an expansion theorem concerning wilson functions and polynomials

Acta Math. Hungar., 135 (4) (2012), 350–382DOI: 10.1007/s10474-012-0194-7

First published online March 7, 2012

AN EXPANSION THEOREM CONCERNINGWILSON FUNCTIONS AND POLYNOMIALS

A. BIRO∗

A. Renyi Institute of Mathematics, Hungarian Academy of Sciences, 1053 Budapest,Realtanoda u. 13-15, Hungary

e-mail: [email protected]

(Received May 3, 2011; revised October 15, 2011; accepted October 19, 2011)

Abstract. We prove that a relatively general even function f(x) (satisfy-ing a vanishing condition, and also some analyticity and growth conditions) onthe real line can be expanded in terms of a certain function series closely relatedto the Wilson functions introduced by Groenevelt in 2003. The coefficients inthe expansion of f will be inner products in a suitable Hilbert space of f andsome polynomials closely related to Wilson polynomials (these are well-knownhypergeometric orthogonal polynomials).

1. Introduction

1.1. Necessary notations. We use the notation Γ(X ± Y ) =Γ(X + Y )Γ(X − Y ), and similarly for more occurrences of the symbol ±,e.g.

Γ(X ± Y ± Z) = Γ(X + Y + Z)Γ(X + Y − Z)Γ(X − Y + Z)Γ(X − Y − Z).

We define

(X)n =Γ(X + n)

Γ(X)and (X ± Y )n = (X + Y )n(X − Y )n.

∗ Research partially supported by the Hungarian National Foundation for Scientific Research(OTKA) Grants No. K72731, K67676, K81658 and ERC-AdG Grant no. 228005.

Key words and phrases: series of Wilson functions, sums of Wilson polynomials.2010 Mathematics Subject Classification: 33C20, 33C45.

0236-5294/$20.00 c© 2012 Akademiai Kiado, Budapest, Hungary

mailto:[email protected]

WILSON FUNCTIONS AND POLYNOMIALS 351

For the generalized hypergeometric functions AFB see [8]. The Wilsonpolynomials are defined by the formula

Wn(x; a, b, c, d)(a + b)n(a + c)n(a + d)n

= 4F3

(−n, n + a + b + c + d − 1, a + ix, a − ix

a + b, a + c, a + d; 1

).

The Wilson function φλ(x; a, b, c, d) is defined in [4] in formula (3.2), us-ing 7F6-functions. But instead of this definition, we will work with formula(3.3) of [4], which states that φλ(x; a, b, c, d) equals the sum of

Γ(1 − a − d)4F3

(a + ix, a − ix, a + iλ, a − iλ

a + b, a + c, a + d; 1

)

Γ(a + b)Γ(a + c)Γ(1 − d ± ix)Γ(1 − d ± iλ)

and

Γ(a + d − 1)4F3

(1 − d + ix, 1 − d − ix, 1 − d + iλ, 1 − d − iλ

1 + b − d, 1 + c − d, 2 − a − d; 1

)

Γ(1 + b − d)Γ(1 + c − d)Γ(a ± ix)Γ(a ± iλ),

where

a =a + b + c + d − 1

2, d =

a − b − c + d + 12

.

We can take it as the definition of φλ(x; a, b, c, d) in the case when a + d isnot an integer, and this will be true in our cases.

Let t1 and t2 be two given positive numbers; they will be fixed through-out the paper. We write s1 = 1

2 + it1, s2 = 12 + it2.

Introduce the notations

N+(S, t) =Γ(S ± it1 + it2)Γ

(14 + it2 ± it

)sin π(2it2)

×(

sin πs1 + sin π

(12

− it2 − S

))φ+

i( 12

−S)(t),

N −(S, t) =Γ(S ± it1 − it2)Γ

(14 − it2 ± it

)sin π(−2it2)

×(

sin πs1 + sin π

(12

+ it2 − S

))φ−

i( 12

−S)(t),

Acta Mathematica Hungarica 135, 2012

352 A. BIRO

φ+λ (x) = φλ

(x;

34

+ it2,14

+ it1,14

− it1,34

− it2

),

φ−λ (x) = φλ

(x;

34

− it2,14

+ it1,14

− it1,34

+ it2

),

and let

N(S, t) = N+(S, t) + N −(S, t).

For any a use the notation

Wj

(x

2; a +

±it1 ± it22

)(1.1)

for the Wilson polynomial

Wj

(x

2; a + i

t1 + t22

, a + it1 − t2

2, a − i

t1 + t22

, a − it1 − t2

2

).

For integers N � 1 define

V (N, x) =N∑

j=1

(1 − N)j−1(12 + N)

j−1

( 12)j

( 12 ± it1)j

( 12 ± it2)j

Wj

(x

2;14

+±it1 ± it2

2

).

The notation∫(σ) for a real number σ will mean that we integrate over the

line with real part σ, i.e. from σ − i∞ to σ + i∞.

1.2. Statement of the result.

Theorem. There is a positive constant K depending only on t1 and t2such that the following statements are true. If f is an even holomorphicfunction on the domain | Imx| < K satisfying that |f(x)e−π|x|(1 + |x|

)K | isbounded on this domain, and

∫ ∞

− ∞f(y)Γ

(14

± it12

± it22

± iy

2

)dy = 0,(1.2)

then

∞∑n=1

(−1)n Γ(12 + n)Γ(n)

(n − 1

4

) Γ(12 ± it2 + n)

Γ(±it2 + n)Hf (n)N

(12

+ ix, i

(14

− n

))(1.3)



equals

22ixΓ(14 ± it1

2 ± it22 + ix

2 )Γ3(12 ± it2)

Γ(12 ± ix)

πf(x)(1.4)

for every real x, where

Hf (n) =∫ ∞

− ∞f(y)Γ

(14

± it12

± it22

± iy

2

)V (n, y) dy.

We have

Hf (n) = Of,t1,t2

(n−4)(1.5)

as n → ∞, and the sum in (1.3) is absolutely convergent for every real x.

Remark 1. The Wilson functions have a close connection to automor-phic forms (see [2]). We will show in a forthcoming paper that the functionseries N(S, i( 1

4 − n)) also has a relation to automorphic forms. However,the present paper belongs entirely to the subject of special functions.

We now discuss briefly these connections to automorphic forms, but sincewe do not work with automorphic forms in the present paper, this discussionwill be very sketchy.

In [2] we proved a Poisson-type summation formula for every given pairu1, u2 of Maass cusp forms of weight 0 for the full modular group. So, on theone hand, let u1 and u2 be two fixed such forms with Laplace-eigenvalues− 1

4 − t21 and − 14 − t22, respectively. On the other hand, let uj,1/2 (j � 1) be

a complete orthonormal system of cusp forms of weight 1/2 for the groupΓ0(4) (the congruence subgroup of level 4) with Laplace-eigenvalues − 1

4 − T 2j

with some real numbers Tj . Let

D+ ={

i

(14

− n

): n � 1 is an integer

}.

The formula of [2] states that there are some weights Uu1,u2(j), Vu1,u2(r) andWu1,u2(n) such that if F is a “nice” function on R ∪ D+, even on R, thenthe expression

∞∑j=1

Uu1,u2(j)F (Tj) +∫ ∞

− ∞Vu1,u2(r)F (r) dr +

∞∑n=1

Wu1,u2(n)F(

i

(14

− n

))

remains unchanged if we write u2 in place of u1, u1 in place of u2, and wereplace F by G, where G is obtained from F by applying a certain inte-gral transform which maps functions on R ∪ D+, even on R again to such


354 A. BIRO

functions: this integral transform is a Wilson function transform of type II,which was introduced by Groenevelt in [4]. This integral transform playsthe role what the Fourier transform plays in the case of Poisson’s formula.

The weights U , V , W in the above formula are interesting automorphicquantities, their concrete form can be found in [2].

The functions N(S, i( 14 − n)) of the present paper are related to the

weights Wu1,u2(n) in the above formula. It will be shown in a forthcomingpaper that these weights essentially equal

12πi

∫( 1

2 )ζ(2S)Lu1,u2(S)fn(S) dS,

where ζ is the Riemann zeta-function, Lu1,u2(S) is the Rankin–Selberg L-function of u1 and u2, and fn(S) equals N(S, i( 1

4 − n)) up to some simplefactor.

Remark 2. We will also show in a forthcoming paper that N(S,

i( 14 − n)) is closely related to continuous Hahn polynomials.

