le 0 (6)
If 119891 and 119892 are two differentiable functions on [119886 119887] with|1198911015840
(119909)| le 119872 and |1198921015840(119909)| le 119877 for 119909 isin [119886 119887] and 119901 is a positiveintegrable function on [119886 119887] then we have
1003816100381610038161003816119879 (119891 119892 119901)1003816100381610038161003816 le 119872119877119879 (119909 minus 119886 119909 minus 119886 119901) le 0 (7)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 579260 11 pageshttpdxdoiorg1011552014579260
2 Abstract and Applied Analysis
The functional (1) has attracted many researchersrsquo atten-tion due mainly to diverse applications in numerical quadra-ture transform theory and probability and statistical prob-lems Among those applications the functional (1) has alsobeen employed to yield a number of integral inequalities (seeeg [5ndash12] for a very recent work see also [13])
Here in this paper we aim at establishing certain pre-sumably new inequalities involving Saigo fractional integraloperator (8) whose special cases are shown to yield cor-responding inequalities associated with Riemann-Liouvillefractional integral operator (11) and Erdelyi-Kober fractionalintegral operator (12) We also investigate and present 119902-extensions of our results and some other presumably newones Relevant connections of some of the results presentedhere with those earlier ones are also pointed out
For our purpose we also need to recall the followingdefinitions and some earlier works
Definition 1 A real-valued function 119891(119905) (119905 gt 0) is said to bein the space 119862
120583
(120583 isin R) if there exists a real number 119901 gt 120583
such that 119891(119905) = 119905119901
120601(119905) where 120601(119905) isin 119862(0infin)
Definition 2 A function119891(119905) (119905 gt 0) is said to be in the space119862119899
120583
(119899 isin R) if 119891(119899) isin 119862120583
Definition 3 Let 120572 gt 0 and 120573 120578 isin R Then the Saigo frac-tional integral 119868120572120573120578
0119905
(in terms of the Gauss hypergeometricfunction) of order 120572 for a real-valued continuous function119891(119905) is defined by (see [14] see also [15])
119868120572120573120578
0119905
119891 (119905) =119905minus120572minus120573
Γ (120572)int
119905
0
(119905 minus 120591)120572minus1
2
1198651
times (120572 + 120573 minus120578 120572 1 minus120591
119905)119891 (120591) 119889120591
(8)
where the function2
1198651
(sdot) is the Gaussian hypergeometricfunction defined by (see eg [16 Section 15])
2
1198651
(119886 119887 119888 119905) =
infin
sum
119899=0
(119886)119899
(119887)119899
(119888)119899
119905119899
119899(9)
and Γ(120572) is the familiar Gamma function Here (119886)119899
is thePochhammer symbol defined for 119886 isin C by (see eg [16 p2 and pp 4ndash6])
(119886)119899
= 1 (119899 = 0)
119886 (119886 + 1) sdot sdot sdot (119886 + 119899 minus 1) (119899 isin N)
=Γ (119886 + 119899)
Γ (119886)(119886 isin C Z
minus
0
)
(10)
It is noted that Saigo fractional integral operator 1198681205721205731205780119905
includes both Riemann-Liouville and Erdelyi-Kober frac-tional integral operators respectively given by the followingrelationships (see eg [17]) for 119891 isin 119862
120583
(120583 ge minus1)
119868120572
0119905
119891 (119905) (= 119868120572minus120572120578
0119905
119891 (119905))
=1
Γ (120572)int
119905
0
(119905 minus 120591)120572minus1
119891 (120591) 119889120591 (120572 gt 0)
(11)
119868120572120578
0119905
119891 (119905) (= 1198681205720120578
0119905
119891 (119905))
=119905minus120572minus120578
Γ (120572)int
119905
0
(119905 minus 120591)120572minus1
120591120578
119891 (120591) 119889120591
(120572 gt 0 120578 isin R)
(12)
For our purpose we also recall the following definitions(see eg [16 Section 6]) and some earlier works
The 119902-shifted factorial (119886 119902)119899
is defined by
(119886 119902)119899
=
1 (119899 = 0)
119899minus1
prod
119896=0
(1 minus 119886119902119896
) (119899 isin N) (13)
where 119886 119902 isin C and it is assumed that 119886 = 119902minus119898
(119898 isin N0
)The 119902-shifted factorial for negative subscript is defined by
(119886 119902)minus119899
=1
(1 minus 119886119902minus1) (1 minus 119886119902minus2) sdot sdot sdot (1 minus 119886119902minus119899)
(119899 isin N0
)
(14)
We also write
(119886 119902)infin
=
infin
prod
119896=0
(1 minus 119886119902119896
) (119886 119902 isin C10038161003816100381610038161199021003816100381610038161003816 lt 1) (15)
It follows from (13) (14) and (15) that
(119886 119902)119899
=(119886 119902)infin
(119886119902119899 119902)infin
(119899 isin Z) (16)
which can be extended to 119899 = 120572 isin C as follows
(119886 119902)120572
=(119886 119902)infin
(119886119902120572 119902)infin
(120572 isin C10038161003816100381610038161199021003816100381610038161003816 lt 1) (17)
where the principal value of 119902120572 is takenWe begin by noting that F J Jackson was the first to
develop 119902-calculus in a systematic way The 119902-derivative of afunction 119891(119905) is defined by
119863119902
119891 (119905) =119889119902
119889119902
119905119891 (119905) =
119891 (119902119905) minus 119891 (119905)
(119902 minus 1) 119905 (18)
It is noted that
lim119902rarr1
119863119902
119891 (119905) =119889
119889119905119891 (119905) (19)
if 119891(119905) is differentiable
Abstract and Applied Analysis 3
The function 119865(119905) is a 119902-antiderivative of 119891(119905) if119863119902
119865(119905) = 119891(119905) It is denoted by
int119891 (119905) 119889119902
119905 (20)
The Jackson integral of 119891(119905) is thus defined formally by
int119891 (119905) 119889119902
119905 = (1 minus 119902) 119905
infin
sum
119895=0
119902119895
119891 (119902119895
119905) (21)
which can be easily generalized as follows
int119891 (119905) 119889119902
119892 (119905) =
infin
sum
119895=0
119891 (119902119895
119905) (119892 (119902119895
119905) minus 119892 (119902119895+1
119905)) (22)
Suppose that 0 lt 119886 lt 119887 The definite 119902-integral is definedas follows
int
119887
0
119891 (119905) 119889119902
119905 = (1 minus 119902) 119887
infin
sum
119895=0
119902119895
119891 (119902119895
119887) (23)
int
119887
119886
119891 (119905) 119889119902
119905 = int
119887
0
119891 (119905) 119889119902
119905 minus int
119886
0
119891 (119905) 119889119902
119905 (24)
A more general version of (23) is given by
int
119887
0
119891 (119905) 119889119902
119892 (119905) =
infin
sum
119895=0
119891 (119902119895
119887) (119892 (119902119895
119887) minus 119892 (119902119895+1
119887)) (25)
The classical Gamma function Γ(119911) (see eg [16 Section11]) was found by Euler while he was trying to extend thefactorial 119899 = Γ(119899 + 1) ( 119899 isin N
0
) to real numbers The 119902-factorial function [119899]
119902
(119899 isin N0
) of 119899 defined by
[119899]119902
= 1 if 119899 = 0
[119899]119902
[119899 minus 1]119902
sdot sdot sdot [2]119902
[1]119902
if 119899 isin N(26)
can be rewritten as follows
(1 minus 119902)minus119899
infin
prod
119896=0
(1 minus 119902119896+1
)
(1 minus 119902119896+1+119899)=
(119902 119902)infin
(119902119899+1 119902)infin
(1 minus 119902)minus119899
= Γ119902
(119899 + 1) (0 lt 119902 lt 1)
(27)
By replacing 119899 by 119886 minus 1 in (27) Jackson [18] defined the 119902-Gamma function Γ
119902
(119886) by
Γ119902
(119886) =(119902 119902)infin
(119902119886 119902)infin
(1 minus 119902)1minus119886
(0 lt 119902 lt 1) (28)
The 119902-analogue of (119905 minus 119886)119899 is defined by the polynomial
(119905 minus 119886)119899
119902
= 1 (119899 = 0)
(119905 minus 119886) (119905 minus 119902119886) sdot sdot sdot (119905 minus 119902119899minus1
119886) (119899 isin N)
= 119905119899
(119886
119905 119902)119899
(119899 isin N0
)
(29)
Definition 4 Let R(120572) gt 0 let 120573 and 120578 be real or complexnumbers Then a 119902-analogue of Saigorsquos fractional integral119868120572120573120578
119902
is given for |120591119905| lt 1 by (see [19 p 172 Equation (21)])
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times int
119905
0
(119902120591
119905 119902)120572minus1
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
sdot 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119891 (120591) 119889119902
120591
(30)
The integral operator 119868120572120573120578119902
includes both the 119902-analoguesof the Riemann-Liouville and Erdelyi-Kober fractional inte-gral operators given by the following definitions
Definition 5 A 119902-analogue of Riemann-Liouville fractionalintegral operator of a function 119891(119905) of an order 120572 is given by(see [20])
119868120572
119902
119891 (119905) (= 119868120572minus120572120578
119902
119891 (119905))
=119905120572minus1
Γ119902
(120572)int
119905
0
(119902120591
119905 119902)120572minus1
119891 (120591) 119889119902
120591
(120572 gt 0 0 lt 119902 lt 1)
(31)
where (119886 119902)120572
is given by (17)
Definition 6 A 119902-analogue of the Erdelyi-Kober fractionalintegral operator for 120572 gt 0 120578 isin R and 0 lt 119902 lt 1 is givenby (see [20])
119868120578120572
119902
119891 (119905) (= 1198681205720120578
119902
119891 (119905))
=119905minus120578minus1
Γ119902
(120572)int
119905
0
(119902120591
119905 119902)120572minus1
120591120578
119891 (120591) 119889119902
120591
(120572 gt 0 0 lt 119902 lt 1)
(32)
2 Certain Inequalities Involving SaigoFractional Integral Operator
Here we start with presenting an inequality involving Saigofractional integral (8) asserted by the following lemma
Lemma 7 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119906 (119905) 119868120572120573120578
0119905
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119868120572120573120578
0119905
V (119905) 119892 (119905)
+ 119868120572120573120578
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(33)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
4 Abstract and Applied Analysis
Proof Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) Then for all 120591 120588 isin (0 119905) with 119905 gt 0we have
(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) ge 0 (34)
