Journal of Computational and Applied Mathematics 233 (2010) 2149–2160

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Journal of Computational and AppliedMathematics

journal homepage: www.elsevier.com/locate/cam

Complete monotonicity of some functions involvingpolygamma functionsFeng Qi a,∗, Senlin Guo b, Bai-Ni Guo aa Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin City, 300160, Chinab Department of Mathematics, Zhongyuan University of Technology, Zhengzhou City, Henan Province, 450007, China

a r t i c l e i n f o

Article history:Received 5 January 2009Received in revised form 25 September2009

MSC:26A4826A5126D2033B1033B1544A10

Keywords:MonotonicityCompletely monotonic functionPolygamma functionInfinite seriesBernoulli numbers

a b s t r a c t

In the present paper, we establish necessary and sufficient conditions for the functionsxα |ψ (i)(x + β)| and α|ψ (i)(x + β)| − x|ψ (i+1)(x + β)| respectively to be monotonic andcompletely monotonic on (0,∞), where i ∈ N, α > 0 and β ≥ 0 are scalars, and ψ (i)(x)are polygamma functions.

© 2009 Elsevier B.V. All rights reserved.

1. Introduction and main results

1.1

Recall [1, Chapter XIII] and [2, Chapter IV] that a function f (x) is said to be completely monotonic on an interval I ⊆ R iff (x) has derivatives of all orders on I and

0 ≤ (−1)kf (k)(x) <∞ (1.1)

holds for all k ≥ 0 on I . This definition was introduced in 1921 in [3], who called such functions ‘‘total monoton’’.The celebrated Bernstein–Widder Theorem [2, p. 161] states that a function f (x) is completely monotonic on (0,∞) if

and only if

f (x) =∫∞

0e−xsdµ(s), (1.2)

∗ Corresponding author.E-mail addresses: qifeng618@gmail.com, qifeng618@hotmail.com, qifeng618@qq.com (F. Qi), sguo@hotmail.com, senlinguo@gmail.com (S. Guo),

bai.ni.guo@gmail.com, bai.ni.guo@hotmail.com (B.-N. Guo).URLs: http://qifeng618.spaces.live.com (F. Qi), http://guobaini.spaces.live.com (B.-N. Guo).

0377-0427/$ – see front matter© 2009 Elsevier B.V. All rights reserved.doi:10.1016/j.cam.2009.09.044

2150 F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160

whereµ is a nonnegative measure on [0,∞) such that the integral (1.2) converges for all x > 0. This means that a functionf (x) is completely monotonic on (0,∞) if and only if it is a Laplace transform of the measure µ.The most important properties of completely monotonic functions can be found in [1, Chapter XIII], [2, Chapter IV], [4,5]

and the related references therein.The completely monotonic functions have applications in different branches of mathematical sciences. For example,

they play some role in combinatorics [6], numerical and asymptotic analysis [7,8], physics [9,10], potential theory [11], andprobability theory [12,10,13].

1.2

It is well known [14–16] that the classical Euler gamma function may be defined for x > 0 by

0(x) =∫∞

0tx−1e−tdt. (1.3)

The logarithmic derivative of 0(x), denoted by ψ(x) = 0′(x)0(x) , is called psi function, and ψ

(k)(x) for k ∈ N are calledpolygamma functions.It should be common knowledge [14–16] that the special functions 0(x), ψ(x) and ψ (i)(x) for i ∈ N are important and

basic and that they have much extensive applications in mathematical sciences.

1.3

In [17, Lemma 1], it was shown that the functions xc∣∣ψ (k)(x)

∣∣ for k ∈ N and c ∈ R are strictly decreasing (or strictlyincreasing, respectively) on (0,∞) if and only if c ≤ k (or c ≥ k+ 1, respectively).In [18, Theorem 4.14], it was obtained that the function xc

∣∣ψ (k)(x)∣∣ for k ∈ N and c ∈ R is strictly convex on (0,∞) if

and only if either c ≤ k, or c = k+ 1, or c ≥ k+ 2. In [18, Remark 4.15], it was pointed out that there does not exist a realnumber c such that the function xc

∣∣ψ (k)(x)∣∣ for k ∈ N is concave on (0,∞).