1.3. Structure of the paper and a convention. The main resultof Section 2 will be the proof of (1.5), but the lemmas of Section 2 alsoprepare the proof of the expansion formula given in Section 3. In ordernot to interrupt the reasoning in Sections 2 and 3 we postpone the proof ofsome important lemmas to Section 4, and we just quote them in the earliersections.

Convention. Since the positive numbers t1 and t2 are fixed, we willnot denote the dependence on t1 and t2 in the sequel. In particular, if wespeak about an absolute constant, this will mean that this constant dependsonly on t1 and t2.

2. Estimation of the coefficients in the expansion formula

In order to overcome some convergence problems during the computa-tions in Section 3, we need to introduce a parameter 0 � c � 1. The case weare really interested in is c = 0.

Lemma 2.1. Define F (x, z) as

12πi

∫( 1

8 )

Γ(s ± ix2 )Γ( 1

4 − i t1−t22 − s)Γ( 1

4 ± i( t1+t22 ) − s)Γ( 5

4 + i t1−t22 + s + z)

Γ( 14 + i t1−t2

2 + s)Γ( 74 − i t1−t2

2 − s + z)ds

(2.1)



for real x and Re z > −11/8, and write

Wc(N, x) :=1

2πi

∫(− 1

2 )

Γ( 52 + z)Γ( 1

2 + N + z)Γ(1 + z)Γ(−z)F (x, z)

Γ( 52 + c + z)Γ( 3

2 + it1 + z)Γ( 32 − it2 + z)Γ(N − z)

dz

(2.2)

for real x, 0 � c � 1 and integers N � 1. Define also

Vc(N, x) =Γ(N)Γ(1

2)Γ(12 ± it1)Γ(1

2 ± it2)Γ(1

2 + N)Γ(14 − i t1−t2

2 ± ix2 )Γ(1

4 ± i( t1+t22 ) ± ix

2 )Wc(N,x).

(2.3)

(i) The double integral in s and z obtained by substituting (2.1) into (2.2)is absolutely convergent.

(ii) There is an absolute constant K0 > 0 such that the following state-ments are true. We have

∣∣Vc(N,x)∣∣ � K0e

π|x|(1 + |x| + N)K0(2.4)

for real x, 0 � c � 1 and integer N � 1. For every ε > 0 there is a Cε > 0such that for real x and Re z � − 1

2 we have

∣∣F (x, z)∣∣ � Cε

(1 + |z|

) − 12+ε

e− π

2|x|(1 + |x|

)K0 .(2.5)

(iii) For real x and integers j,N � 1 we have that (1.1) with a = 14 there

equals

Γ(12 + j)Γ(1

2 − it1 + j)Γ(12 + it2 + j)

Γ(14 − i t1−t2

2 ± ix2 )Γ(1

4 ± i( t1+t22 ) ± ix

2 )F (x, j − 1)(2.6)

and

V (N,x) = V0(N,x).(2.7)

(iv) Let K be a real number which is larger than a suitable positive absoluteconstant. Assume that f is an even holomorphic function on the domain| Imx| < K satisfying that |f(x)e−π|x|(1+ |x|

)K | is bounded on this domain.For 0 < Re s < K

4 define

gf (s) =1

Γ(14 + i t1−t2

2 + s)

∫ ∞

− ∞f(x)Γ

(14

+ it1 − t2

2± ix

2

)Γ

(s ± ix

2

)dx.

(2.8)


356 A. BIRO

Then for integers N � 1 and 0 � c � 1,∫ ∞

− ∞f(x)Γ

(14

± it12

± it22

± ix

2

)Vc(N,x) dx

equals the product of

Γ(N)Γ(12 ± it1)Γ(1

2 ± it2)(1

2)N

(2.9)

and

12πi

∫(− 1

2 )

Γ(52 + z)Γ(1

2 + N + z)Γ(1 + z)Γ(−z)Gf (z)

Γ(52 + c + z)Γ(3

2 + it1 + z)Γ(32 − it2 + z)Γ(N − z)

dz,(2.10)

where we define

Gf (z) :=1

2πi(2.11)

×∫( 1

8 )

gf (s)Γ(14 − i t1−t2

2 − s)Γ(14 ± i( t1+t2

2 ) − s)Γ(54 + i t1−t2

2 + s + z)Γ(7

4 − i t1−t22 − s + z)

ds

for Re z > −11/8. The triple integral in s, x and z obtained by substituting(2.11) and (2.8) into (2.10) is absolutely convergent.

Proof. The statements in (i) and (ii) are more or less straightforwardby Stirling’s formula, we only note that for the proof of (2.5) we have toshift the line of integration in (2.1) from Re s = 1/8 to Re s = δ with a smallδ > 0. We get (2.6) by Lemma 4.1, and then (2.7) follows by shifting theline of integration to the right in (2.2). The proof of (iv) is easy (we use onlythe upper bound for f on the real line, we do not need here the holomor-phicity, we imposed that condition only because we will use later only suchfunctions f ). �

Lemma 2.2. Assume that f satisfies the conditions given in Lemma2.1(iv) with some K, and let gf (s) be as in (2.8). Let k > 0 be arbitrary.If K in Lemma 2.1(iv) is large enough depending on k, then gf (s) extendsmeromorphically to | Re s| < k, it is holomorphic there except the points − j

2with integers j satisfying 0 � j < 2k, it has at most simple poles at thesepoints, and it satisfies that

|gf (s)e− π

2|s|(1 + |s|

)k |is bounded for | Re s| < k, | Im s| � 1.



Proof. We first remark that if

f(x) =(

14

+ it1 − t2

2± ix

2

)j

(2.12)

with some given integer j � 0, then the statement is true. Indeed, if (2.12)holds, then we have

gf (s) = 2πΓ( 1

2 + it1 − it2 + 2j)Γ2( 14 + i t1−t2

2 + s + j)Γ(2s)

Γ( 12 + it1 − it2 + 2s + 2j)Γ( 1

4 + i t1−t22 + s)

by [7], p. 644, 6.412, and every required property of g can be easily checked.Since every even polynomial P can be written as a linear combination of

functions of the form ( 14 + i t1−t2

2 ± ix2 )

j, so it is enough to prove the state-

ment for gf −P in place of gf with some fixed even polynomial P (which maydepend on K). It is clear that there is an even polynomial P (dependingonly on K) such that

D(x) :=(f(x) − P (x)

)Γ

(12

± ix

2

)Γ

(14

+ it1 − t2

2± ix

2

)

is holomorphic on the domain | Im x| < K. It is also clear by the conditionon f that |D(x)

(1 + |x|

)K | is bounded on this domain. By these propertiesof D it follows that

Y −K/10∫ ∞

− ∞D(x)(Y

ix

2 + Y − ix

2 ) dx(2.13)

(say) is bounded for 0 < Y � 1.We have that

gf −P (s) =1

Γ(14 + i t1−t2

2 + s)

∫ ∞

− ∞D(x)

Γ(s ± ix2 )

Γ(12 ± ix

2 )dx.(2.14)

Assume that 0 < Re s < 12 . Then we have

Γ(s ± ix2 )

Γ(12 ± ix

2 )=

12Γ(1 − 2s) sin πs

∫ 1

0Y s−1(Y

ix

2 + Y − ix

2 )(1 − Y )−2s dY,

(2.15)


358 A. BIRO

this follows easily from [7], pp. 898–899, 8.380.1 and 8.384.1, and from theidentity

1Γ(1

2 ± ix2 )

=1

2 sin πs

(1

Γ(1 − s + ix2 )Γ(s − ix

2 )+

1Γ(1 − s − ix

2 )Γ(s + ix2 )

).

This identity follows easily from the fact that for fixed x the right-hand sideis a bounded entire function of s, hence it is constant, and its value at s = 1

2is just the left-hand side.