or equivalently
119891 (120591) 119892 (120591) + 119891 (120588) 119892 (120588) ge 119891 (120591) 119892 (120588) + 119891 (120588) 119892 (120591) (35)
Consider the following function 119865(119905 120591) defined by
119865 (119905 120591) =119905minus120572minus120573
(119905 minus 120591)120572minus1
Γ (120572)2
1198651
(120572 + 120573 minus120578 120572 1 minus120591
119905)
(120591 isin (0 119905) 119905 gt 0)
=1
Γ (120572)
(119905 minus 120591)120572minus1
119905120572+120573+(120572 + 120573) (minus120578)
Γ (120572 + 1)
(119905 minus 120591)120572
119905120572+120573+1
+(120572+120573) (120572+120573+1) (minus120578) (minus120578+1)
Γ (120572+2)
(119905minus120591)120572+1
119905120572+120573+2+sdot sdot sdot
(36)
We observe that each term of the above series is nonnegativeunder the conditions in Lemma 7 and hence the function119865(119905 120591) remains nonnegative for all 120591 isin (0 119905) (119905 gt 0)
Now by multiplying both sides of (35) by 119865(119905 120591)119906(120591)
defined by (36) integrating the resulting inequality withrespect to 120591 from 0 to 119905 and using (8) we get
119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
0119905
119906 (119905)
ge 119892 (120588) 119868120572120573120578
0119905
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(37)
Next bymultiplying both sides of (37) by119865(119905 120588)V(120588) (120588 isin
(0 119905) 119905 gt 0) where 119865(119905 120588) is given when 120591 is replaced by 120588
in (36) integrating the resulting inequality with respect to 120588
from 0 to 119905 and using (8) we are led to the desired result (33)
Theorem 8 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(38)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in Lemma 7 we get
119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(39)
Since 119868120572120573120578
0119905
119897(119905) ≧ 0 under the given conditions bymultiplying both sides of (39) by 119868120572120573120578
0119905
119897(119905) we have
119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(40)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897119898 respectivelyin (33) and then multiplying both sides of the resultinginequalities by 119868120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) both of which arenonnegative under the given assumptions respectively weget the following inequalities
119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
Abstract and Applied Analysis 5
119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(41)
Finally by adding (40) and (41) side by side we arrive at thedesired result (38)
We present another inequality involving the Saigo frac-tional integral operator in (8) asserted by the followinglemma
Lemma 9 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120574120575120577
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119906 (119905)
ge 119868120574120575120577
0119905
V (119905) 119892 (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(42)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By multiplying both sides of (37) by
119905minus120574minus120575
(119905 minus 120588)120574minus1
Γ (120574)2
1198651
(120574 + 120575 minus120577 120574 1 minus120588
119905) V (120588)
(120588 isin (0 119905) 119905 gt 0)
(43)
which remains nonnegative under the conditions in (42)integrating the resulting inequality with respect to 120588 from 0
to 119905 and using (8) we get the desired result (42)
Theorem 10 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119897 (119905) [2119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905)]
(44)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in (42) we have
119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)
ge 119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(45)
By multiplying both sides of (45) by 1198681205721205731205780119905
119897(119905) after a littlesimplification we get
119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(46)
Now by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (42)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) respectively weget the following two inequalities
119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
6 Abstract and Applied Analysis
119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(47)
Finally we find that the inequality (44) follows by adding theinequalities (46) and (47) side by side
Remark 11 It may be noted that the inequalities (38) and(44) in Theorems 8 and 10 respectively are reversed if thefunctions are asynchronous on [0infin) The special case of(44) in Theorem 10 when 120572 = 120574 120573 = 120575 and 120578 = 120577 is easilyseen to yield the inequality (38) in Theorem 8
Here we derive certain presumably new integralinequalities by setting 120573 = 0 in (38) and 120573 = 0 = 120575 in (44)respectively and applying the integral operator (12) to theresulting inequalities we obtain two integral inequalitiesinvolving Erdelyi-Kober fractional integral operators statedin Corollaries 12 and 13 below
Corollary 12 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (t) 1198681205721205780119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
(48)
for all 119905 gt 0 120572 gt 0 and 120578 isin R
Corollary 13 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120578
0119905
119897 (119905) [2119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120577
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905)
+119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905)]
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119898 (t) 119891 (119905)]
(49)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Remark 14 The special cases 120573 = minus120572 in Theorem 8 and 120573 =
minus120572 and 120575 = minus120574 in Theorem 10 are seen to yield the knownfractional integral inequalities due to Dahmani [21]
3 Saigo Fractional 119902-Integral Inequalities
We establish certain 119902-integral inequalities which are the 119902-analogues (or extensions) of the results derived in the preced-ing section some of which are presumably new ones For ourpurpose we begin with providing comparison properties forthe fractional 119902-integral operators asserted by the followinglemma
Lemma 15 Let 0 lt 119902 lt 1 and 119891 [0infin) rarr R be continuousfunctions with 119891(119905) ge 0 for all 119905 isin [0infin) Then we have thefollowing inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (30)
119868120572120573120578
119902
119891 (119905) ge 0 (50)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0(ii) the 119902-analogue of Riemann-Liouville fractional integral
operator of the function 119891(119905) of an order 120572 in (31)
119868120572
119902
119891 (119905) ge 0 (51)
for all 120572 gt 0
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
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OptimizationJournal of
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International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
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Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
The functional (1) has attracted many researchersrsquo atten-tion due mainly to diverse applications in numerical quadra-ture transform theory and probability and statistical prob-lems Among those applications the functional (1) has alsobeen employed to yield a number of integral inequalities (seeeg [5ndash12] for a very recent work see also [13])
Here in this paper we aim at establishing certain pre-sumably new inequalities involving Saigo fractional integraloperator (8) whose special cases are shown to yield cor-responding inequalities associated with Riemann-Liouvillefractional integral operator (11) and Erdelyi-Kober fractionalintegral operator (12) We also investigate and present 119902-extensions of our results and some other presumably newones Relevant connections of some of the results presentedhere with those earlier ones are also pointed out
For our purpose we also need to recall the followingdefinitions and some earlier works
Definition 1 A real-valued function 119891(119905) (119905 gt 0) is said to bein the space 119862
120583
(120583 isin R) if there exists a real number 119901 gt 120583
such that 119891(119905) = 119905119901
120601(119905) where 120601(119905) isin 119862(0infin)
Definition 2 A function119891(119905) (119905 gt 0) is said to be in the space119862119899
120583
(119899 isin R) if 119891(119899) isin 119862120583
Definition 3 Let 120572 gt 0 and 120573 120578 isin R Then the Saigo frac-tional integral 119868120572120573120578
0119905
(in terms of the Gauss hypergeometricfunction) of order 120572 for a real-valued continuous function119891(119905) is defined by (see [14] see also [15])
119868120572120573120578
0119905
119891 (119905) =119905minus120572minus120573
Γ (120572)int
119905
0
(119905 minus 120591)120572minus1
2
1198651
times (120572 + 120573 minus120578 120572 1 minus120591
119905)119891 (120591) 119889120591
(8)
where the function2
1198651
(sdot) is the Gaussian hypergeometricfunction defined by (see eg [16 Section 15])
2
1198651
(119886 119887 119888 119905) =
infin
sum
119899=0
(119886)119899
(119887)119899
(119888)119899
119905119899
119899(9)
and Γ(120572) is the familiar Gamma function Here (119886)119899
is thePochhammer symbol defined for 119886 isin C by (see eg [16 p2 and pp 4ndash6])
(119886)119899
= 1 (119899 = 0)
119886 (119886 + 1) sdot sdot sdot (119886 + 119899 minus 1) (119899 isin N)
=Γ (119886 + 119899)
Γ (119886)(119886 isin C Z
minus
0
)
(10)
It is noted that Saigo fractional integral operator 1198681205721205731205780119905
includes both Riemann-Liouville and Erdelyi-Kober frac-tional integral operators respectively given by the followingrelationships (see eg [17]) for 119891 isin 119862
120583
(120583 ge minus1)
119868120572
0119905
119891 (119905) (= 119868120572minus120572120578
0119905
119891 (119905))
=1
Γ (120572)int
119905
0
(119905 minus 120591)120572minus1
119891 (120591) 119889120591 (120572 gt 0)
(11)
119868120572120578
0119905
119891 (119905) (= 1198681205720120578
0119905
119891 (119905))
=119905minus120572minus120578
Γ (120572)int