In [18, Lemma 2.2] and [19, Lemma 5], the functions xα∣∣ψ (i)(x + 1)

∣∣ for i ∈ N are proved to be strictly increasing (orstrictly decreasing, respectively) on (0,∞) if and only if α ≥ i (or α ≤ 0, respectively).In [20, Lemma 2.1], the function xψ ′(x+ a) is proved to be strictly increasing on [0,∞) for a ≥ 1.Motivated by the above results, the first and third authors considered in [21] themonotonicity of amore general function

xα∣∣ψ (i)(x+ β)

∣∣ and the complete monotonicity of several related functions as follows: For i ∈ N, α > 0 and β ≥ 0,

(1) the function xα∣∣ψ (i)(x+β)

∣∣ is strictly increasing on (0,∞) if (α, β) ∈ {α ≥ i, 12 ≤ β < 1}∪{α ≥ i, β ≥ α−i+12

}∪{α ≥

i+ 1, β ≤ α−i+12

}and only if α ≥ i;

(2) the function αx

∣∣ψ (i)(x)∣∣− ∣∣ψ (i+1)(x)

∣∣ is completely monotonic on (0,∞) if and only if α ≥ i+ 1;(3) the function

∣∣ψ (i+1)(x)∣∣− α

x

∣∣ψ (i)(x)∣∣ is completely monotonic on (0,∞) if and only if 0 < α ≤ i;

(4) the function αx

∣∣ψ (i)(x+ 1)∣∣−∣∣ψ (i+1)(x+ 1)

∣∣ is completely monotonic on (0,∞) if and only if α ≥ i;(5) the function αx

∣∣ψ (i)(x+β)∣∣−∣∣ψ (i+1)(x+β)

∣∣ on (0,∞) is completelymonotonic if (α, β) ∈ {α ≥ i+1, β ≤ α−i+12

}∪{i ≤

α ≤ i+ 1, β ≥ α−i+12

}∪{i ≤ α ≤ (i+1)(i+4β−2)

i+2β , 12 ≤ β < 1}and only if α ≥ i;

(6) the function α∣∣ψ (i)(x + β)

∣∣−x∣∣ψ (i+1)(x + β)∣∣ is completely monotonic on (0,∞) if (α, β) ∈ {i ≤ α ≤ i + 1, β ≥

α−i+12

}∪{α ≥ i+ 1, β ≤ α−i+1

2

}and only if α ≥ i.

1.4

The first aimof this paper is to present necessary and sufficient conditions for the function xα∣∣ψ (i)(x+β)

∣∣ to bemonotonicon (0,∞), which can be summarized as the following Theorem 1.

Theorem 1. Let i ∈ N, α ∈ R and β ≥ 0.

(1) The function xα∣∣ψ (i)(x)

∣∣ is strictly increasing (or strictly decreasing, respectively) on (0,∞) if and only if α ≥ i+1 (or α ≤ i,respectively).

(2) For β ≥ 12 , the function x

α∣∣ψ (i)(x+ β)

∣∣ is strictly increasing on [0,∞) if and only if α ≥ i.(3) Let δ : (0,∞)→

(0, 12

)be defined by

δ(t) =et(t − 1)+ 1(et − 1)2

(1.4)

F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160 2151

and δ−1 :(0, 12

)→ (0,∞) stand for the inverse function of δ. If 0 < β < 1

2 and

α ≥ i+ 1−[eδ−1(β)

eδ−1(β) − 1+ β − 1

]δ−1(β), (1.5)

then the function xα∣∣ψ (i)(x+ β)

∣∣ is strictly increasing on (0,∞).As a by-product of the proof of Theorem1, lower and upper bounds for infinite serieswhose coefficients involve Bernoulli

numbers may be derived as follows.

Corollary 1. Let 0 < β < 12 and δ

−1 be the inverse function of δ defined by (1.4). Then the following inequalities hold:

12>

∞∑k=1

B2kt2k−1

(2k− 1)!> 0, (1.6)

t2>

∞∑k=0

B2k+2t2k+2

(2k+ 2)!> max

{0,t2− 1

}, (1.7)

∞∑k=0

B2k+2t2k+2

(2k+ 2)!>

(12− β

)t +

[eδ−1(β)

eδ−1(β) − 1− β + 1

]δ−1(β)− 1, (1.8)

where t ∈ (0,∞) and Bn for n ≥ 0 represent Bernoulli numbers which may be defined [14–16] by

xex − 1

=

∞∑n=0

Bnn!xn = 1−

x2+

∞∑j=1

B2jx2j

(2j)!, |x| < 2π. (1.9)

1.5

The second aim of this paper is to establish necessary and sufficient conditions for the function α∣∣ψ (i)(x + β)

∣∣−x∣∣ψ (i+1)(x+ β)

∣∣ to be completely monotonic on (0,∞), which may be stated as the following Theorem 2.Theorem 2. Let i ∈ N, α ∈ R and β ≥ 0.