From (2.14), (2.15) and the boundedness of (2.13) we can easily see thatgf −P (s) sin πs extends holomorphically to − K

10 < Re s � 14 , and it satisfies

that

|gf −P (s)e− π

2|s|(1 + |s|

) − Re s|(2.16)

is bounded for − K20 � Re s � 1

4 , | Im s| � 1. Using the original definition(2.8) and the upper bound for f to estimate gf −P (s) for Re s � 1

4 , by(2.16) and the Phragmen–Lindelof principle we obtain the required esti-mate for gf −P (s). (Note that first we have good estimates for Re s = − K

20

and for Re s � 14 , and then we apply the Phragmen–Lindelof principle in the

form of Lindelof’s Theorem (see [3], p. 184) to get a good estimate also for− K

20 � Re s � 14 .) �

Lemma 2.3. Let f satisfy the conditions given in Lemma 2.1(iv) withsome K, and assume that K is larger than a suitable positive absolute con-stant. Assume also that (1.2) is true. Then (see (2.11) for Gf (z))

Gf (z)Γ(3

2 + it1 + z)Γ(32 − it2 + z)

(2.17)

extends regularly to Re z > − 94 and

∣∣Gf (z)|z|A∣∣(2.18)

is bounded with the exponent A = 14 for

− 178

� Re z � − 12, | Im z| � t1 + t2 + 1.(2.19)



There are some constants cj,f such that for Re z = − 178 the expression

|z|154

∣∣∣∣∣∣Gf (z) −3∑

j=0

cj,f

Γ(54 + i t1−t2

2 − j2 + z)

Γ(74 − i t1−t2

2 + j2 + z)

∣∣∣∣∣∣(2.20)

is bounded. Furthermore, Gf (−1) = 0.

Proof. By using (2.11) and that K is large (applying for Re s = 18 the

upper bound given for gf (s) in Lemma 2.2) we see that (2.18) is true forRe z = − 1

2 .By shifting the line of integration in (2.11) to Re s = 1 we see that Gf (z)

equals the sum of

gf

(14

− it1 − t2

2

)Γ(it1)Γ(−it2),(2.21)

gf

(14

+ it1 + t2

2

)Γ(−it1)Γ

(− i(t1 + t2)

) Γ(32 + it1 + z)

Γ(32 − it1 + z)

,(2.22)

gf

(14

− it1 + t2

2

)Γ(it2)Γ

(i(t1 + t2)

) Γ(32 − it2 + z)

Γ(32 + it2 + z)

(2.23)

and

12πi

∫(1)

gf (s)Γ(14 − i t1−t2

2 − s)Γ(14 ± i( t1+t2

2 ) − s)Γ(54 + i t1−t2

2 + s + z)Γ(7

4 − i t1−t22 − s + z)

ds

(2.24)

in the case Re z > −11/8. We can then see the analytic continuation of(2.17) to the domain Re z > − 9

4 , and we also get (similarly as above) theestimate (2.18) for every z in (2.19) with the weak exponent A = −3/2.

Let us consider now Gf (z) for the case Re z = − 178 . We now shift the

line of integration in (2.24) to Re s = − 138 . The residues coming from the

poles at

s =14

− it1 − t2

2, s =

14

+ it1 + t2

2, s =

14

− it1 + t2

2

cancel out with (2.21), (2.22) and (2.23). Using also Lemma 2.2 and thatK is large enough, we see that the product of |z|

154 and the obtained integral


360 A. BIRO

on Re s = − 138 is bounded in z. We see in the same way that the residues

coming from the poles of

Γ(

54

+ it1 − t2

2+ s + z

)

are even smaller, if K is large. Hence we are left with the poles of gf (s), i.e.with s = 0, − 1

2 , −1, − 32 . Formula (2.20) is proved. In particular, this implies

that (2.18) is true for Re z = − 178 .

Since we proved that (2.18) is true for Re z = − 12 and Re z = − 17

8 , andwe also proved it for every z in (2.19) with a weaker exponent, hence (2.18)follows by applying again Lindelof’s Theorem ([3], p. 184).

We finally prove that Gf (−1) = 0. By (2.11) and (2.8) we have

Gf (−1) =∫ ∞

− ∞f(x)Γ

(14

+ it1 − t2

2± ix

2

)A(x) dx(2.25)

with

A(x) :=1

2πi

∫( 1

8 )

Γ(s ± ix2 )Γ( 1

4 − i t1−t22 − s)Γ( 1

4 ± i( t1+t22 ) − s)

Γ( 34 − i t1−t2

2 − s)ds.

By [8], (4.2.2.2) we have

A(x) =Γ( 1

4 − i t1−t22 ± ix

2 )Γ( 14 ± i( t1+t2

2 ) ± ix2 )

Γ( 12)Γ( 1

2 + it2)Γ( 12 − it1)

,

so we get Gf (−1) = 0 by (1.2) and (2.25). �

Lemma 2.4. Let K be a real number which is larger than a suitable pos-itive absolute constant. Assume that f is an even holomorphic function onthe domain | Im x| < K such that |f(x)e−π|x|(1 + |x|

)K | is bounded on thisdomain. Let f satisfy (1.2). Then there is a constant df such that for everyinteger N � 1 the difference

∫ ∞

− ∞f(x)Γ

(14

± it12

± it22

± ix

2

)Vc(N,x) dx

− dfΓ(1 + c)

Γ(32 + c + N)Γ(2 + c − N)

is a continuous function for 0 � c � 1, and its maximum is Of

(N −4

). In

particular, (1.5) is true.



Proof. We first note that using (2.7) the case c = 0 really implies (1.5).The continuity follows at once from Lemma 2.2(iv) and (2.18), (2.19).Let N � 2. By Lemma 2.3 we see that the difference of

12πi

∫(− 1

2 )

Γ(52 + z)Γ(1

2 + N + z)Γ(1 + z)Γ(−z)Gf (z)

Γ(52 + c + z)Γ(3

2 + it1 + z)Γ(32 − it2 + z)Γ(N − z)

dz(2.26)

and

12πi

∫(− 17

8 )

Γ(52 + z)Γ(1

2 + N + z)Γ(1 + z)Γ(−z)Gf (z)

Γ(52 + c + z)Γ(3

2 + it1 + z)Γ(32 − it2 + z)Γ(N − z)

dz(2.27)

equals the residue at z = −2, hence the maximum of this difference for 0 � c

� 1 is Of

(N −7/2

). For the estimation of (2.27) we will use (2.20). First

remark that there are constants dj,f such that

3∑j=0

cj,f

Γ( 54 + i t1−t2

2 − j2 + z)

Γ( 74 − i t1−t2

2 + j2 + z)

equals

3∑j=0

dj,f

Γ( 32 + it1 + z)Γ( 3

2 − it2 + z)Γ(−j − z)

Γ(1 + z)Γ( 52 + z)Γ(−z)

+ Of

(|z| −9/2) ;

indeed, this follows from Stirling’s formula. Hence (2.20) implies that

Gf (z) =3∑

j=0

dj,f

Γ(32 + it1 + z)Γ(3

2 − it2 + z)Γ(−j − z)

Γ(1 + z)Γ(52 + z)Γ(−z)

+ Of

(|z| −15/4)

(2.28)

for Re z = − 178 . We substitute this relation into (2.27). The Of

(|z| −15/4)

part and the j = 3 term of the sum give there Of

(N −15/4

).

The 0 � j � 2 part of the sum in (2.28) gives

2∑j=0

dj,f

2πi

∫(− 17

8 )

Γ(12 + N + z)Γ(−j − z)

Γ(52 + c + z)Γ(N − z)

dz(2.29)

in (2.27). The integrals can be computed, so (2.29) equals

2∑j=0

dj,f

Γ( 12 + N − j)

Γ( 52 + c − j)Γ(N + j)

F

(12

+ N − j, 1 − N − j,52

+ c − j; 1)

,


362 A. BIRO

and by [8], (1.7.7) this equals

2∑j=0

dj,f

Γ( 12 + N − j)Γ(1 + c + j)

Γ(N + j)Γ( 32 + c + N)Γ(2 + c − N)

.

The terms j = 1, 2 are continuous functions for 0 � c � 1 with maximumOf

(N −4

). Considering (2.26), (2.9) and (2.10) the reasoning above proves

the present lemma. �

3. Proof of the expansion formula

Similarly to the parameter c, we introduce now another parameter C tohandle the convergence problems. The relevant case will be C = 1.