119905
0
(119905 minus 120591)120572minus1
120591120578
119891 (120591) 119889120591
(120572 gt 0 120578 isin R)
(12)
For our purpose we also recall the following definitions(see eg [16 Section 6]) and some earlier works
The 119902-shifted factorial (119886 119902)119899
is defined by
(119886 119902)119899
=
1 (119899 = 0)
119899minus1
prod
119896=0
(1 minus 119886119902119896
) (119899 isin N) (13)
where 119886 119902 isin C and it is assumed that 119886 = 119902minus119898
(119898 isin N0
)The 119902-shifted factorial for negative subscript is defined by
(119886 119902)minus119899
=1
(1 minus 119886119902minus1) (1 minus 119886119902minus2) sdot sdot sdot (1 minus 119886119902minus119899)
(119899 isin N0
)
(14)
We also write
(119886 119902)infin
=
infin
prod
119896=0
(1 minus 119886119902119896
) (119886 119902 isin C10038161003816100381610038161199021003816100381610038161003816 lt 1) (15)
It follows from (13) (14) and (15) that
(119886 119902)119899
=(119886 119902)infin
(119886119902119899 119902)infin
(119899 isin Z) (16)
which can be extended to 119899 = 120572 isin C as follows
(119886 119902)120572
=(119886 119902)infin
(119886119902120572 119902)infin
(120572 isin C10038161003816100381610038161199021003816100381610038161003816 lt 1) (17)
where the principal value of 119902120572 is takenWe begin by noting that F J Jackson was the first to
develop 119902-calculus in a systematic way The 119902-derivative of afunction 119891(119905) is defined by
119863119902
119891 (119905) =119889119902
119889119902
119905119891 (119905) =
119891 (119902119905) minus 119891 (119905)
(119902 minus 1) 119905 (18)
It is noted that
lim119902rarr1
119863119902
119891 (119905) =119889
119889119905119891 (119905) (19)
if 119891(119905) is differentiable
Abstract and Applied Analysis 3
The function 119865(119905) is a 119902-antiderivative of 119891(119905) if119863119902
119865(119905) = 119891(119905) It is denoted by
int119891 (119905) 119889119902
119905 (20)
The Jackson integral of 119891(119905) is thus defined formally by
int119891 (119905) 119889119902
119905 = (1 minus 119902) 119905
infin
sum
119895=0
119902119895
119891 (119902119895
119905) (21)
which can be easily generalized as follows
int119891 (119905) 119889119902
119892 (119905) =
infin
sum
119895=0
119891 (119902119895
119905) (119892 (119902119895
119905) minus 119892 (119902119895+1
119905)) (22)
Suppose that 0 lt 119886 lt 119887 The definite 119902-integral is definedas follows
int
119887
0
119891 (119905) 119889119902
119905 = (1 minus 119902) 119887
infin
sum
119895=0
119902119895
119891 (119902119895
119887) (23)
int
119887
119886
119891 (119905) 119889119902
119905 = int
119887
0
119891 (119905) 119889119902
119905 minus int
119886
0
119891 (119905) 119889119902
119905 (24)
A more general version of (23) is given by
int
119887
0
119891 (119905) 119889119902
119892 (119905) =
infin
sum
119895=0
119891 (119902119895
119887) (119892 (119902119895
119887) minus 119892 (119902119895+1
119887)) (25)
The classical Gamma function Γ(119911) (see eg [16 Section11]) was found by Euler while he was trying to extend thefactorial 119899 = Γ(119899 + 1) ( 119899 isin N
0
) to real numbers The 119902-factorial function [119899]
119902
(119899 isin N0
) of 119899 defined by
[119899]119902
= 1 if 119899 = 0
[119899]119902
[119899 minus 1]119902
sdot sdot sdot [2]119902
[1]119902
if 119899 isin N(26)
can be rewritten as follows
(1 minus 119902)minus119899
infin
prod
119896=0
(1 minus 119902119896+1
)
(1 minus 119902119896+1+119899)=
(119902 119902)infin
(119902119899+1 119902)infin
(1 minus 119902)minus119899
= Γ119902
(119899 + 1) (0 lt 119902 lt 1)
(27)
By replacing 119899 by 119886 minus 1 in (27) Jackson [18] defined the 119902-Gamma function Γ
119902
(119886) by
Γ119902
(119886) =(119902 119902)infin
(119902119886 119902)infin
(1 minus 119902)1minus119886
(0 lt 119902 lt 1) (28)
The 119902-analogue of (119905 minus 119886)119899 is defined by the polynomial
(119905 minus 119886)119899
119902
= 1 (119899 = 0)
(119905 minus 119886) (119905 minus 119902119886) sdot sdot sdot (119905 minus 119902119899minus1
119886) (119899 isin N)
= 119905119899
(119886
119905 119902)119899
(119899 isin N0
)
(29)
Definition 4 Let R(120572) gt 0 let 120573 and 120578 be real or complexnumbers Then a 119902-analogue of Saigorsquos fractional integral119868120572120573120578
119902
is given for |120591119905| lt 1 by (see [19 p 172 Equation (21)])
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times int
119905
0
(119902120591
119905 119902)120572minus1
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
sdot 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119891 (120591) 119889119902
120591
(30)
The integral operator 119868120572120573120578119902
includes both the 119902-analoguesof the Riemann-Liouville and Erdelyi-Kober fractional inte-gral operators given by the following definitions
Definition 5 A 119902-analogue of Riemann-Liouville fractionalintegral operator of a function 119891(119905) of an order 120572 is given by(see [20])
119868120572
119902
119891 (119905) (= 119868120572minus120572120578
119902
119891 (119905))
=119905120572minus1
Γ119902
(120572)int
119905
0
(119902120591
119905 119902)120572minus1
119891 (120591) 119889119902
120591
(120572 gt 0 0 lt 119902 lt 1)
(31)
where (119886 119902)120572
is given by (17)
Definition 6 A 119902-analogue of the Erdelyi-Kober fractionalintegral operator for 120572 gt 0 120578 isin R and 0 lt 119902 lt 1 is givenby (see [20])
119868120578120572
119902
119891 (119905) (= 1198681205720120578
119902
119891 (119905))
=119905minus120578minus1
Γ119902
(120572)int
119905
0
(119902120591
119905 119902)120572minus1
120591120578
119891 (120591) 119889119902
120591
(120572 gt 0 0 lt 119902 lt 1)
(32)
2 Certain Inequalities Involving SaigoFractional Integral Operator
Here we start with presenting an inequality involving Saigofractional integral (8) asserted by the following lemma
Lemma 7 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119906 (119905) 119868120572120573120578
0119905
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119868120572120573120578
0119905
V (119905) 119892 (119905)
+ 119868120572120573120578
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(33)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
4 Abstract and Applied Analysis
Proof Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) Then for all 120591 120588 isin (0 119905) with 119905 gt 0we have
(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) ge 0 (34)
or equivalently
119891 (120591) 119892 (120591) + 119891 (120588) 119892 (120588) ge 119891 (120591) 119892 (120588) + 119891 (120588) 119892 (120591) (35)
Consider the following function 119865(119905 120591) defined by
119865 (119905 120591) =119905minus120572minus120573
(119905 minus 120591)120572minus1
Γ (120572)2
1198651
(120572 + 120573 minus120578 120572 1 minus120591
119905)
(120591 isin (0 119905) 119905 gt 0)
=1
Γ (120572)
(119905 minus 120591)120572minus1
119905120572+120573+(120572 + 120573) (minus120578)
Γ (120572 + 1)
(119905 minus 120591)120572
119905120572+120573+1
+(120572+120573) (120572+120573+1) (minus120578) (minus120578+1)
Γ (120572+2)
(119905minus120591)120572+1
119905120572+120573+2+sdot sdot sdot
(36)
We observe that each term of the above series is nonnegativeunder the conditions in Lemma 7 and hence the function119865(119905 120591) remains nonnegative for all 120591 isin (0 119905) (119905 gt 0)
Now by multiplying both sides of (35) by 119865(119905 120591)119906(120591)
defined by (36) integrating the resulting inequality withrespect to 120591 from 0 to 119905 and using (8) we get
119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
0119905
119906 (119905)
ge 119892 (120588) 119868120572120573120578
0119905
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(37)
Next bymultiplying both sides of (37) by119865(119905 120588)V(120588) (120588 isin
(0 119905) 119905 gt 0) where 119865(119905 120588) is given when 120591 is replaced by 120588
in (36) integrating the resulting inequality with respect to 120588
from 0 to 119905 and using (8) we are led to the desired result (33)
Theorem 8 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(38)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in Lemma 7 we get
119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(39)
Since 119868120572120573120578
0119905
119897(119905) ≧ 0 under the given conditions bymultiplying both sides of (39) by 119868120572120573120578
0119905
119897(119905) we have
119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(40)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897119898 respectivelyin (33) and then multiplying both sides of the resultinginequalities by 119868120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) both of which arenonnegative under the given assumptions respectively weget the following inequalities
119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
Abstract and Applied Analysis 5
119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(41)
Finally by adding (40) and (41) side by side we arrive at thedesired result (38)
We present another inequality involving the Saigo frac-tional integral operator in (8) asserted by the followinglemma
Lemma 9 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120574120575120577
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119906 (119905)
ge 119868120574120575120577
0119905
V (119905) 119892 (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(42)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By multiplying both sides of (37) by
119905minus120574minus120575
(119905 minus 120588)120574minus1
Γ (120574)2
1198651
(120574 + 120575 minus120577 120574 1 minus120588
119905) V (120588)
(120588 isin (0 119905) 119905 gt 0)
(43)
which remains nonnegative under the conditions in (42)integrating the resulting inequality with respect to 120588 from 0
to 119905 and using (8) we get the desired result (42)
Theorem 10 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119897 (119905) [2119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905)]