(1) The function

α∣∣ψ (i)(x)

∣∣−x∣∣ψ (i+1)(x)∣∣ (1.10)

is completely monotonic on (0,∞) if and only if α ≥ i+ 1.(2) The negative of the function (1.10) is completely monotonic on (0,∞) if and only if α ≤ i.(3) If β ≥ 1

2 , then the function

α∣∣ψ (i)(x+ β)

∣∣−x∣∣ψ (i+1)(x+ β)∣∣ (1.11)

is completely monotonic on (0,∞) if and only if α ≥ i.(4) If 0 < β < 1

2 and the inequality (1.5) is valid, then the function (1.11) is completely monotonic on (0,∞).

As immediate consequences of Theorem 2, the following corollary is obtained.

Corollary 2. Let i ∈ N, α ∈ R and β ≥ 0.

(1) The functionα

x

∣∣ψ (i)(x)∣∣−∣∣ψ (i+1)(x)

∣∣ (1.12)

is completely monotonic on (0,∞) if and only if α ≥ i+ 1.(2) The negative of the function (1.12) is completely monotonic on (0,∞) if and only if α ≤ i.(3) If β ≥ 1

2 , then the function

α

x

∣∣ψ (i)(x+ β)∣∣−∣∣ψ (i+1)(x+ β)

∣∣ (1.13)

is completely monotonic on (0,∞) if and only if α ≥ i.(4) If 0 < β < 1

2 and the inequality (1.5) holds true, then the function (1.13) is completely monotonic on (0,∞).

2152 F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160

2. Remarks and applications

Before proving the above theorems and corollaries, we give several remarks about Theorem 1, Theorem 2 and theirapplications.

Remark 1. Since limβ→0+ [βδ−1(β)] = 0 and that the inverse function δ−1 is decreasing from(0, 12

)onto (0,∞), we claim

that

0 <[eδ−1(β)

eδ−1(β) − 1+ β − 1

]δ−1(β) < 1, β ∈ (0, 1). (2.1)

Indeed, if replacing δ−1(β) by s, the middle term in (2.1) becomes s2es

(es−1)2which is decreasing from (0,∞) onto (0, 1). This

implies that the condition (1.5) is not only sufficient but also necessary in Theorems 1 and 2.

Remark 2. As mentioned in Section 1.3, some conclusions in Theorem 1 have been applied in nearby fields.

(1) The first conclusion in Theorem1wasutilized in [17, Theorem2] to obtain a functional inequality concerningpolygammafunctions: For k ≥ 1 and n ≥ 2, the inequality∣∣ψ (k)(M [r]n (xν; pν))∣∣ ≤ M [s]n (∣∣ψ (k)(xν)

∣∣; pν) (2.2)

holds if and only if either r ≥ 0 and s ≥ − rk+1 or r < 0 and s ≥ −

rk , where xν > 0 and pν > 0 with

∑nν=1 pν = 1, and

M [t]n (xν; pν) =

( n∑ν=1

pνxtν

)1/t, t 6= 0

n∏ν=1

xpνν , t = 0(2.3)

stands for the discrete weighted power means.(2) Some applications of the first two conclusions in Theorem 1 were carried out in [18] as follows.(a) The first conclusion in Theorem 1 was applied in [18, Theorem 4.9] to obtain that the inequalities(

1+α

x+ s

)n<ψ (n)(x+ s)ψ (n)(x+ 1)

<

(1+

β

x+ s

)n(2.4)

hold for n ∈ N and s ∈ (0, 1)with the best possible constants

α = 1− s and β = s[ψ (n)(s)ψ (n)(1)

]1/n− s. (2.5)

(b) The special cases for β = 1 of the third conclusion in Theorem 1 was employed in [18, Theorem 4.8] to derive thatthe inequalities

(n− 1)! exp[α

x− nψ(x)

]<∣∣ψ (n)(x)

∣∣ < (n− 1)! exp[β

x− nψ(x)

](2.6)

hold for n ∈ N and x > 0 if and only if α ≤ −n and β ≥ 0.(c) Moreover, the convexity of xc

∣∣ψ (k)(x)∣∣was used in [18, Theorem 4.16] to establish that the double inequalities

α

(1x−1y

)< xn

∣∣ψ (n)(x)∣∣− yn∣∣ψ (n)(y)

∣∣ < β

(1x−1y

)(2.7)

hold for n ∈ N and y > x > 0 with the best possible constants α = n!2 and β = n!.

(3) The special cases for β = 1 of the third conclusion in Theorem 1, the monotonic properties of the functions

x2iψ (2i)(1+ x), x2i+1ψ (2i)(1+ x),x2i−1ψ (2i−1)(1+ x) and x2iψ (2i−1)(1+ x)

on [0,∞), were used in [19,22] to establish the monotonic, logarithmically convex, completely monotonic propertiesof the functions

[0(1+ x)]y

0(1+ xy)and

0(1+ y)[0(1+ x)]y

0(1+ xy)(2.8)

or their first and second logarithmic derivatives.(4) The special case for β ≥ 1 and i = 1 of the third conclusion in Theorem 1 was employed in [20, Theorem 3.1] to revealthe subadditive property of the function ψ(a+ ex) on (−∞,∞).