Lemma 3.1. Let f and S be given such that f is a function satisfyingthe conditions of Lemma 2.4 with some K, and S �= 1

2 is a complex numberwith Re S = 1

2 . For integers n � 1 and real numbers 0 � c � 1 write

Hf,c(n) =∫ ∞

− ∞f(y)Γ(

14

± it12

± it22

± iy

2 )Vc(n, y) dy.(3.1)

Introduce the notations

I(n,C) =Γ2(1

2 + n)(C)n−1

Γ2(n)Γ(32 + n − C)

(n − 1

4

) Γ(12 ± it2 + n)

Γ(±it2 + n)(3.2)

and

Σ+f,c(S,C) =

∞∑n=1

(−1)nI(n,C)Hf,c(n)N+(

S, i

(14

− n

)).(3.3)

(i) Σ+f,c(S, 1) is absolutely convergent for every 0 � c � 1, and

limc→0+0

Σ+f,c(S, 1) = Σ+

f,0(S, 1).(3.4)

(ii) For every fixed 0 < c � 1 the function Σ+f,c(S,C) is a regular function

of C on a domain containing the half-plane Re C � 1.(iii) There is a positive absolute constant K1 > 0 such that if the condi-

tions of Lemma 2.4 are true for f with a K satisfying K � K1, then forevery C � −K1, for every S �= 1

2 with ReS = 12 and for every 0 < c � 1 the

following statement is true: the sum

Σ+c (y, S, C) :=

∞∑n=1

(−1)nI(n,C)Vc(n, y)N+(

S, i

(14

− n

))(3.5)



is absolutely convergent for every real y, moreover,

Σ+f,c(S,C) =

∫ ∞

− ∞f(y)Γ

(14

± it12

± it22

± iy

2

)Σ+

c (y, S, C) dy,(3.6)

and the integral on the right-hand side of (3.6) is also absolutely convergent.

Proof. Statements (i) and (ii) follow from (4.29), (3.2), (3.1) andLemma 2.4. For the proof of (3.4) we use also the fact that the Riemann zeta-function is regular at the points 1 ± 2ix. Statement (iii) follows from (2.4).�

Let K1 be as in Lemma 3.1(iii), and let us consider now (3.5) with givenc, y, S and C such that 0 < c � 1, y is real, S satisfies (4.23) with x �= 0,and C � −K1. We apply (2.3), (2.2) and formulas (4.24)–(4.27), and getthat (3.5) equals

E(x, y)(

12πi

)2 ∫(− 1

2 )

∫γε

Uc(z)F (y, z)Φ(x, R)J(z,R,C) dR dz,(3.7)

where

Uc(z) :=Γ(5

2 + z)Γ(1 + z)Γ(−z)

Γ(52 + c + z)Γ(3

2 + it1 + z)Γ(32 − it2 + z)

,(3.8)

E(x, y) :=E(x)Γ(1

2)Γ(12 ± it1)Γ(1

2 ± it2)Γ(1

4 − i t1−t22 ± iy

2 )Γ(14 ± i( t1+t2

2 ) ± iy2 )

,(3.9)

J(z,R,C) :=∞∑

n=1

(−1)nI(n,C)Γ(n)Γ(1

2 + n + z)Γ(1

2 + n)Γ(n − z)G(n,R),(3.10)

and (taking into account (2.5)) the double integral in R and z is absolutelyconvergent if K1 is large enough.

By (4.27) and (3.2) we see that J(z,R,C) equals the product of

Γ(32 + z)

Γ(1 − z)Γ(3

2)Γ(5

2 − C)

(34

) Γ(32 + it2)

Γ(1 − it2)Γ(1

2 ± it2)Γ(3

2 + it2 + R)Γ(it2 + R)(3.11)

and

6F5

(74 , 3

2 , C, 32 + it2,

32 + z, 1 − it2 − R

34 , 5

2 − C, 1 − it2, 1 − z, 32 + it2 + R

; −1

).(3.12)


364 A. BIRO

By applying [8], (6.2.3), writing a = b = 32 there, we see that if K1 is large

enough, then (3.12) equals

Γ(1 − z)Γ( 32 + it2 + R)

Γ( 52)Γ(−z + it2 + R) 3F2

(1 − C − it2,

32 + z, 1 − it2 − R

52 − C, 1 − it2

; 1)

.

By [8], (2.3.3.7) we see that this (and so (3.12) also) equals

Γ(1 − z)Γ(52 − C)Γ(1 − it2)

Γ(52)Γ(1 − it2 − R)Γ(1 − C + R − z)

(3.13)

× 3F2

(32 − C + it2 + R,R, R + it2 − z

32 + R + it2, 1 − C + R − z

; 1)

.

Let K1, f , C, S and c be as in Lemma 3.1(iii). Substituting (3.5) into(3.6), and expressing (3.5) as (3.7), we get a triple integral in y, z and Rfor Σ+

f,c(S, C). Estimate here F (y, z) by (2.5), where we may choose ε tobe the same small positive number as in (4.28). If ε is small enough (interms of c), then we see by (3.8), (3.9), (3.11), (3.12), (3.13), Lemma 4.7and (4.26) that the triple integral in y, z and R expressing Σ+

f,c(S, C) isabsolutely convergent. We also see by the same estimates that if f, c and Sare fixed, then this triple integral is a regular function of C for Re C < 1,| Im C| < 1, and it is continuous for Re C < 1, | Im C| � 1. Since we haveseen the same statement for Σ+

f,c(S,C) in Lemma 3.1(ii), and we have seenthe equality for C � −K1, it follows that this triple integral is also valid forΣ+

f,c(S, 1). Hence we proved the following lemma.

Lemma 3.2. Let f satisfy the conditions of Lemma 2.4 with a largeenough absolute constant K, and let S �= 1

2 satisfying (4.23) and also 0 < c

� 1 be fixed. Then Σ+f,c(S, 1) equals

Γ(12 ± it2)Γ(3

2 + it2)2

(1

2πi

)2

(3.14)

×∫(− 1

2 )

∫γε

U ∗c (z)Vf (x, z)Φ∗(x, R)F3,2(z,R) dR dz,

where

U ∗c (z) := Γ

(32

+ z

)Uc(z),(3.15)

Φ∗(x,R) :=Φ(x, R)

Γ(32 + it2 + R)Γ(it2 + R)Γ(1 − it2 − R)

,(3.16)



F3,2(z,R) :=1

Γ(R − z)3F2

(12 + it2 + R, R, R + it2 − z

32 + R + it2, R − z

; 1)

,(3.17)

Vf (x, z) :=∫ ∞

− ∞f(y)Γ

(14

± it12

± it22

± iy

2

)E(x, y)F (y, z) dy,(3.18)

and the triple integral in y, z and R is absolutely convergent. �Let us compute in (3.15) the z-integral for a fixed R ∈ γε, i.e. the integral

12πi

∫(− 1

2 )U ∗

c (z)Vf (x, z)F3,2(z,R) dz.(3.19)

We estimate here F3,2(z,R) by Lemma 4.8, and we estimate Vf (x, z) using(3.18), (3.9), the upper bound for f given in Lemma 2.4, and (2.5) with an εwhich is small in terms of c. Taking into account also (3.15), (3.8), (3.17) and(2.1), we see that we can compute (3.19) by shifting the z-integration to theright there: since we apply Lemma 4.8, we replace the line of integration tothe lines ( − 1

2 +N) with positive integers N , and we tend to infinity with N .We get that (3.19) equals the sum of the residues at the nonnegative integers,i.e. at the points z = j − 1, where j is a positive integer.

We have seen in Lemma 2.1(iii) that (2.6) equals (1.1) with a = 14 there,

therefore we see for every j � 1 by (3.18) and (3.9) that

Vf (x, j − 1) =E(x)Γ( 1

2 + it1)Γ( 12 − it2)

( 12)j(

12 − it1)j(

12 + it2)j

If (j)

with

If (j) :=∫ ∞

− ∞f(y)Γ

(14

± it12

± it22

± iy

2

)Wj

(y

2;14

+±it1 ± it2

2

)dy.

(3.20)

By estimating If (j), using again the equality of (1.1) and (2.6), and usingalso (2.5), we see for every ε > 0 that

If (j)(1

2)j(1

2 − it1)j(1

2 + it2)j

= Of,ε

(jε− 1

2

).(3.21)

On the other hand, by (3.17) and by Corollary 3.3.5 of [1],

F3,2(z,R) =Γ( 3

2 + R + it2)Γ(1 − R − it2)

Γ( 32)Γ(R − z) 3F2

(12 + it2 + R, −z, −it2

32 , R − z

; 1)

.


366 A. BIRO

Hence if z is a nonnegative integer, then by [8], (2.3.1.4)

F3,2(z,R) =Γ( 3

2 + z + it2)Γ(1 − R)Γ(1 − R − it2 + z)Γ( 32 + R + it2)

Γ(1 − R + z)Γ( 32 + z)Γ( 3

2 + it2)Γ(R − z).