(44)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in (42) we have
119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)
ge 119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(45)
By multiplying both sides of (45) by 1198681205721205731205780119905
119897(119905) after a littlesimplification we get
119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(46)
Now by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (42)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) respectively weget the following two inequalities
119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
6 Abstract and Applied Analysis
119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(47)
Finally we find that the inequality (44) follows by adding theinequalities (46) and (47) side by side
Remark 11 It may be noted that the inequalities (38) and(44) in Theorems 8 and 10 respectively are reversed if thefunctions are asynchronous on [0infin) The special case of(44) in Theorem 10 when 120572 = 120574 120573 = 120575 and 120578 = 120577 is easilyseen to yield the inequality (38) in Theorem 8
Here we derive certain presumably new integralinequalities by setting 120573 = 0 in (38) and 120573 = 0 = 120575 in (44)respectively and applying the integral operator (12) to theresulting inequalities we obtain two integral inequalitiesinvolving Erdelyi-Kober fractional integral operators statedin Corollaries 12 and 13 below
Corollary 12 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (t) 1198681205721205780119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
(48)
for all 119905 gt 0 120572 gt 0 and 120578 isin R
Corollary 13 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120578
0119905
119897 (119905) [2119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120577
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905)
+119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905)]
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119898 (t) 119891 (119905)]
(49)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Remark 14 The special cases 120573 = minus120572 in Theorem 8 and 120573 =
minus120572 and 120575 = minus120574 in Theorem 10 are seen to yield the knownfractional integral inequalities due to Dahmani [21]
3 Saigo Fractional 119902-Integral Inequalities
We establish certain 119902-integral inequalities which are the 119902-analogues (or extensions) of the results derived in the preced-ing section some of which are presumably new ones For ourpurpose we begin with providing comparison properties forthe fractional 119902-integral operators asserted by the followinglemma
Lemma 15 Let 0 lt 119902 lt 1 and 119891 [0infin) rarr R be continuousfunctions with 119891(119905) ge 0 for all 119905 isin [0infin) Then we have thefollowing inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (30)
119868120572120573120578
119902
119891 (119905) ge 0 (50)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0(ii) the 119902-analogue of Riemann-Liouville fractional integral
operator of the function 119891(119905) of an order 120572 in (31)
119868120572
119902
119891 (119905) ge 0 (51)
for all 120572 gt 0
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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OptimizationJournal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
The function 119865(119905) is a 119902-antiderivative of 119891(119905) if119863119902
119865(119905) = 119891(119905) It is denoted by
int119891 (119905) 119889119902
119905 (20)
The Jackson integral of 119891(119905) is thus defined formally by
int119891 (119905) 119889119902
119905 = (1 minus 119902) 119905
infin
sum
119895=0
119902119895
119891 (119902119895
119905) (21)
which can be easily generalized as follows
int119891 (119905) 119889119902
119892 (119905) =
infin
sum
119895=0
119891 (119902119895
119905) (119892 (119902119895
119905) minus 119892 (119902119895+1
119905)) (22)
Suppose that 0 lt 119886 lt 119887 The definite 119902-integral is definedas follows
int
119887
0
119891 (119905) 119889119902
119905 = (1 minus 119902) 119887
infin
sum
119895=0
119902119895
119891 (119902119895
119887) (23)
int
119887
119886
119891 (119905) 119889119902
119905 = int
119887
0
119891 (119905) 119889119902
119905 minus int
119886
0
119891 (119905) 119889119902
119905 (24)
A more general version of (23) is given by
int
119887
0
119891 (119905) 119889119902
119892 (119905) =
infin
sum
119895=0
119891 (119902119895
119887) (119892 (119902119895
119887) minus 119892 (119902119895+1
119887)) (25)
The classical Gamma function Γ(119911) (see eg [16 Section11]) was found by Euler while he was trying to extend thefactorial 119899 = Γ(119899 + 1) ( 119899 isin N
0
) to real numbers The 119902-factorial function [119899]
119902
(119899 isin N0
) of 119899 defined by
[119899]119902
= 1 if 119899 = 0
[119899]119902
[119899 minus 1]119902
sdot sdot sdot [2]119902
[1]119902
if 119899 isin N(26)
can be rewritten as follows
(1 minus 119902)minus119899
infin
prod
119896=0
(1 minus 119902119896+1
)
(1 minus 119902119896+1+119899)=
(119902 119902)infin
(119902119899+1 119902)infin
(1 minus 119902)minus119899
= Γ119902
(119899 + 1) (0 lt 119902 lt 1)
(27)
By replacing 119899 by 119886 minus 1 in (27) Jackson [18] defined the 119902-Gamma function Γ
119902
(119886) by
Γ119902
(119886) =(119902 119902)infin
(119902119886 119902)infin
(1 minus 119902)1minus119886
(0 lt 119902 lt 1) (28)
The 119902-analogue of (119905 minus 119886)119899 is defined by the polynomial
(119905 minus 119886)119899
119902
= 1 (119899 = 0)
(119905 minus 119886) (119905 minus 119902119886) sdot sdot sdot (119905 minus 119902119899minus1
119886) (119899 isin N)
= 119905119899
(119886
119905 119902)119899
(119899 isin N0
)
(29)
Definition 4 Let R(120572) gt 0 let 120573 and 120578 be real or complexnumbers Then a 119902-analogue of Saigorsquos fractional integral119868120572120573120578
119902
is given for |120591119905| lt 1 by (see [19 p 172 Equation (21)])
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times int
119905
0
(119902120591
119905 119902)120572minus1
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
sdot 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119891 (120591) 119889119902
120591
(30)
The integral operator 119868120572120573120578119902
includes both the 119902-analoguesof the Riemann-Liouville and Erdelyi-Kober fractional inte-gral operators given by the following definitions
Definition 5 A 119902-analogue of Riemann-Liouville fractionalintegral operator of a function 119891(119905) of an order 120572 is given by(see [20])
119868120572
119902
119891 (119905) (= 119868120572minus120572120578
119902
119891 (119905))
=119905120572minus1
Γ119902
(120572)int
119905
0
(119902120591
119905 119902)120572minus1
119891 (120591) 119889119902
120591
(120572 gt 0 0 lt 119902 lt 1)
(31)
where (119886 119902)120572
is given by (17)
Definition 6 A 119902-analogue of the Erdelyi-Kober fractionalintegral operator for 120572 gt 0 120578 isin R and 0 lt 119902 lt 1 is givenby (see [20])
119868120578120572
119902
119891 (119905) (= 1198681205720120578
119902
119891 (119905))
=119905minus120578minus1
Γ119902
(120572)int
119905
0
(119902120591
119905 119902)120572minus1
120591120578
119891 (120591) 119889119902
120591
(120572 gt 0 0 lt 119902 lt 1)
(32)
2 Certain Inequalities Involving SaigoFractional Integral Operator
Here we start with presenting an inequality involving Saigofractional integral (8) asserted by the following lemma
Lemma 7 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119906 (119905) 119868120572120573120578
0119905
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119868120572120573120578
0119905
V (119905) 119892 (119905)
+ 119868120572120573120578
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(33)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
4 Abstract and Applied Analysis
Proof Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) Then for all 120591 120588 isin (0 119905) with 119905 gt 0we have
(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) ge 0 (34)
or equivalently
119891 (120591) 119892 (120591) + 119891 (120588) 119892 (120588) ge 119891 (120591) 119892 (120588) + 119891 (120588) 119892 (120591) (35)
Consider the following function 119865(119905 120591) defined by
119865 (119905 120591) =119905minus120572minus120573
(119905 minus 120591)120572minus1
Γ (120572)2
1198651
(120572 + 120573 minus120578 120572 1 minus120591
119905)
(120591 isin (0 119905) 119905 gt 0)
=1
Γ (120572)
(119905 minus 120591)120572minus1
119905120572+120573+(120572 + 120573) (minus120578)
Γ (120572 + 1)
(119905 minus 120591)120572
119905120572+120573+1
+(120572+120573) (120572+120573+1) (minus120578) (minus120578+1)
Γ (120572+2)
(119905minus120591)120572+1
119905120572+120573+2+sdot sdot sdot
(36)
We observe that each term of the above series is nonnegativeunder the conditions in Lemma 7 and hence the function119865(119905 120591) remains nonnegative for all 120591 isin (0 119905) (119905 gt 0)
Now by multiplying both sides of (35) by 119865(119905 120591)119906(120591)
defined by (36) integrating the resulting inequality withrespect to 120591 from 0 to 119905 and using (8) we get
119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
0119905
119906 (119905)
ge 119892 (120588) 119868120572120573120578
0119905
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(37)
Next bymultiplying both sides of (37) by119865(119905 120588)V(120588) (120588 isin
(0 119905) 119905 gt 0) where 119865(119905 120588) is given when 120591 is replaced by 120588
in (36) integrating the resulting inequality with respect to 120588
from 0 to 119905 and using (8) we are led to the desired result (33)
Theorem 8 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(38)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in Lemma 7 we get
119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(39)
Since 119868120572120573120578
0119905
119897(119905) ≧ 0 under the given conditions bymultiplying both sides of (39) by 119868120572120573120578
0119905
119897(119905) we have