Remark 3. Let us recall the following four definitions.

F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160 2153

Definition 1 ([23, Definition 1.1]). Let f be a positive functionwhich has derivatives of all orders on an interval I . If [ln f (x)](k)for some nonnegative integer k is completely monotonic on I , but [ln f (x)](k−1) is not completely monotonic on I , then f issaid to be a logarithmically completely monotonic function of kth order on I .

Definition 2 ([23, Definition 1.2]). Let f be a positive functionwhich has derivatives of all orders on an interval I . If [ln f (x)](k)for some nonnegative integer k is absolutely monotonic on I , but [ln f (x)](k−1) is not absolutely monotonic on I , then f issaid to be a logarithmically absolutely monotonic function of kth order on I .

Definition 3 ([23, Definition 1.3]). A positive function f which has derivatives of all orders on an interval I is said to belogarithmically absolutely convex on I if [ln f (x)](2k) ≥ 0 on I for k ∈ N.

Definition 4 ([24,25]). A k-times differentiable function f (x) > 0 is said to be k-log-convex (or k-log-concave, respectively)on an interval I , with k ∈ N, if and only if [ln f (x)](k) ≥ 0 (or [ln f (x)](k) ≤ 0, respectively) on I .

The former two conclusions in Theorem 1, which were ever circulated in the preprint [26], have been employed in theproofs of [23, Theorem 1.4] and [25, Theorem 2.2 and Theorem 2.3].(1) The special cases for α = i and β = 1 of the second conclusion in Theorem 1 were made use of to procure the followingtheorem.

Theorem 3 ([23, Theorem 1.4]). The function

Gs,t(x) =[0(1+ tx)]s

[0(1+ sx)]t(2.9)

for x, s, t ∈ R such that 1+ sx > 0 and 1+ tx > 0 with s 6= t has the following properties:(a) For t > s > 0 and x ∈ (0,∞), Gs,t(x) is an increasing function and a logarithmically completely monotonic function ofsecond order in x;

(b) For t > s > 0 and x ∈(−1t , 0

), Gs,t(x) is a logarithmically completely monotonic function in x;

(c) For s < t < 0 and x ∈ (−∞, 0), Gs,t(x) is a decreasing function and a logarithmically absolutely monotonic function ofsecond order in x;

(d) For s < t < 0 and x ∈(0,− 1s

), Gs,t(x) is a logarithmically completely monotonic function in x;

(e) For s < 0 < t and x ∈(−1t , 0

), Gt,s(x) is an increasing function and a logarithmically absolutely convex function in x;

(f) For s < 0 < t and x ∈(0,− 1s

), Gt,s(x) is a decreasing function and a logarithmically absolutely convex function in x.

(2) The first two conclusions in Theorem1were hired in [25], a simplified version of the preprint [24], to derive the followingtwo theorems.

Theorem 4 ([25, Theorem 2.2]). For b > a > 0 and i ∈ N, the function [0(bx)]a

[0(ax)]bis (2i+1)-log-convex and (2i)-log-concave with

respect to x ∈ (0,∞).

Theorem 5 ([25, Theorem 2.3]). For b > a > 0, i ∈ N and β ≥ 12 , the function

[0(bx+β)]a

[0(ax+β)]bis (2i+ 1)-log-concave and (2i)-log-

convex with respect to x ∈ (0,∞).

Remark 4. The former two conclusions in Theorem 2, which were also issued in the preprint [26], have also been appliedin the proofs of [27, Theorem 1] and [28, Theorem 1].(1) The first result in Theorem 2 was utilized in [27, Theorem 1] to procure upper bounds for the ratio of two gammafunctions and the divided differences of the psi and polygamma functions as follows.

Theorem 6 ([27, Theorem 1]). For a > 0 and b > 0 with a 6= b, inequalities[0(a)0(b)

]1/(a−b)≤ eψ(I(a,b)) (2.10)

and

(−1)n[ψ (n−1)(a)− ψ (n−1)(b)

]a− b

≤ (−1)nψ (n)(I(a, b)) (2.11)

hold true, where n ∈ N and

I(a, b) =1e

(bb

aa

)1/(b−a)(2.12)

represents the identric or exponential mean.

2154 F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160

(2) The first two results in Theorem2were employed in [28, Theorem1] to acquire lower bounds for the ratio of two gammafunctions and the divided differences of the psi and polygamma functions and to refine the inequality (2.11).