Therefore, shifting the integration to the right and using the above expres-sions for Vf (x, z) and F3,2(z,R) at z = j − 1, we finally get that (3.19) equalsthe product of

E(x)Γ(1

2 + it1)Γ(12 ± it2)Γ(3

2 + R + it2)Γ(3

2 + it2)Γ(R)(3.22)

and∞∑

j=1

Γ(32 + j)Γ(−R − it2 + j)

Γ(32 + c + j)Γ(1

2 + it1 + j)Γ(12 − it2 + j)(1

2)j(1

2 − it1)j

If (j).(3.23)

Using the estimate (3.21) it is clear that substituting in place of (3.19) theproduct of (3.22) and (3.23) into (3.14), the sum in j and the integral inR are together absolutely convergent. Hence we may change the order ofsummation and integration, so we may compute the R-integral for everygiven j. By (3.14), (4.26), (3.16) we see that we then have to compute theintegral

12πi

∫γε

Γ(12 + it2 ± it1 + R)Γ(1

2 + R)Γ( − R ± ix)Γ(−R − it2 + j)Γ(it2 + R)Γ(1 − it2 − R)

dR

(3.24)

for any fixed integer j � 1. Since

Γ(−R − it2 + j)Γ(it2 + R)Γ(1 − it2 − R)

= (−1)j−1 1Γ(R + it2 + 1 − j)

,

we see by [8], (4.2.2.1) that (3.24) equals

(−1)j−1 Γ(12 + it2 ± ix ± it1)Γ(1

2 + ix)Γ(1 + 2it2)

W (it1),(3.25)

where we write

W (Z) :=1

Γ(it2 + ix + 1 − j)(3.26)

× 3F2

(12 + it2 + ix + Z, 1

2 + it2 + ix − Z, it2 + 12 − j

1 + 2it2, it2 + ix + 1 − j; 1

).



It is clear that for fixed real t2, x and integer j � 1 the function W is entire.By Corollary 3.3.5 of [1], W (Z) equals

Γ(12 − ix)

Γ(12 − Z − j)Γ(it2 + 1 + Z)

(3.27)

× 3F2

(12 + it2 + ix + Z, 1

2 + it2 − ix + Z, it2 + 12 + j

1 + 2it2, it2 + 1 + Z; 1

)

for Re Z < 12 − j. We can then compute (3.24) using (3.25), (3.27) and

Lemma 4.5, and the result is that (3.24) equals

−πΓ(1

2 + it2 ± ix ± it1)Γ(1 + it2)Γ(12 + it2)Γ(1

2 ± ix)Γ(1 + 2it2)Γ(1

4 + it22 ± it1

2 ± ix2 )Γ(1

2 + j)Γ(12 + it2 + j)

Uj

(x

2

),(3.28)

where Uj(x2) denotes

Γ(14 + it2

2 ± it12 ± ix

2 )Γ(3

4 + it22 ± it1

2 ± ix2 )

Wj

(x

2;14

+±it1 ± it2

2

)(3.29)

− jWj−1

(x

2;34

+±it1 ± it2

2

)

with the notation (1.1). Hence using the expression (3.28) obtained in thisway for (3.24), together with (3.14), (3.19), (3.22), (3.23), (4.26) and (3.16)we get that (3.14) equals the product of

−πE(x)Γ( 1

2 ± it1)Γ2( 12 ± it2)Γ( 1

2 + it2 ± ix ± it1)Γ(1 + it2)Γ( 12 + it2)Γ( 1

2 ± ix)2Γ(1 + 2it2)Γ( 1

4 + it22 ± it1

2 ± ix2 )

(3.30)

and

∞∑j=1

Γ(32 + j)

Γ(32 + c + j)Γ(1

2 ± it2 + j)(12)j

Γ(12 + j)Γ(1

2 ± it1 + j)If (j)Uj

(x

2

),

(3.31)

and the sum in (3.31) is absolutely convergent.To simplify (3.30) we use the following two identities. On the one hand,

sin πs1 + sin π

(12

− it2 − S

)=

2π2

Γ(1±(S+it2)±it12 )

;(3.32)


368 A. BIRO

this follows from [7], p. 29, 1.314.1. On the other hand,

Γ( 12 + it2 + ix ± it1)Γ(1 + it2)Γ( 1

2 + it2)Γ(1 + 2it2)Γ( 1

4 + it22 ± it1

2 + ix2 )Γ( 3

4 + ix+it2±it12 )

equals 22ix−1√

π; this follows from (4.41). Using (4.25) and the two identities

above we get that (3.30) equals

−π5/2Γ(1

2 ± it1)Γ2(12 ± it2)22ix−1

Γ(14 ± it2

2 ± it12 − ix

2 )Γ(±ix) sinπ(2it2).(3.33)

Hence by Lemma 3.2 and the fact proved above that (3.14) equals the prod-uct of (3.30) and (3.31), we get that if f , c and S satisfy the conditions ofLemma 3.2, then Σ+

f,c(S, 1) equals the product of (3.33) and (3.31). Let usassume now that f is real on the real line. Then If (j) is real by (3.20), and

Re

(−21−2ix

Γ(±ix)Γ(14 ± it2

2 ± it12 − ix

2 )Γ(1

2 ± it1)Γ2(12 ± it2)π5/2

Σ+f,c(S, 1)

)(3.34)

equals

∞∑j=1

Γ(32 + j)If (j)

Γ(32 + c + j)Γ(1

2 ± it2 + j)(12)j

Γ(12 + j)Γ(1

2 ± it1 + j)(3.35)

× Re

(Uj(x

2)sin π(2it2)

).

To compute the real part in (3.35) using (3.29) we note the identity that

Γ(14 + it2

2 ± it12 ± ix

2 )Γ(3

4 + it22 ± it1

2 ± ix2 ) sin π(2it2)

+Γ(1

4 − it22 ± it1

2 ± ix2 )

Γ(34 − it2

2 ± it12 ± ix

2 ) sin π(−2it2)

(3.36)

equals

−Γ(1

4 ± it22 ± it1

2 ± ix2 )Γ(1

2 ± it2)2π3Γ(1

2 ± ix)Γ(12 ± it1)

.(3.37)

This follows if we consider the quotient of (3.36) and (3.37) as a functionof t2. Indeed, it can be easily shown that this quotient is a bounded entirefunction of t2 (the boundedness is true only for the sum of the two terms,



and not separately!), and the value of the quotient is indeed 1 at the pointit2 = it1 + ix + 3

2 . By (3.29), (3.36) and (3.37) we then get that (3.35) (andso (3.34) also) equals the product of (3.37) and

∞∑j=1

Γ(32 + j)Wj(x

2 ; 14 + ±it1±it2

2 )2Γ(3

2 + c + j)Γ(12 ± it2 + j)(1

2)jΓ(1

2 + j)Γ(12 ± it1 + j)

If (j).(3.38)

For fixed f and x we see by (3.4) that the limit of (3.34) as c → 0 + 0 isthe same expression with c = 0. Applying Lemma 4.3, taking into account(1.2) and (3.21) we get that the limit of (3.38) as c → 0 + 0 (along a subse-quence) is

2π3/2Γ(±ix)f(x)

for almost all real x. Hence by (3.34), (3.37) and (3.38),

Re

(21−2ix

Γ(14 ± it2

2 ± it12 − ix

2 )Γ2(1

2 ± it2)πΣ+

f,0(S, 1)

)(3.39)

equals

Γ(14 ± it2

2 ± it12 ± ix

2 )Γ(12 ± it2)

Γ(12 ± ix)

f(x)(3.40)

for almost all real x. The functions in (3.39) and (3.40) are continuous func-tions of x on the real line, hence we have equality for every real x. Thecontinuity is clear in the case of (3.40), and it follows from formulas (3.1)–(3.3), Lemma 4.4(ii) and Lemma 2.4 in the case of (3.39).

We now compute the real part in (3.39). We claim that

Re(

2−2ixΓ(

14

± it22

± it12

− ix

2

)N+

(S, i

(14

− n

)))(3.41)

equals

2−2ix−1Γ(

14

± it22

± it12

− ix

2

) (N+

(S, i

(14

− n

))(3.42)

+ N −(

S, i

(14

− n

))).

Indeed, by the identity

Γ(S ± it1 + it2) =22S+2it2

4πΓ

(S ± it1 + it2

2

)Γ

(S + 1 ± it1 + it2

2

)


370 A. BIRO

(which follows by (4.41)) and by (3.32) we see that for the function

Rt2(S) :=Γ(S ± it1 + it2)( sin πs1 + sin π( 1

2 − it2 − S))22SΓ(S±it1±it2

2 )

we have

Rt2(S) = Rt2(1 − S),

and together with the definition of N+(S, i( 14 − n)), N −(S, i( 1

4 − n)) andthe Wilson function (see [4], (3.3)) this proves the equality of (3.41) and(3.42).