119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(40)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897119898 respectivelyin (33) and then multiplying both sides of the resultinginequalities by 119868120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) both of which arenonnegative under the given assumptions respectively weget the following inequalities
119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
Abstract and Applied Analysis 5
119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(41)
Finally by adding (40) and (41) side by side we arrive at thedesired result (38)
We present another inequality involving the Saigo frac-tional integral operator in (8) asserted by the followinglemma
Lemma 9 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120574120575120577
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119906 (119905)
ge 119868120574120575120577
0119905
V (119905) 119892 (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(42)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By multiplying both sides of (37) by
119905minus120574minus120575
(119905 minus 120588)120574minus1
Γ (120574)2
1198651
(120574 + 120575 minus120577 120574 1 minus120588
119905) V (120588)
(120588 isin (0 119905) 119905 gt 0)
(43)
which remains nonnegative under the conditions in (42)integrating the resulting inequality with respect to 120588 from 0
to 119905 and using (8) we get the desired result (42)
Theorem 10 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119897 (119905) [2119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905)]
(44)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in (42) we have
119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)
ge 119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(45)
By multiplying both sides of (45) by 1198681205721205731205780119905
119897(119905) after a littlesimplification we get
119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(46)
Now by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (42)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) respectively weget the following two inequalities
119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
6 Abstract and Applied Analysis
119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(47)
Finally we find that the inequality (44) follows by adding theinequalities (46) and (47) side by side
Remark 11 It may be noted that the inequalities (38) and(44) in Theorems 8 and 10 respectively are reversed if thefunctions are asynchronous on [0infin) The special case of(44) in Theorem 10 when 120572 = 120574 120573 = 120575 and 120578 = 120577 is easilyseen to yield the inequality (38) in Theorem 8
Here we derive certain presumably new integralinequalities by setting 120573 = 0 in (38) and 120573 = 0 = 120575 in (44)respectively and applying the integral operator (12) to theresulting inequalities we obtain two integral inequalitiesinvolving Erdelyi-Kober fractional integral operators statedin Corollaries 12 and 13 below
Corollary 12 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (t) 1198681205721205780119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
(48)
for all 119905 gt 0 120572 gt 0 and 120578 isin R
Corollary 13 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120578
0119905
119897 (119905) [2119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120577
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905)
+119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905)]
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119898 (t) 119891 (119905)]
(49)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Remark 14 The special cases 120573 = minus120572 in Theorem 8 and 120573 =
minus120572 and 120575 = minus120574 in Theorem 10 are seen to yield the knownfractional integral inequalities due to Dahmani [21]
3 Saigo Fractional 119902-Integral Inequalities
We establish certain 119902-integral inequalities which are the 119902-analogues (or extensions) of the results derived in the preced-ing section some of which are presumably new ones For ourpurpose we begin with providing comparison properties forthe fractional 119902-integral operators asserted by the followinglemma
Lemma 15 Let 0 lt 119902 lt 1 and 119891 [0infin) rarr R be continuousfunctions with 119891(119905) ge 0 for all 119905 isin [0infin) Then we have thefollowing inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (30)
119868120572120573120578
119902
119891 (119905) ge 0 (50)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0(ii) the 119902-analogue of Riemann-Liouville fractional integral
operator of the function 119891(119905) of an order 120572 in (31)
119868120572
119902
119891 (119905) ge 0 (51)
for all 120572 gt 0
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
Proof Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) Then for all 120591 120588 isin (0 119905) with 119905 gt 0we have
(119891 (120591) minus 119891 (120588)) (119892 (120591) minus 119892 (120588)) ge 0 (34)
or equivalently
119891 (120591) 119892 (120591) + 119891 (120588) 119892 (120588) ge 119891 (120591) 119892 (120588) + 119891 (120588) 119892 (120591) (35)
Consider the following function 119865(119905 120591) defined by
119865 (119905 120591) =119905minus120572minus120573
(119905 minus 120591)120572minus1
Γ (120572)2
1198651
(120572 + 120573 minus120578 120572 1 minus120591
119905)
(120591 isin (0 119905) 119905 gt 0)
=1
Γ (120572)
(119905 minus 120591)120572minus1
119905120572+120573+(120572 + 120573) (minus120578)
Γ (120572 + 1)
(119905 minus 120591)120572
119905120572+120573+1
+(120572+120573) (120572+120573+1) (minus120578) (minus120578+1)
Γ (120572+2)
(119905minus120591)120572+1
119905120572+120573+2+sdot sdot sdot
(36)
We observe that each term of the above series is nonnegativeunder the conditions in Lemma 7 and hence the function119865(119905 120591) remains nonnegative for all 120591 isin (0 119905) (119905 gt 0)
Now by multiplying both sides of (35) by 119865(119905 120591)119906(120591)
defined by (36) integrating the resulting inequality withrespect to 120591 from 0 to 119905 and using (8) we get
119868120572120573120578
0119905
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
0119905
119906 (119905)
ge 119892 (120588) 119868120572120573120578
0119905
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(37)
Next bymultiplying both sides of (37) by119865(119905 120588)V(120588) (120588 isin
(0 119905) 119905 gt 0) where 119865(119905 120588) is given when 120591 is replaced by 120588
in (36) integrating the resulting inequality with respect to 120588
from 0 to 119905 and using (8) we are led to the desired result (33)
Theorem 8 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(38)
for all 119905 gt 0 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 ≧ 0 and 120578 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in Lemma 7 we get
119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(39)
Since 119868120572120573120578
0119905
119897(119905) ≧ 0 under the given conditions bymultiplying both sides of (39) by 119868120572120573120578
0119905
119897(119905) we have
119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(40)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897119898 respectivelyin (33) and then multiplying both sides of the resultinginequalities by 119868120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) both of which arenonnegative under the given assumptions respectively weget the following inequalities
119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
Abstract and Applied Analysis 5
119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(41)
Finally by adding (40) and (41) side by side we arrive at thedesired result (38)
We present another inequality involving the Saigo frac-tional integral operator in (8) asserted by the followinglemma
Lemma 9 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120574120575120577
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119906 (119905)
ge 119868120574120575120577
0119905
V (119905) 119892 (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(42)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By multiplying both sides of (37) by
119905minus120574minus120575
(119905 minus 120588)120574minus1
Γ (120574)2
1198651
(120574 + 120575 minus120577 120574 1 minus120588
119905) V (120588)
(120588 isin (0 119905) 119905 gt 0)
(43)
which remains nonnegative under the conditions in (42)integrating the resulting inequality with respect to 120588 from 0
to 119905 and using (8) we get the desired result (42)
Theorem 10 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119897 (119905) [2119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905)]
(44)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in (42) we have
119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)
ge 119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(45)
By multiplying both sides of (45) by 1198681205721205731205780119905
119897(119905) after a littlesimplification we get
119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(46)
Now by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (42)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) respectively weget the following two inequalities
119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
6 Abstract and Applied Analysis
119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(47)
Finally we find that the inequality (44) follows by adding theinequalities (46) and (47) side by side
Remark 11 It may be noted that the inequalities (38) and(44) in Theorems 8 and 10 respectively are reversed if thefunctions are asynchronous on [0infin) The special case of(44) in Theorem 10 when 120572 = 120574 120573 = 120575 and 120578 = 120577 is easilyseen to yield the inequality (38) in Theorem 8
Here we derive certain presumably new integralinequalities by setting 120573 = 0 in (38) and 120573 = 0 = 120575 in (44)respectively and applying the integral operator (12) to theresulting inequalities we obtain two integral inequalitiesinvolving Erdelyi-Kober fractional integral operators statedin Corollaries 12 and 13 below
Corollary 12 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (t) 1198681205721205780119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