Theorem 7 ([28, Theorem 1]). For a > 0 and b > 0 with a 6= b, the inequality

(−1)iψ (i)(Lα(a, b)) ≤(−1)i

b− a

∫ b

aψ (i)(u)du ≤ (−1)iψ (i)(Lβ(a, b)) (2.13)

holds if α ≤ −i− 1 and β ≥ −i, where i is a nonnegative integer and

Lp(a, b) =

[bp+1 − ap+1

(p+ 1)(b− a)

]1/p, p 6= −1, 0

b− aln b− ln a

, p = −1

I(a, b), p = 0

(2.14)

stands for the generalized logarithmic mean of order p ∈ R.

Remark 5. Recall [29,30,20] that a function f (x) is said to be subadditive on I if the inequality

f (x+ y) ≤ f (x)+ f (y) (2.15)

holds for all x, y ∈ I with x+ y ∈ I . If the inequality (2.15) is reversed, then f (x) is called superadditive on I .Some subadditive or superadditive properties of the gamma, psi and polygamma functions have been discovered as

follows.In [31], the function ψ(a+ x) is proved to be sub-multiplicative with respect to x ∈ [0,∞) if and only if a ≥ a0, where

a0 denotes the only positive real number which satisfies ψ(a0) = 1.In [30], the function [0(x)]α was proved to be subadditive on (0,∞) if and only if ln 2ln∆ ≤ α ≤ 0, where∆ = minx≥0

0(2x)0(x) .

In [18, Lemma 2.4], the function ψ(ex)was proved to be strictly concave on R.In [20, Theorem 3.1], the function ψ(a + ex) is proved to be subadditive on (−∞,∞) if and only if a ≥ c0, where c0 is

the only positive zero of ψ(x).In [32, Theorem 1], among other things, it was presented that the function ψ (k)(ex) for k ∈ N is concave (or convex,

respectively) on R if k = 2n− 2 (or k = 2n− 1, respectively) for n ∈ N.By the aid of the monotonicity of the function xα

∣∣ψ (i)(x+β)∣∣ in Theorem 1, the following subadditive and superadditive

properties of the function∣∣ψ (i)(ex)

∣∣ for i ∈ Nwere acquired recently.

Theorem 8 ([33,34]). For i ∈ N, the function∣∣ψ (i)(ex)

∣∣ is superadditive on (−∞, ln θ0) and subadditive on (ln θ0,∞), whereθ0 ∈ (0, 1) is the unique root of the equation 2

∣∣ψ (i)(θ)∣∣= ∣∣ψ (i)(θ2)

∣∣.Remark 6. The second conclusion in Theorem 2 was cited in [35, Lemma 2.4] and [36, Remark 2.3].

Remark 7. Theorem 8 and the facts mentioned in Remarks 3–6 show the potential applicability of Theorems 1 and 2convincingly.

Remark 8. In passing, we recollect the notion ‘‘logarithmically completely monotonic function’’ which is equivalent tothe logarithmically completely monotonic function of 0th order mentioned in Remark 3. A function f (x) is said to belogarithmically completely monotonic on an interval I ⊆ R if it has derivatives of all orders on I and its logarithm ln fsatisfies

0 ≤ (−1)k[ln f (x)](k) <∞ (2.16)

for k ∈ N on I . By looking through the database MathSciNet, we find that this phrase was first used in [37], but with nowords to explicitly define it. Thereafter, it seems to have been ignored by the mathematical community. In early 2004,this terminology was recovered in [38] (the preprint of [39]) and it was immediately referenced in [40], the preprint ofthe paper [41]. A natural question that one may ask is: Whether is this notion trivial or not? In [38, Theorem 4], it wasproved that all logarithmically completely monotonic functions are also completely monotonic, but not conversely. Thisresult was formally published when revising [42]. Hereafter, this conclusion and its proofs were dug in [43–46] once andagain. Furthermore, in the paper [43], the logarithmically completely monotonic functions on (0,∞) were characterizedas the infinitely divisible completely monotonic functions studied in [47] and all Stieltjes transforms were proved to belogarithmically completely monotonic on (0,∞), where a function f (x) defined on (0,∞) is called a Stieltjes transform ifit can be of the form

f (x) = a+∫∞

0

1s+ x

dµ(s) (2.17)

F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160 2155

for some nonnegative number a and some nonnegative measureµ on [0,∞) satisfying∫∞

011+sdµ(s) <∞. For more infor-

mation, please refer to [43].It is remarked that many completely monotonic functions founded in a lot of literature such as [48,49,5], [1, Chapter XIII]

and the related references therein are actually logarithmically completely monotonic.

3. Lemmas

In order to verify Theorem 1, Theorem 2 and their corollaries in Sections 1.4 and 1.5, we need the following lemmas,in which Lemma 1 is simple but has been validated in [50–54] to be especially effectual in proving the monotonicity and(logarithmically) complete monotonicity of functions involving the gamma, psi and polygamma functions.