On the other hand, we have by (3.1) and (2.7) that

Hf,0(n) =∫ ∞

− ∞f(y)Γ

(14

± it12

± it22

± iy

2

)V (n, y) dy.

Hence Hf,0(n) is real for every positive integer. Therefore by formulas (3.2),(3.3), (3.39)–(3.42) we finally get the equality of (1.3) and (1.4) stated inthe Theorem. We proved this identity under the condition that f is real onthe real line, but in view of the decomposition

f(x) =f(x) + f

(x)

2+

f(x) − f(x)

2

this condition can be dropped. �

4. Necessary lemmas

4.1. On Wilson polynomials. Note that the next lemma could bededuced from formulas (4.7.1.3) and (2.4.1.1) of [8]. The proof below isessentially equivalent to this reasoning, but we think that it is technicallysimpler to quote formulas of [4] and [5].

Lemma 4.1. Let a, b, c, d be complex numbers with positive realparts, let x be a real number, and let j � 0 be an integer. Let 0 < Re γ <Re b,Re c, Re d, then

12πi

∫(γ)

Γ(s ± ix)Γ(b − s)Γ(c − s)Γ(d − s)Γ(a + s + j)Γ(a + s)Γ(b + c + d − s + j)

ds(4.1)

equals

Γ(c ± ix)Γ(d ± ix)Γ(b ± ix)Γ(c + d + j)Γ(b + c + j)Γ(b + d + j)

Wj(x; a, b, c, d).(4.2)



Proof. By shifting the integration to the left, (4.1) equals

Γ(c ± ix)Γ(d ± ix)Γ(b ± ix)(F+ + F−),(4.3)

where

F+ :=Γ(−2ix)(a − ix)j4F3

(b + ix, c + ix, d + ix, 1 − a + ix

1 − a + ix − j, b + c + d + j + ix, 1 + 2ix; 1)

Γ(b − ix)Γ(c − ix)Γ(d − ix)Γ(b + c + d + ix + j),

(4.4)

F− :=Γ(2ix)(a + ix)j4F3

(b − ix, c − ix, d − ix, 1 − a − ix

1 − a − ix − j, b + c + d + j − ix, 1 − 2ix; 1)

Γ(b + ix)Γ(c + ix)Γ(d + ix)Γ(b + c + d − ix + j).

(4.5)

By (4.4), (4.5) and [4], (3.3) (quoted in detail in our Subsection 1.1),

F+ + F− = (−1)jΓ(1 − a ± ix)φ,(4.6)

where

φ := φ a+d−12i

(b − c

2i;b + c

2+ ix, 1 − a − b + c

2− j,

b + c

2+ d + j, 1 − b + c

2+ ix

).

(4.7)

By (4.7) and [5], Lemma 5.3 (ii),

φ = φ a+d−12i

(b − c

2i;b + c

2+ d + j,

b + c

2− ix,

b + c

2+ ix,

b + c

2+ a + j

),

(4.8)

and then (4.8) and [5], Lemma 5.3 (i) show that

φ = φ a+b+c+d−12

+j(x; , d, b, c, a).(4.9)

By (4.9) and [5], Lemma 5.3 (iii) we see that

φ = (−1)j Wj(x; d, b, c, a)Γ(c + d + j)Γ(b + c + j)Γ(b + d + j)Γ(1 − a ± ix)

.(4.10)

By (4.3), (4.6), (4.10) and the well-known fact that the Wilson polynomi-als are symmetric in the parameters a, b, c, d (as mentioned also in [5],preceding Lemma 5.3), we obtain that (4.1) equals (4.2). �


372 A. BIRO

Lemma 4.2. Let g be an element of the Hilbert space

L2

([0, ∞),

Γ(14 ± i t1

2 ± i t22 ± ix

2 )Γ(±ix)

dx

),(4.11)

and define

αg(j) =12π

∫ ∞

0g(x)

Γ( 14 ± i t1

2 ± i t22 ± ix

2 )Γ(±ix)

Wj

(x

2;14

+±it1 ± it2

2

)dx.

Then

limJ →∞

J∑j=0

Wj(x2 ; 1

4 + ±it1±it22 )αg(j)(1 − δ0,j

2 )Γ(1

2 ± it2 + j)Γ2(12 + j)Γ(1

2 ± it1 + j)= g(x)(4.12)

in the Hilbert space (4.11), where δ0,j is Kronecker’s symbol.

Proof. It follows easily from the orthogonality properties of the Wil-son polynomials ([1], (6.10.4)) and Bessel’s inequality that the left-hand sideof (4.12) is convergent in the Hilbert space (4.11), and if h(x) denotes itslimit, then for the difference

d(x) := g(

|x|)

− h(

|x|)

we have that

∫ ∞

− ∞d(x)P (x)

Γ( 14 ± i t1

2 ± i t22 ± ix

2 )Γ(±ix)

dx = 0

for every polynomial P (note that for odd P this is trivial). It follows by [1],Theorem 6.5.2 that d vanishes almost everywhere. �

Lemma 4.3. Let f be a smooth even function on the real line such thatthere is a positive integer M such that |f(y)e−π|y|(1 + |y|

) −M | is boundedon the real line, and (3.21) holds for every ε > 0 and j � 1, where If (j) isdefined in (3.20) for every j � 0. Then the sum

∞∑j=0

Γ(32 + j)Wj(x

2 ; 14 + ±it1±it2

2 )If (j)(1 − δ0,j

2 )Γ(3

2 + c + j)Γ(12 ± it2 + j)Γ2(1

2 + j)Γ(12 ± it1 + j)

(4.13)



is absolutely convergent for every c > 0 and every real x, and there is astrictly decreasing series {cm}∞

m=1 of positive numbers such that cm → 0 and

limm→∞

∞∑j=0

Γ(32 + j)Wj(x

2 ; 14 + ±it1±it2

2 )If (j)(1 − δ0,j

2 )Γ(3

2 + cm + j)Γ(12 ± it2 + j)Γ2(1

2 + j)Γ(12 ± it1 + j)

(4.14)

= 4πf(x)Γ(±ix)

for almost all real x (in the sense of Lebesgue measure).

Proof. We have seen in Lemma 2.1(iii) that (1.1) with a = 14 there

equals (2.6), and by (2.5) we get the absolute convergence of (4.13) (we takeinto account also (3.21)). We also note that by shifting the integration tothe left in (2.1) we get that if the real number x �= 0 is fixed, then

F (x, j − 1) = j− 12+i(t1−t2)

(C+(x)jix + C−(x)j−ix + O

(1j

))(4.15)

as j → ∞, where C+(x) and C−(x) depend only on x (besides t1 and t2).We first show that (4.14) is true in the case f(0) = 0. Define g(x) =

f(x)Γ(±ix). Then g is an element of the Hilbert space (4.11), and it followseasily from Lemma 4.2 that the sum (4.13) is convergent in the space (4.11)(here we claim L2-convergence, and not pointwise convergence) for everyc � 0, and denoting this L2-sum by Gc(x), we have that

G0(x) = 4πg(x), limc→0+0

Gc(x) = G0(x)(4.16)

in the space (4.11). Since (4.13) is convergent not only in the L2-sense butalso pointwise, so for every fixed c > 0 the pointwise sum of (4.13) equalsGc(x) for almost all x. Hence (4.16) implies that the relation (4.14) holds inthe L2-sense (for any sequence cm). But then we can choose a subsequencecm for which (4.14) holds for almost all x. So we proved (4.14) in the casef(0) = 0.

Define

fD(y) :=18π

Γ(D ± iy2 )

Γ(14 − i t1

2 − i t22 ± iy

2 )Γ(±iy).(4.17)

By Wilson’s formula ([1], Theorem 3.6.2), for every nonnegative integer jand any D � 0, IfD

(j) equals the product of

Γ(

12

+ j

)Γ

(14

+ D ± it1 − t2

2

)Γ

(12

+ it1 + j

)Γ

(12

+ it2 + j

)(4.18)


374 A. BIRO

andj∑

k=0

(−j)k(j)kΓ( 14 + i t1

2 + i t22 + D + k)

k!( 12)k

Γ( 34 + i t1

2 + i t22 + D + k)

.