(48)
for all 119905 gt 0 120572 gt 0 and 120578 isin R
Corollary 13 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120578
0119905
119897 (119905) [2119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120577
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905)
+119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905)]
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119898 (t) 119891 (119905)]
(49)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Remark 14 The special cases 120573 = minus120572 in Theorem 8 and 120573 =
minus120572 and 120575 = minus120574 in Theorem 10 are seen to yield the knownfractional integral inequalities due to Dahmani [21]
3 Saigo Fractional 119902-Integral Inequalities
We establish certain 119902-integral inequalities which are the 119902-analogues (or extensions) of the results derived in the preced-ing section some of which are presumably new ones For ourpurpose we begin with providing comparison properties forthe fractional 119902-integral operators asserted by the followinglemma
Lemma 15 Let 0 lt 119902 lt 1 and 119891 [0infin) rarr R be continuousfunctions with 119891(119905) ge 0 for all 119905 isin [0infin) Then we have thefollowing inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (30)
119868120572120573120578
119902
119891 (119905) ge 0 (50)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0(ii) the 119902-analogue of Riemann-Liouville fractional integral
operator of the function 119891(119905) of an order 120572 in (31)
119868120572
119902
119891 (119905) ge 0 (51)
for all 120572 gt 0
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(41)
Finally by adding (40) and (41) side by side we arrive at thedesired result (38)
We present another inequality involving the Saigo frac-tional integral operator in (8) asserted by the followinglemma
Lemma 9 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119906 V [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120574120575120577
0119905
V (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119906 (119905)
ge 119868120574120575120577
0119905
V (119905) 119892 (119905) 1198681205721205731205780119905
119906 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
V (119905) 119891 (119905) 119868120572120573120578
0119905
119906 (119905) 119892 (119905)
(42)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By multiplying both sides of (37) by
119905minus120574minus120575
(119905 minus 120588)120574minus1
Γ (120574)2
1198651
(120574 + 120575 minus120577 120574 1 minus120588
119905) V (120588)
(120588 isin (0 119905) 119905 gt 0)
(43)
which remains nonnegative under the conditions in (42)integrating the resulting inequality with respect to 120588 from 0
to 119905 and using (8) we get the desired result (42)
Theorem 10 Let 119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120573120578
0119905
119897 (119905) [2119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
0119905
119898 (119905) 119868120574120575120577
0119905
119899 (119905)
+119868120572120573120578
0119905
119899 (119905) 119868120574120575120577
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119898 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119898 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119899 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
0119905
119899 (119905) [119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119868120574120575120577
0119905
119898 (119905) 119892 (119905)
+119868120572120573120578
0119905
119897 (119905) 119892 (119905) 119868120574120575120577
0119905
119898 (119905) 119891 (119905)]
(44)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin R with 120572 + 120573 ≧ 0120578 ≦ 0 120574 + 120575 ≧ 0 and 120577 ≦ 0
Proof By setting 119906 = 119898 and V = 119899 in (42) we have
119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)
ge 119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+ 119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)
(45)
By multiplying both sides of (45) by 1198681205721205731205780119905
119897(119905) after a littlesimplification we get
119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905)]
ge 119868120572120573120578
0119905
119897 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119898 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119898 (119905) 119892 (119905)]
(46)
Now by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (42)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
0119905
119898(119905) and 119868120572120573120578
0119905
119899(119905) respectively weget the following two inequalities
119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119898 (119905) [119868120574120575120577
0119905
119899 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119899 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
6 Abstract and Applied Analysis
119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(47)
Finally we find that the inequality (44) follows by adding theinequalities (46) and (47) side by side
Remark 11 It may be noted that the inequalities (38) and(44) in Theorems 8 and 10 respectively are reversed if thefunctions are asynchronous on [0infin) The special case of(44) in Theorem 10 when 120572 = 120574 120573 = 120575 and 120578 = 120577 is easilyseen to yield the inequality (38) in Theorem 8
Here we derive certain presumably new integralinequalities by setting 120573 = 0 in (38) and 120573 = 0 = 120575 in (44)respectively and applying the integral operator (12) to theresulting inequalities we obtain two integral inequalitiesinvolving Erdelyi-Kober fractional integral operators statedin Corollaries 12 and 13 below
Corollary 12 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (t) 1198681205721205780119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
(48)
for all 119905 gt 0 120572 gt 0 and 120578 isin R
Corollary 13 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120578
0119905
119897 (119905) [2119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120577
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905)
+119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905)]
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119898 (t) 119891 (119905)]
(49)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Remark 14 The special cases 120573 = minus120572 in Theorem 8 and 120573 =
minus120572 and 120575 = minus120574 in Theorem 10 are seen to yield the knownfractional integral inequalities due to Dahmani [21]
3 Saigo Fractional 119902-Integral Inequalities
We establish certain 119902-integral inequalities which are the 119902-analogues (or extensions) of the results derived in the preced-ing section some of which are presumably new ones For ourpurpose we begin with providing comparison properties forthe fractional 119902-integral operators asserted by the followinglemma
Lemma 15 Let 0 lt 119902 lt 1 and 119891 [0infin) rarr R be continuousfunctions with 119891(119905) ge 0 for all 119905 isin [0infin) Then we have thefollowing inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (30)
119868120572120573120578
119902
119891 (119905) ge 0 (50)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0(ii) the 119902-analogue of Riemann-Liouville fractional integral
operator of the function 119891(119905) of an order 120572 in (31)
119868120572
119902
119891 (119905) ge 0 (51)
for all 120572 gt 0
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
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OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905)]
ge 119868120572120573120578
0119905
119899 (119905) [119868120574120575120577
0119905
119898 (119905) 119892 (119905) 119868120572120573120578
0119905
119897 (119905) 119891 (119905)
+119868120574120575120577
0119905
119898 (119905) 119891 (119905) 119868120572120573120578
0119905
119897 (119905) 119892 (119905)]
(47)
Finally we find that the inequality (44) follows by adding theinequalities (46) and (47) side by side
Remark 11 It may be noted that the inequalities (38) and(44) in Theorems 8 and 10 respectively are reversed if thefunctions are asynchronous on [0infin) The special case of(44) in Theorem 10 when 120572 = 120574 120573 = 120575 and 120578 = 120577 is easilyseen to yield the inequality (38) in Theorem 8
Here we derive certain presumably new integralinequalities by setting 120573 = 0 in (38) and 120573 = 0 = 120575 in (44)respectively and applying the integral operator (12) to theresulting inequalities we obtain two integral inequalitiesinvolving Erdelyi-Kober fractional integral operators statedin Corollaries 12 and 13 below
Corollary 12 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
2119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120578
0119905
119898 (119905) 119868120572120578
0119905
119899 (119905) 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119898 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120572120578
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119899 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (t) 1198681205721205780119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120572120578
0119905
119897 (119905) 119892 (119905)]
(48)
for all 119905 gt 0 120572 gt 0 and 120578 isin R
Corollary 13 Let119891 and 119892 be two continuous and synchronousfunctions on [0infin) and let 119897 119898 119899 [0infin) rarr [0infin) becontinuous functions Then the following inequality holds true
119868120572120578
0119905
119897 (119905) [2119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120577
0119905
119899 (119905) 119868120572120578
0119905
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120578
0119905
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120578
0119905
119898 (119905) 119868120574120577
0119905
119899 (119905)
+119868120572120578
0119905
119899 (119905) 119868120574120577
0119905
119898 (119905)]
ge 119868120572120578
0119905
119897 (119905) [119868120572120578
0119905
119898 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119898 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119898 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119899 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119899 (119905) 119891 (119905)]