Lemma 1. Let f (x) be a function defined on an infinite interval I whose right end is∞. If limx→∞ f (x) = δ and f (x)−f (x+ε) > 0hold true for some given scalar ε > 0 and all x ∈ I , then f (x) > δ.

Proof. By mathematical induction, for all x ∈ I , we have

f (x) > f (x+ ε) > f (x+ 2ε) > · · · > f (x+ kε)→ δ

as k→∞. The proof of Lemma 1 is complete. �

Lemma 2 ([14–16]). The polygamma functions ψ (k)(x)may be expressed for x > 0 and k ∈ N as

ψ (k)(x) = (−1)k+1∫∞

0

tke−xt

1− e−tdt. (3.1)

For x > 0 and r > 0,

1xr=

10(r)

∫∞

0t r−1e−xtdt. (3.2)

For i ∈ N and x > 0,

ψ (i−1)(x+ 1) = ψ (i−1)(x)+(−1)i−1(i− 1)!

xi. (3.3)

Lemma 3. For k ∈ N, the double inequality

(k− 1)!xk

+k!2xk+1

< (−1)k+1ψ (k)(x) <(k− 1)!xk

+k!xk+1

(3.4)

holds on (0,∞).

Proof. In [48, Theorem 2.1] and [53, Lemma 1.3], the function ψ(x)− ln x+ αx was proved to be completely monotonic on

(0,∞), i.e.,

(−1)i[ψ(x)− ln x+

α

x

](i)≥ 0 (3.5)

for i ≥ 0, if and only if α ≥ 1, so is its negative, i.e., the inequality (3.5) is reversed, if and only if α ≤ 12 . In [55] and

[49, Theorem 2.1], the function ex0(x)xx−α was proved to be logarithmically completely monotonic on (0,∞), i.e.,

(−1)k[lnex0(x)xx−α

](k)≥ 0 (3.6)

for k ∈ N, if and only if α ≥ 1, so is its reciprocal, i.e., the inequality (3.6) is reversed, if and only if α ≤ 12 . Considering the

fact [41, p. 82] that a completely monotonic function which is non-identically zero cannot vanish at any point on (0,∞)and rearranging either (3.5) for i ∈ N or (3.6) for k ≥ 2 leads to the double inequality (3.4) immediately. �

4. Proofs of theorems and corollaries

Proof of Theorem 1. It is a standard argument to obtain that the function δ(t) is strictly decreasing from (0,∞) onto(0, 12

).

2156 F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160

Let gi,α,β(x) = xα∣∣ψ (i)(x+ β)

∣∣ on (0,∞). The direct calculation and rearrangement yieldsg ′i,α,β(x)

xα−1= α

∣∣ψ (i)(x+ β)∣∣−x∣∣ψ (i+1)(x+ β)

∣∣= (−1)i+1

[αψ (i)(x+ β)+ xψ (i+1)(x+ β)

]. (4.1)

Making use of (3.4) in (4.1) gives

limx→∞

g ′i,α,β(x)

xα−1= 0 (4.2)

for i ∈ N, α ∈ R and β ≥ 0. By virtue of formulas (3.3), (3.2) and (3.1) in sequence, a straightforward computation reveals

g ′i,α,β(x)

xα−1−g ′i,α,β(x+ 1)

(x+ 1)α−1= (−1)i+1

{α[ψ (i)(x+ β)− ψ (i)(x+ β + 1)

]+ x[ψ (i+1)(x+ β)− ψ (i+1)(x+ β + 1)

]− ψ (i+1)(x+ β + 1)

}=

i!α(x+ β)i+1

−(i+ 1)!x(x+ β)i+2

−(i+ 1)!(x+ β)i+2

+ (−1)i+2ψ (i+1)(x+ β)

= (−1)i+2ψ (i+1)(x+ β)+i!(α − i− 1)(x+ β)i+1

+(i+ 1)!(β − 1)(x+ β)i+2

=

∫∞

0

[t

1− e−t+ (β − 1)t + α − i− 1

]t ie−(x+β)tdt

,

∫∞

0hi,α,β(t)t ie−(x+β)tdt. (4.3)

For β = 0, easy differentiation shows that h′i,α,0(t) = −δ(t) < 0, and so the function hi,α,0(t) is strictly decreasing from(0,∞) onto (α − i− 1, α − i). Thus, if α ≥ i+ 1, the functions hi,α,0(t) and

g ′i,α,0(x)

xα−1−g ′i,α,0(x+ 1)

(x+ 1)α−1

are positive on (0,∞). Combining this with (4.2) and considering Lemma 1, it is deduced that the functionsg ′i,α,0(x)

xα−1and

g ′i,α,0(x) are positive on (0,∞). Hence, the function gi,α,0(x) is strictly increasing on (0,∞) for α ≥ i + 1. Similarly, forα ≤ i, the function gi,α,0(x) is strictly decreasing on (0,∞).For β > 0, it is easy to see that h′i,α,β(t) = −δ(t) + β , and h

′

i,α,β(t) is strictly increasing from (0,∞) onto(β − 1

2 , β).