By [8], (2.3.1.4) this last sum equals

Γ(14 + i t1

2 + i t22 + D)Γ(3

4 + i t12 + i t2

2 + D)Γ(12 − j)Γ(1

2 + j)Γ2(1

2)Γ(34 + i t1

2 + i t22 + D ± j)

.(4.19)

So IfD(j) equals the product of (4.18) and (4.19).

Let the real number x �= 0 be fixed. By (4.18), (4.19), (2.6), (4.15),Lemma 4.6 and the fact that the Riemann zeta-function is regular at thepoints 1 ± ix,

LD(x)(4.20)

:= limc→0+0

∞∑j=0

Γ(32 + j)Wj(x

2 ; 14 + ±it1±it2

2 )(1 − δ0,j

2 )Γ(3

2 + c + j)Γ(12 ± it2 + j)Γ2(1

2 + j)Γ(12 ± it1 + j)

IfD(j)

exists for every D � 0 and every nonzero real x, and

limD→0+0

LD(x) = L0(x)(4.21)

for every nonzero real x.Since fD(0) = 0 for D > 0, so we know from the result already proved

that for every fixed D > 0

LD(x) = 4πfD(x)Γ(±ix)

for almost all real x. We then get by (4.21) and (4.17) that

L0(x) = 4πf0(x)Γ(±ix)(4.22)

for almost all real x.We can now prove (4.14) for the general case. Let us write (using that

f0(0) �= 0)

f(x) =f(0)f0(0)

f0(x) + g(x).

Then for the function g defined in this way we have g(0) = 0, hence we al-ready know (4.14) by writing g in place of f there. By (4.20) and (4.22) weget (4.14) also for f . �



4.2. On Wilson functions.

Lemma 4.4. Let

S =12

+ ix, x ∈ R.(4.23)

(i) If x �= 0, then for any integer n � 1 we have

N+(

S, i

(14

− n

))= E(x)

12πi

∫γε

Φ(x, R)G(n,R) dR,(4.24)

where

E(x) =sin πs1 + sin π(1

2 − it2 − S)Γ( ± ix)Γ(1

2 ± ix)Γ(12 + it2 ± it1 − ix) sin π(2it2)

,(4.25)

Φ(x, R) = Γ(

12

+ it2 ± it1 + R

)Γ(R)Γ

(12

+ R

)Γ(−R ± ix),(4.26)

G(n,R) =Γ(n + it2)Γ(1

2 − n + it2)Γ(1

2 + it2 + n + R)Γ(1 + it2 − n + R),(4.27)

ε > 0 is a fixed small number, and γε is a curve connecting −i∞ and i∞ suchthat the poles of the function Γ(R) lie to the left of the path of integration,and the poles of the functions Γ(ix − R), Γ(−ix − R) lie to the right of it,moreover,

| Re R| < ε for R ∈ γε.(4.28)

(ii) There are two continuous functions, a1(x) and a2(x) on the real linesuch that for any X > 0 we have

N+(

S, i

(14

− n

))= a1(x)n−1+2ix + a2(x)n−1−2ix + OX

(n−2)(4.29)

for any integer n � 1 and any S satisfying (4.23) with |x| � X .

Proof. By the definition of N+(S, t) we see that N+(S, i( 14 − n)) equals

the product of

Γ(S ± it1 + it2)Γ(n + it2)Γ( 12 − n + it2)

sin π(2it2)

(sin πs1 + sin π

(12

− it2 − S

))

and

φx

(i

(14

− n

);34

+ it2,14

+ it1,14

− it1,34

− it2

).(4.30)


376 A. BIRO

By Lemma 5.3(i) of [5], (4.30) equals

φ1/4i

(t1;

12

+ it2 + ix, n,12

− n,12

− it2 + ix

).(4.31)

We use that (4.31) equals the product of

1Γ(±ix)Γ(1

2 ± ix)Γ( 12 + it2 ± it1 ± ix)

and

12πi

∫γε

Γ( 12 + it2 ± it1 + R)Γ(R)Γ( 1

2 + R)Γ(−R ± ix)

Γ( 12 + it2 + n + R)Γ(1 + it2 − n + R)

dR.

One can see this integral representation by shifting the line of integrationto the right in the above integral, and then using [4], formula (3.3) (whichformula was quoted in detail in our Subsection 1.1). So statement (i) follows.For the proof of (ii) for any x �= 0 we replace the curve of integration from γε

to ReR = 12 in (4.24). We see by continuity that the obtained expression for

N+(S, i( 14 − n)) is valid also for x = 0. Statement (ii) follows, the lemma

is proved.

4.3. A generalization of Watson’s theorem. The following lemmacould be deduced from [6], but we decided to prove it directly. The casej = 0 (what we will not need) is Watson’s theorem, see [8], (2.3.3.13).

Lemma 4.5. Let j � 1 be an integer, x a real number, and let z, Z becomplex numbers such that Re Z < 1

2 − j and Re z � 0. Then for j � 1

1πΓ(1 + z + Z)3F2

(12 + z + ix + Z, 1

2 + z − ix + Z, z + 12 + j

1 + 2z, 1 + z + Z; 1

)(4.32)

equals the sum of

Γ(12 − Z)Γ(1 + z)

Γ(12)Γ(3

4 + z2 ± Z

2 ± ix2 )

4F3

(−j, j, 1

4 + z+Z+ix2 , 1

4 + z+Z−ix2

12 + z, 1

2 + Z, 12

; 1

)(4.33)

and

jΓ( − 12 − Z)Γ(1 + z)Γ(1

2 + z)Γ(3

2)Γ(32 + z)Γ(1

4 + z2 ± Z

2 ± ix2 )

(4.34)

× 4F3

(1 − j, 1 + j, 3

4 + z+Z+ix2 , 3

4 + z+Z−ix2

32 + z, 3

2 + Z, 32

; 1

).



Proof. We first prove the special case

z = j − 12.(4.35)

In this case (4.32) equals

Γ(−z − Z)πΓ(1

2 ± ix)(4.36)

by [8], (1.7.6). We can compute also (4.33) and (4.34), using [8], (2.3.1.4).By that formula (and (4.35)) we see that (4.33) equals

Γ(12 − Z)Γ(1

2 + Z)Γ(1 + z)Γ(−z)

Γ(1 + z + Z)Γ2(12)Γ(3

4 + z2 − Z

2 ± ix2 )Γ(1

4 − z2 + Z

2 ± ix2 )

.(4.37)

Similarly, (4.34) equals

Γ(12 − Z)Γ(1

2 + Z)Γ(1 + z)Γ(−z)

Γ(1 + z + Z)Γ2(12)Γ(1

4 + z2 − Z

2 ± ix2 )Γ(3

4 − z2 + Z

2 ± ix2 )

.(4.38)

So we have to prove that (4.36) equals the sum of (4.37) and (4.38). Using(4.35) it is easy to see that

Γ(−z − Z) =Γ( 1

2 − Z)Γ( 12 + Z)Γ(1 + z)Γ(−z)

Γ(1 + z + Z)Γ2( 12)

,

hence it is enough to prove that

πΓ(12 ± ix)

Γ(34 + z

2 − Z2 ± ix

2 )Γ(14 − z

2 + Z2 ± ix

2 )

+πΓ(1

2 ± ix)Γ(1

4 + z2 − Z

2 ± ix2 )Γ(3

4 − z2 + Z

2 ± ix2 )

= 1.

Considering the left-hand side as a function of x, the left-hand side is an en-tire function, since it is also regular at the points where 1

2 + ix is an integer.By Stirling’s formula and periodicity it is easy to check that this functionis bounded, hence it is a constant. At the point x = 0 the identity followsfrom sin2 + cos2 = 1, so we proved the case (4.35).

We now show that the general case can be deduced from this specialcase. Let j and Z be given such that j � 1 is an integer, and

−j < Re Z <12

− j.(4.39)


378 A. BIRO

It is clear that we may assume (4.39). For a complex number z with

Re z � j − 12

(4.40)

denote by C(z) the statement that (4.32) equals the sum of (4.33) and (4.34)for every real x (and for the given j and Z). Hence we proved above thatC(j − 1

2) is true.Let us denote (4.32) by F1(x), (4.33) by F2(x), finally (4.34) by F3(x).