+ 119868120572120578
0119905
119899 (119905) [119868120572120578
0119905
119897 (119905) 119891 (119905) 119868120574120577
0119905
119898 (119905) 119892 (119905)
+119868120572120578
0119905
119897 (119905) 119892 (119905) 119868120574120577
0119905
119898 (t) 119891 (119905)]
(49)
for all 119905 gt 0 120572 gt 0 120574 gt 0 and 120578 120577 isin R
Remark 14 The special cases 120573 = minus120572 in Theorem 8 and 120573 =
minus120572 and 120575 = minus120574 in Theorem 10 are seen to yield the knownfractional integral inequalities due to Dahmani [21]
3 Saigo Fractional 119902-Integral Inequalities
We establish certain 119902-integral inequalities which are the 119902-analogues (or extensions) of the results derived in the preced-ing section some of which are presumably new ones For ourpurpose we begin with providing comparison properties forthe fractional 119902-integral operators asserted by the followinglemma
Lemma 15 Let 0 lt 119902 lt 1 and 119891 [0infin) rarr R be continuousfunctions with 119891(119905) ge 0 for all 119905 isin [0infin) Then we have thefollowing inequalities
(i) the Saigo fractional 119902-integral operator of the function119891(119905) in (30)
119868120572120573120578
119902
119891 (119905) ge 0 (50)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0(ii) the 119902-analogue of Riemann-Liouville fractional integral
operator of the function 119891(119905) of an order 120572 in (31)
119868120572
119902
119891 (119905) ge 0 (51)
for all 120572 gt 0
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
(iii) the 119902-analogue of Erdelyi-Kober fractional integraloperator of the function 119891(119905) in (32)
119868120578120572
119902
119891 (119905) ge 0 (52)
for all 120572 gt 0 and 120578 isin R
Proof By applying (23) to the 119902-integral in (30) we have
119868120572120573120578
119902
119891 (119905) =119905minus120573minus1
Γ119902
(120572)
times
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)
sdot (1minus119902) 119905
infin
sum
119895=0
119902119895
(119902119895+1
119902)120572minus1
(119902119895
minus1)119898
119902
119891 (119902119895
119905)
(53)
It is easy to see that
Γ119902
(120572) gt 0 (119902120572
119902)119898
gt 0
(119902119895+1
119902)120572minus1
=
(119902119895+1
119902)infin
(119902120572+119895 119902)infin
gt 0
(54)
for all 120572 gt 0 and 119895119898 isin N0
Next for simplicity
ℎ (119895119898 119902) = (minus1)119898
(119902119895
minus 1)119898
119902
(119895 119898 isin N0
) (55)
Then we claim that ℎ(119895 119898 119902) ge 0 for all 119895119898 isin N0
We findfrom (29) that
ℎ (119895119898 119902) = (minus1)119898
119902119895119898
(1
119902119895 119902)
119898
= 119902119895119898
(minus1)119898
119898minus1
prod
119896=0
(1 minus 119902119896minus119895
)
(56)
It is easy to see that if119898 gt 119895 then ℎ(119895119898 119902) = 0 On the otherhand if119898 le 119895 then we have
ℎ (119895119898 119902) = 119902119895119898
119898minus1
prod
119896=0
(119902119896minus119895
minus 1) gt 0 (57)
since 119896 minus 119895 lt 0 for all 119896 with 0 le 119896 le 119898 minus 1 lt 119898 le 119895Finally we find that under the given conditions each termin the double series of (53) is nonnegativeThis completes theproof of (50) The other two inequalities in (51) and (52) maybe easily proved
For convenience and simplicity we define the followingfunctionH by
H (120591 119905 119906 (120591) 120572 120573 120578 119902)
=119905minus120573minus1
Γ119902
(120572)(119902120591
119905 119902)120572minus1
sdot
infin
sum
119898=0
(119902120572+120573
119902)119898
(119902minus120578
119902)119898
(119902120572 119902)119898
(119902 119902)119898
times 119902(120578minus120573)119898
(minus1)119898
119902minus(119898
2
)(120591
119905minus 1)
119898
119902
119906 (120591)
(58)
where 119905 gt 0 0 le 120591 le 119905 120572 gt 0 120573 120578 isin R with 120572 + 120573 gt 0
and 120578 lt 0 0 lt 119902 lt 1 119906 [0infin) rarr [0infin) is a continuousfunction As in the process of Lemma 15 it is seen that
H (120591 119905 119906 (120591) 120572 120573 120578 119902) ge 0 (59)
under the conditions given in (58)Here we present two 119902-integral inequalities involving the
Saigo fractional 119902-integral operator (30) stated in Lemmas 16and 17 below
Lemma 16 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 [0infin) rarr [0infin) be acontinuous function Then the following inequality holds truefor 119905 isin (0infin)
119868120572120573120578
119902
119906 (119905) 119868120572120573120578
119902
V (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119906 (119905) 119891 (119905) 119868120572120573120578
119902
V (119905) 119892 (119905)
+ 119868120572120573120578
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(60)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof Since 119891 and 119892 are two synchronous functions on[0infin) for all 120591 120588 ge 0 the inequality (35) is satisfiedBy multiplying both sides of (35) by H(120591 119905 119906(120591) 120572 120573 120578 119902)
in (58) together with (59) and taking 119902-integration of theresulting inequality with respect to 120591 from 0 to 119905 with the aidof Definition 4 we get
119868120572120573120578
119902
119906 (119905) 119891 (119905) 119892 (119905) + 119891 (120588) 119892 (120588) 119868120572120573120578
119902
119906 (119905)
ge 119892 (120588) 119868120572120573120578
119902
119906 (119905) 119891 (119905) + 119891 (120588) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(61)
Next by multiplying both sides of (61) byH(120588 119905 V(120588) 120572 120573 120578 119902) in (58) together with (59) taking119902-integration of the last resulting inequality with respect to 120588from 0 to 119905 and using Definition 4 we are led to the desiredresult (60)
Lemma 17 Let 0 lt 119902 lt 1 let 119891 and 119892 be two continuous andsynchronous functions on [0infin) let 119906 V [0infin) rarr [0infin)
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014
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Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
be continuous functions Then the following inequality holdstrue for 119905 isin (0infin)
119868120574120575120577
119902
V (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119906 (119905)
ge 119868120574120575120577
119902
V (119905) 119892 (119905) 119868120572120573120578119902
119906 (119905) 119891 (119905)
+ 119868120574120575120577
119902
V (119905) 119891 (119905) 119868120572120573120578
119902
119906 (119905) 119892 (119905)
(62)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By multiplying both sides of (61) byH(120588 119905 V(120588) 120574 120575 120577 119902) in (58) together with (59) andtaking the 119902-integration of the resulting inequality withrespect to 120588 from 0 to 119905 with the aid of Definition 4 we getthe desired result (62)
Theorem 18 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (t)] (63)
for all 120572 gt 0 and 120573 120578 isin R with 120572 + 120573 gt 0 and 120578 lt 0
Proof We start with (60) in Lemma 16 by putting 119906 = 119898 andV = 119899 we get
119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
ge 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(64)
By multiplying both sides of (64) by 119868120572120573120578119902
119897(119905) we have
119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(65)
Similarly by replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60)respectively and then multiplying both sides of the resultinginequalities by 119868
120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively weget
119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(66)
Finally by adding (65) and (66) side by side we arrive at thedesired result in Theorem 18
Theorem 19 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572120573120578
119902
119897 (119905) [2119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572120573120578
119902
119898 (119905) 119868120574120575120577
119902
119899 (119905)
+119868120572120573120578
119902
119899 (119905) 119868120574120575120577
119902
119898 (119905)]
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
ge 119868120572120573120578
119902
119897 (119905) [119868120572120573120578
119902
119898 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119898 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119898 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119899 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119899 (119905) 119891 (119905)]
+ 119868120572120573120578
119902
119899 (119905) [119868120572120573120578
119902
119897 (119905) 119891 (119905) 119868120574120575120577
119902
119898 (119905) 119892 (119905)
+119868120572120573120578
119902
119897 (119905) 119892 (119905) 119868120574120575120577
119902
119898 (119905) 119891 (119905)]
(67)
for all 120572 gt 0 120574 gt 0 and 120573 120578 120575 120577 isin Rwith 120572+120573 gt 0 120574+120575 gt 0120578 lt 0 and 120577 lt 0
Proof By setting 119906 = 119898 and V = 119899 in (62) we have
119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)
ge 119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+ 119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)
(68)
By multiplying both sides of (68) by 119868120572120573120578119902
119897(119905) after a littlesimplification we get
119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905)]
ge 119868120572120573120578
119902
119897 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119898 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119898 (119905) 119892 (119905)]
(69)
By replacing 119906 V by 119897 119899 and 119906 V by 119897 119898 in (60) andthen multiplying both sides of the resulting inequalities by119868120572120573120578
119902
119898(119905) and 119868120572120573120578
119902
119899(119905) respectively we get
119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119898 (119905) [119868120574120575120577
119902
119899 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119899 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905) 119892 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905)]