Consequently, if β ≥ 12 , the function h

′

i,α,β(t) is positive and hi,α,β(t) is strictly increasing from (0,∞) onto (α − i,∞).Accordingly, if α ≥ i and β ≥ 1

2 , the function hi,α,β(t) is positive on (0,∞), that is,

g ′i,α,β(x)

xα−1−g ′i,α,β(x+ 1)

(x+ 1)α−1> 0 (4.4)

on (0,∞). Combining this with Lemma 1 results in the positivity of g ′i,α,β(x) on (0,∞). Therefore, for α ≥ i and β ≥12 , the

function gi,α,β(x) is strictly increasing on (0,∞).For 0 < β < 1

2 , since h′

i,α,β(t) is strictly increasing from (0,∞) onto(β − 1

2 , β), the function hi,α,β(t) attains its unique

minimum at some point t0 ∈ (0,∞)with δ(t0) = β . As a result, the unique minimum of hi,α,β(t) equals

δ−1(β)eδ−1(β)

eδ−1(β) − 1+ (β − 1)δ−1(β)+ α − i− 1,

where δ−1 is the inverse function of δ and is strictly decreasing from (0, 12 ) onto (0,∞). Consequently, when the inequal-ity (1.5) holds for 0 < β < 1

2 , the function hi,α,β(t) is positive on (0,∞), which means that the inequality (4.4) holds true.Accordingly, making use of the limit (4.2) and Lemma 1 again yields that the function gi,α,β(x) is strictly increasing on (0,∞)if 0 < β < 1

2 and the inequality (1.5) is valid. The sufficiency is proved.If gi,α,0(x) is strictly decreasing on (0,∞), then

xi+1−αg ′i,α,0(x) = αxi∣∣ψ (i)(x)

∣∣−xi+1∣∣ψ (i+1)(x)∣∣< 0. (4.5)

F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160 2157

Applying (3.4) in (4.5) and letting x→∞ lead to

0 ≥ limx→∞

xi+1−αg ′i,α,0(x)

≥ α limx→∞

xi[(i− 1)!xi+

i!2xi+1

]− limx→∞

xi+1[i!xi+1+(i+ 1)!xi+2

]= (i− 1)!(α − i),

which means α ≤ i.If gi,α,0(x) is strictly increasing on (0,∞), then

xi+2−αg ′i,α,0(x) = αxi+1∣∣ψ (i)(x)

∣∣−xi+2∣∣ψ (i+1)(x)∣∣> 0. (4.6)

Employing (3.3) and (3.4) in (4.6) and taking x→∞ results in

0 ≤ limx→0+

xi+2−αg ′i,α,0(x)

= limx→0+

{αxi+1

∣∣ψ (i)(x)∣∣−xi+2[∣∣ψ (i+1)(x+ 1)

∣∣+ (i+ 1)!xi+2

]}= α lim

x→0+xi+1

∣∣ψ (i)(x)∣∣−(i+ 1)! − lim

x→0+xi+2

∣∣ψ (i+1)(x+ 1)∣∣

≤ α limx→0+

xi+1[(i− 1)!xi+i!xi+1

]− (i+ 1)! − lim

x→0+xi+2

[i!

(x+ 1)i+1+

(i+ 1)!2(x+ 1)i+2

]= i!(α − i− 1),

thus, the necessary condition α ≥ i+ 1 follows.If the function gi,α,β(x) is strictly increasing on (0,∞) for β > 0, then

xi+1−αg ′i,α,β(x) = αxi∣∣ψ (i)(x+ β)

∣∣−xi+1∣∣ψ (i+1)(x+ β)∣∣> 0. (4.7)

Utilizing (3.4) in (4.7) and taking limit gives

0 ≤ limx→∞

xi+1−αg ′i,α,β(x)

≤ α limx→∞

xi[(i− 1)!(x+ β)i

+i!

(x+ β)i+1

]− limx→∞

xi+1[

i!(x+ β)i+1

+(i+ 1)!2(x+ β)i+2

]= (i− 1)!(α − i),

which is equivalent to α ≥ i. The proof of Theorem 1 is thus completed. �

Remark 9. The first two conclusions in Theorem 1 were ever proved by virtue of the convolution theorem for Laplacetransforms in [17, Lemma 1] and [18, Lemma 2.2], so we supply a new and unified proof for them here.