Write

IJ(A)

=∫ ∞

− ∞

Γ( 14 + z

2 ± Z2 ± ix

2 )Γ( 34 + z

2 ± Z2 ± ix

2 )Γ(A ± ix2 )Γ( 1

2 + A ± ix2 )

Γ(±ix)FJ (x) dx

for J = 1,2,3 and ReA > 0. By giving trivial upper bounds for the functionsFJ(x) and using the identity

Γ(s)Γ(

s +12

)= Γ(2s)

√π21−2s(4.41)

twice, it is easy to see that C(z) is true if and only if

I1(A) = I2(A) + I3(A)(4.42)

for every ReA > 0. (Note that already the cases A = m2 of (4.42) with pos-

itive integers m imply the statement C(z).)By Wilson’s formula ([1], Theorem 3.6.2), taking into account (4.39) and

(4.40), we see that

I2(A) = Γ(12

+ z)Γ(1 + z)Γ(

12

+ 2A)

I2,1

(A +

z

2

)I2,2

(A +

z

2

),(4.43)

where

I2,1(B) = 8πΓ(1

2 − Z)Γ(B + 14 − Z

2 )Γ(B + 34 − Z

2 )Γ(1

2),

I2,2(B) =j∑

k=0

(−j)k(j)kΓ(B + 14 + Z

2 + k)Γ(B + 34 + Z

2 + k)k!(1

2 + Z)k(12)k

Γ(1 + 2B + k).

Similarly, we have

I3(A) = Γ(

12

+ z

)Γ(1 + z)Γ

(12

+ 2A)

I3,1

(A +

z

2

)I3,2

(A +

z

2

),

(4.44)



where

I3,1(B) = 8πjΓ( − 1

2 − Z)Γ(34 + B − Z

2 )Γ(54 + B − Z

2 )Γ(3

2),

I3,2(B) =j−1∑k=0

(1 − j)k(1 + j)kΓ(34 + B + Z

2 + k)Γ(54 + B + Z

2 + k)k!(3

2 + Z)k(32)k

Γ(2 + 2B + k).

Finally, using the identity (4.41) several times, as well as

∞∑k=0

(z + 12 + j)k

Γ(12 + 2A + z + Z + k)

k!Γ(32 + 2z + 2A + k)

=Γ(1

2 + 2A + z + Z)Γ(12 − j − Z)

Γ(1 + z − Z)Γ(1 + z + 2A − j),

which follows from [8], (1.7.6), we get that

I1(A) = Γ(12

+ z)Γ(1 + z)Γ(12

+ 2A)I1,1(A +z

2),(4.45)

where

I1,1(B) = π3/224−4BΓ(

12

+ 2B − Z

) Γ( 12 + 2B + Z)Γ( 1

2 − j − Z)Γ(1 + 2B − j)

.

By (4.43), (4.44) and (4.45) we see that (4.42) (and so C(z)) is true if andonly if the identity

I1,1(B) = I2,1(B)I2,2(B) + I3,1(B)I3,2(B)

holds for every B with Re B > Re z/2. But by analytic continuation thiscondition is independent of z. Since we have seen that C(j − 1

2) is true,hence C(z) is true for every z satisfying (4.40). By analytic continuationin z the lemma follows.

4.4. Some asymptotics and upper bounds.

Lemma 4.6. Let z �= 0 be a complex number, | arg z| < 3π4 . Then, as

|z| → ∞, we have

Γ(b + z)Γ(a + z)

= zb−a

(1 + O

(M2

|z|

)),


380 A. BIRO

uniformly in |a|, |b| � |z|/10, where

M = 1 + max(

|a|, |b|).(4.46)

Proof. The statement follows at once from the relation

log Γ(w) =(

w − 12

)log w − w + C0 +

C1

w+ O

(1

|w|2

),

valid with some absolute constants C0, C1 for | arg w| < π − δ (δ is any fixedpositive number), see [9], Section 13.6. Indeed, we use the difference of thecases w = z + a and w = z + b of this relation, and use the first few termsof the power series expansions. �

Lemma 4.7. Assume that ε > 0 is small enough. Then for every R, C,z satisfying

| Re R| < ε,(4.47)

Re z = − 12 , Re C � 1, | Im C| � 1 we have that

∣∣∣∣ 1Γ(1 − C + R − z)3F2

(32 − C + it2 + R, R, R + it2 − z

32 + R + it2, 1 − C + R − z

; 1)∣∣∣∣

equals

O

⎛⎝

∣∣∣∣∣∣1

Γ( 12 + R)Γ

(32 − C + it2 + R

)Γ(−z + it2 + R)

∣∣∣∣∣∣⎞⎠

with implied absolute constant.

Proof. If ε in (4.47) is small enough, then

∣∣(R)k

∣∣ �∣∣∣∣(

12

+ R

)k

∣∣∣∣(4.48)

for every R satisfying (4.47) and every integer k � 0. Using the well-knownfact that

min− ∞<t<∞

∣∣Γ(σ + it)∣∣ = Γ(σ)

for any σ > 0, we see that

∣∣∣∣(

12

+ R

)k

∣∣∣∣ �Γ(1

2 + Re R + k)|Γ(1

2 + R)| ,(4.49)


WILSON FUNCTIONS AND POLYNOMIALS 381∣∣∣∣(

32

− C + it2 + R

)k

∣∣∣∣ �Γ(3

2 − Re C + Re R + k)|Γ(3

2 − C + it2 + R)|(4.50)

for every R satisfying (4.47), ReC � 1 and every integer k � 0. We obviouslyhave by the same conditions that

1|(3

2 + it2 + R)k

| � 1(3

2 + Re R)k

.(4.51)

Now ∣∣∣∣Γ(−z − i ImC + R + k)Γ(1 − C + R − z + k)

∣∣∣∣ � Γ(Re R − Re z + k)Γ(1 − Re C + Re R − Re z + k)

(4.52)

for R satisfying (4.47), Re z = − 12 , Re C � 1 and every integer k � 0. This

follows by the identity

Γ(A)Γ(1 − Re C)Γ(1 − Re C + A)

=∫ 1

0Y A−1(1 − Y )− Re C dY,

which is valid for every A > 0 and Re C < 1. On the other hand, by Lemma4.6

Γ(−z + it2 + R + k)Γ(−z − i Im C + R + k)

= O(1)(4.53)

with an implied absolute constant (since t2 is fixed and | Im C| � 1). By(4.52) and (4.53)

(R + it2 − z)k

Γ(1 − C + R − z + k)(4.54)

= O

(∣∣∣∣ 1Γ(−z + it2 + R)

∣∣∣∣ Γ(Re R − Re z + k)Γ(1 − Re C + Re R − Re z + k)

)

with an implied absolute constant. By (4.48)–(4.51) and (4.54) we get thelemma.

Lemma 4.8. Assume that ε > 0 is small enough, and let R be fixed suchthat (4.47) holds. Assume that z satisfies at least one of the following twoconditions:

(i) | Im z − Im R| � 1 + t2,(ii) | Re z − Re R − n| � 1/3 for every integer n.Then (for F3,2(z,R) see (3.17))

F3,2(z,R) = OR

(∣∣∣∣ 1Γ(−z + it2 + R)

∣∣∣∣)

.


382 A. BIRO: WILSON FUNCTIONS AND POLYNOMIALS

Proof. By the given conditions, for any nonnegative integer k

Γ(−z + it2 + R + k)Γ(−z + R + k)

= O(1).(4.55)

Indeed, this follows from the fact (implied by Stirling’s formula) that

Γ(w + it2)Γ(w)

= O(1),

if | Imw| � 1 + t2, or if | Rew − n| � 1/3 for every integer n. Formula (4.55)implies the lemma.

Acknowledgement. I am grateful to the referee for valuable remarks.

References

[1] G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge Univ. Press(1999).

[2] A. Biro, A duality relation for certain triple products of automorphic forms, to appearin Israel J. Math.

[3] H. M. Edwards, Riemann’s Zeta Function, Dover (New York, 2001).[4] W. Groenevelt, The Wilson function transform, Int. Math. Res. Not. (2003), no. 52,

2779–2817.[5] W. Groenevelt, Wilson function transforms related to Racah coefficients, Acta Appl.

Math., 91 (2006), 133–191.[6] S. C. Mitra, On certain transformations in generalized hypergeometric series, J. In-

dian Math. Soc., 7 (1943), 102–109.[7] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series and Products, 6th ed.,

Academic Press (2000).[8] L. J. Slater, Generalized Hypergeometric Functions, Cambridge Univ. Press (1966).[9] E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambridge Univ.

Press (1927).


an expansion theorem concerning wilson functions and polynomials

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