ge 119868120572120573120578
119902
119899 (119905) [119868120574120575120577
119902
119898 (119905) 119892 (119905) 119868120572120573120578
119902
119897 (119905) 119891 (119905)
+119868120574120575120577
119902
119898 (119905) 119891 (119905) 119868120572120573120578
119902
119897 (119905) 119892 (119905)]
(70)
We therefore find that the inequality (67) follows by addingthe inequalities (69) and (70) side by side
Remark 20 It is noted that the inequalities in (63) and (67)are reversed if the functions 119891 and 119892 are asynchronous It isalso easily seen that the special cases 120572 = 120574 120573 = 120575 and 120578 = 120577
of (67) inTheorem 19 reduce to those inTheorem 18
Following Garg and Chanchlani [19] the operator (30)would reduce immediately to the extensively investigated119902-analogue of Erdelyi-Kober and Riemann-Liouville typefractional integral operators in (31) and (32) respectively(see also [20]) Indeed by suitably specializing the values ofparameters 120572 (and additionally 120575 in Theorem 19) the resultspresented in this section would find further fractional 119902-integral inequalities involving the 119902-analogues of Erdelyi-Kober and Riemann-Liouville type fractional integral oper-ators in (31) and (32) For example if we set 120573 = minus120572 inTheorem 18 and 120573 = minus120572 and 120575 = minus120574 in Theorem 19respectively andmake use of the relation (31) we are led to thefollowing presumably new 119902-integral inequalities involving119902-Riemann-Liouville fractional integral operators given inCorollaries 21 and 22 below
Corollary 21 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120572
119902
119898 (119905) 119868120572
119902
119899 (119905) 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119899 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120572
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119891 (119905) 119868120572
119902
119897 (119905) 119892 (119905)]
(71)
for all 120572 gt 0
Corollary 22 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120572
119902
119897 (119905) [2119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905) 119891 (119905) 119892 (119905)
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
+119868120574
119902
119899 (119905) 119868120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120572
119902
119898 (119905) 119868120574
119902
119899 (119905)
+119868120572
119902
119899 (119905) 119868120574
119902
119898 (119905)]
ge 119868120572
119902
119897 (119905) [119868120572
119902
119898 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119898 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119898 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119899 (119905) 119891 (119905)]
+ 119868120572
119902
119899 (119905) [119868120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119898 (119905) 119892 (119905)
+119868120572
119902
119897 (119905) 119892 (119905) 119868120574
119902
119898 (119905) 119891 (119905)]
(72)
for all 120572 gt 0 and 120574 gt 0
Furthermore if we set 120573 = 0 in Theorem 18 and 120573 = 0 =
120575 in Theorem 19 we obtain the following presumably newinequalities involving 119902-analogue of the Erdelyi-Kober typefractional integral operators asserted by Corollaries 23 and24 below
Corollary 23 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
2119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119891 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 2119868120578120572
119902
119898 (119905) 119868120578120572
119902
119899 (119905) 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905)
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119899 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120578120572
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119891 (119905) 119868120578120572
119902
119897 (119905) 119892 (119905)]
(73)
for all 120572 gt 0
Corollary 24 Let 0 lt 119902 lt 1 let 119891 119892 [0infin) rarr R betwo continuous synchronous functions let 119897 119898 119899 [0infin) rarr
[0infin) be continuous functions Then the following inequalityholds true for 119905 isin (0infin)
119868120578120572
119902
119897 (119905) [2119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905) 119892 (119905)
+ 119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905) 119892 (119905)
+119868120574120575
119902
119899 (119905) 119868120578120572
119902
119898 (119905) 119891 (119905) 119892 (119905)]
+ 119868120578120572
119902
119897 (119905) 119891 (119905) 119892 (119905) [119868120578120572
119902
119898 (119905) 119868120574120575
119902
119899 (119905)
+119868120578120572
119902
119899 (119905) 119868120574120575
119902
119898 (119905)]
ge 119868120578120572
119902
119897 (119905) [119868120578120572
119902
119898 (119905) 119891 (119905) 119868120574120575
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119898 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119898 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574
119902
119899 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119899 (119905) 119891 (119905)]
+ 119868120578120572
119902
119899 (119905) [119868120578120572
119902
119897 (119905) 119891 (119905) 119868120574120575
119902
119898 (119905) 119892 (119905)
+119868120578120572
119902
119897 (119905) 119892 (119905) 119868120574120575
119902
119898 (119905) 119891 (119905)]
(74)
for all 120572 gt 0 120575 gt 0 and 120574 120578 isin R
4 Concluding Remarks
It is easy to see that
lim119902rarr1
Γ119902
(119886) = Γ (119886)
lim119902rarr1
(119902119886
119902)119899
(1 minus 119902)119899
= (119886)119899
(75)
where (119886)119899
is the Pochhammer symbol given in (10) Takingthe limit of some of the results presented in Section 3 as 119902 rarr
1 with the aid of (75) the resulting inequalities are seen tocorrespond with those results in Section 2 It is noted that theinequalities in (71) and (72) are equal to the known ones byDahmani [21 pp 494ndash496 Equations (8) and (24)]
Representations of the inequalities in terms of the frac-tional integral operators have been investigated by manyresearchers in the existing literature Here we have presentedsome presumably new Saigo type fractional integral inequal-ities and their 119902-analogues by using the one parameters ofdeformation Very recently Baleanu and Agarwal [22] alsogave some new Saigo type 119902-fractional integral inequalitiesby using the two parameters of deformation 119902
1
and 1199022
whichare totally new and different ones from those presented here
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
Acknowledgment
This work was supported by Dongguk University ResearchFund of 2013
References
[1] J C Kuang Applied Inequalities Shandong Sciences and Tech-nologie Press Shandong China 2004 (Chinese)
[2] D S Mitrinovic Analytic Inequalities Springer Berlin Ger-many 1970
[3] P L Chebyshev ldquoSur les expressions approximatives des inte-grales definies par les autres prises entre les memes limitesrdquoProceedings of the Mathematical Society of Kharkov vol 2 pp93ndash98 1882
[4] A M Ostrowski ldquoOn an integral inequalityrdquo AequationesMathematicae vol 4 pp 358ndash373 1970
[5] G A AnastassiouAdvances on Fractional Inequalities SpringerBriefs in Mathematics Springer New York NY USA 2011
[6] S Belarbi and Z Dahmani ldquoOn some new fractional integralinequalitiesrdquo Journal of Inequalities in Pure and Applied Mathe-matics vol 10 no 3 article 86 p 5 2009
[7] Z Dahmani O Mechouar and S Brahami ldquoCertain inequali-ties related to the Chebyshevrsquos functional involving a Riemann-Liouville operatorrdquoBulletin ofMathematical Analysis and Appli-cations vol 3 no 4 pp 38ndash44 2011
[8] S S Dragomir ldquoSome integral inequalities of Gruss typerdquoIndian Journal of Pure and Applied Mathematics vol 31 no 4pp 397ndash415 2000
[9] S L Kalla and A Rao ldquoOn Gruss type inequality for ahypergeometric fractional integralrdquoLeMatematiche vol 66 no1 pp 57ndash64 2011
[10] V Lakshmikantham and A S Vatsala ldquoTheory of fractionaldifferential inequalities and applicationsrdquo Communications inApplied Analysis vol 11 no 3-4 pp 395ndash402 2007
[11] H Ogunmez and U M Ozkan ldquoFractional quantum integralinequalitiesrdquo Journal of Inequalities and Applications vol 2011Article ID 787939 7 pages 2011
[12] W T Sulaiman ldquoSome new fractional integral inequalitiesrdquoJournal of Mathematical Analysis vol 2 no 2 pp 23ndash28 2011
[13] D Baleanu S D Purohit and P Agarwal ldquoOn fractional inte-gral inequalities involving hypergeometric operatorsrdquo ChineseJournal of Mathematics vol 2014 Article ID 609476 5 pages2014
[14] M Saigo ldquoA remark on integral operators involving the Gausshypergeometric functionsrdquo Mathematical Reports of College ofGeneral Education Kyushu University vol 11 no 2 pp 135ndash1431978
[15] V S Kiryakova Generalized Fractional Calculus and Applica-tions vol 301 of Pitman Research Notes in Mathematical SeriesLongman Scientific amp Technical Harlow UK 1994
[16] H M Srivastava and J Choi Zeta and q-Zeta Functions andAssociated Series and Integrals Elsevier Science Publishers NewYork NY USA 2012
[17] S G Samko A A Kilbas and O I Marichev FractionalIntegrals and Derivatives Theory and Applications Gordon andBreach Science Publishers New York NY USA 1993
[18] F H Jackson ldquoOn 119902-definite integralsrdquoTheQuarterly Journal ofPure and Applied Mathematics vol 41 pp 193ndash203 1910
[19] M Garg and L Chanchlani ldquo119902-analogues of Saigorsquos fractionalcalculus operatorsrdquoBulletin ofMathematical Analysis andAppli-cations vol 3 no 4 pp 169ndash179 2011
[20] R P Agarwal ldquoCertain fractional 119902-integrals and 119902-derivativesrdquoMathematical Proceedings of the Cambridge Philosophical Soci-ety vol 66 pp 365ndash370 1969
[21] Z Dahmani ldquoNew inequalities in fractional integralsrdquo Interna-tional Journal of Nonlinear Science vol 9 no 4 pp 493ndash4972010
[22] D Baleanu and P Agarwal ldquoCertain inequalities involving thefractional 119902-integral operatorsrdquo In press
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of