Proof of Corollary 1. It is well known [14–16] that Bernoulli polynomials Bk(x)may be defined by

text

et − 1=

∞∑k=0

Bk(x)tk

k!(4.8)

and that Bernoulli numbers Bk and Bernoulli polynomials Bk(x) are connected by Bk(1) = (−1)kBk(0) = (−1)kBk andB2k+1(0) = B2k+1 = 0 for k ≥ 1. Using these notations, the functions hi,α,β(t) and h′i,α,β(t)may be rewritten as

hi,α,β(t) =tet

et − 1+ (β − 1)t + α − i− 1

= α − i+(β −

12

)t +

∞∑k=2

Bk(1)tk

k!

= α − i+(β −

12

)t +

∞∑k=2

(−1)kBktk

k!

= α − i+(β −

12

)t +

∞∑k=1

(−1)k+1Bk+1tk+1

(k+ 1)!

= α − i+(β −

12

)t +

∞∑k=0

B2k+2t2k+2

(2k+ 2)!

2158 F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160

and

h′i,α,β(t) = β −12+

∞∑k=1

B2kt2k−1

(2k− 1)!.

The proof of Theorem 1 shows that

(1) h′i,α,0(t) < 0 on (0,∞);(2) hi,α,0(t) > 0 on (0,∞) if α ≥ i+ 1;(3) hi,α,0(t) < 0 on (0,∞) if 0 < α ≤ i;(4) h′i,α,β(t) > 0 on (0,∞) if β ≥

12 ;

(5) hi,α,β(t) > 0 on (0,∞) if α ≥ i and β ≥ 12 ;

(6) hi,α,β(t) > 0 on (0,∞) if 0 < β < 12 and inequality (1.5) holds true.

Based on these and by standard argument, Corollary 1 is thus proved. �

Proof of Theorem 2. If hi,α,β(t) ≷ 0 on (0,∞), then the function

±

∫∞

0hi,α,β(t)t ie−(x+β)tdt

is completely monotonic on (−β,∞), and so, by virtue of (4.3), it is derived that

±

[g ′i,α,β(x)xα−1

−g ′i,α,β(x+ 1)

(x+ 1)α−1

]is completely monotonic on (0,∞), that is,

(−1)j[g ′i,α,β(x)xα−1

−g ′i,α,β(x+ 1)

(x+ 1)α−1

](j)= (−1)j

[g ′i,α,β(x)xα−1

](j)− (−1)j

[g ′i,α,β(x+ 1)(x+ 1)α−1

](j)R 0 (4.9)

on (0,∞) for j ≥ 0. Moreover, formulas (3.4) and (4.1) imply

limx→∞

[g ′i,α,β(x)xα−1

](j)= limx→∞

(−1)j[g ′i,α,β(x)xα−1

](j)= 0. (4.10)

Combining (4.9) and (4.10) with Lemma 1 concludes that

(−1)j[g ′i,α,β(x)xα−1

](j)R 0,

that is, the function

±g ′i,α,β(x)

xα−1= ±

[α∣∣ψ (i)(x+ β)

∣∣−x∣∣ψ (i+1)(x+ β)∣∣]

is completely monotonic on (0,∞), if hi,α,β(t) ≷ 0 on (0,∞). In the proof of Theorem 1, we have demonstrated thathi,α,β(t) is positive on (0,∞) if either β = 0 and α ≥ i + 1, or β ≥ 1

2 and α ≥ i, or 0 < β < 12 and the inequality (1.5) is

satisfied, and that hi,α,β(t) is negative on (0,∞) if β = 0 and α ≤ i. As a result, the sufficient conditions for the functionα∣∣ψ (i)(x+ β)

∣∣−x∣∣ψ (i+1)(x+ β)∣∣ to be completely monotonic on (0,∞) follow.

The derivation of necessary conditions is same as in Theorem 1. The proof of Theorem 2 is complete. �

Proof of Corollary 2. It follows easily from Theorem 2 and the facts that

±

[α

x

∣∣ψ (i)(x+ β)∣∣−∣∣ψ (i+1)(x+ β)

∣∣] = ±1x

{α∣∣ψ (i)(x+ β)

∣∣−x∣∣ψ (i+1)(x+ β)∣∣},

that the function 1x is completely monotonic on (0,∞), and that the product of any finite completely monotonic functionsis also completely monotonic on the intersection of their domains. �

Acknowledgements

The authors appreciate the anonymous referees for their helpful and valuable comments on this manuscript. The firstauthor was partially supported by the China Scholarship Council.

F. Qi et al. / Journal of Computational and Applied Mathematics 233 (2010) 2149–2160 2159

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