An analog of the Dougall formula and of the de Branges--Wilson integral
Yury A. Neretin

TL;DR
This paper introduces a new beta-integral over a mixed discrete-continuous domain, extending classical hypergeometric identities using a Lorentz group-related integral transform.
Contribution
It derives a novel beta-integral that generalizes the Dougall and de Branges--Wilson formulas through a Lorentz group-based integral transform.
Findings
Includes $_{10}H_{10}$-summation within the integral
Establishes a new connection between hypergeometric identities and Lorentz group representations
Provides a new integral transform related to the Jacobi transform
Abstract
We derive a beta-integral over , which is a counterpart of the Dougall -formula and of the de Branges--Wilson integral, our integral includes -summation. For a derivation we use a two-dimensional integral transform related to representations of the Lorentz group, this transform is a counterpart of the Olevskii index transform (a synonym: Jacobi transform).
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**An analog of the Dougall formula
and of the de Branges–Wilson integral**
Yury A. Neretin111Supported by the grants FWF, P28421, P31591.
We derive a beta-integral over , which is a counterpart of the Dougall -formula and of the de Branges–Wilson integral, our integral includes -summation. For a derivation we use a two-dimensional integral transform related to representations of the Lorentz group, this transform is a counterpart of the Olevskii index transform (a synonym: Jacobi transform).
1 The statement
1.1. Gamma function of the complex field. Denote by the set of all pairs such that . For nonzero we denote
[TABLE]
Denote by the set of all satisfying the additional condition: is pure imaginary. We have
[TABLE]
Elements of can be represented as
[TABLE]
Let be a complex variable. Denote the Lebesgue measure by
[TABLE]
Following [8], define the gamma function of the complex field by
[TABLE]
Here . The -function has poles at points , where , .
1.2. The statement. We derive the following beta-integral:
Theorem 1.1
Let , , , satisfy the conditions , . Then
[TABLE]
We also can write the left hand side as
[TABLE]
Then the identity with the same right-hand side holds for
[TABLE]
1.3. The de Branges–Wilson integral and the Dougall formula. Recall that the de Branges–Wilson integral is given by
[TABLE]
This formula was obtained by de Branges [4],[5] in 1972, a proof was not published; the formula was rediscovered by Wilson [21], 1980; see also [2]. The Dougall formula is
[TABLE]
Setting , we get a series of type , this result was contained in a family of identities obtained by Dougall [7], 1906. It seems that the general bilateral formula was obtained by Bailey [3], 1936; see also [2].
Let us explain a similarity of such formulas. Denote by the integrand in (1.5) and extend it into a complex domain writing
[TABLE]
Next, consider the sum . We have
[TABLE]
and we get the left-hand side of (1.6) with . Also, we get the same right-hand sides.
In [19] (see formula (1.28)) there was derived a one-dimensional hybrid of (1.5) and (1.6) including both integration over the real axis and a summation over a lattice on the imaginary axis.
Our integral (1.2) can be obtained by formal replacing -factors in the integrand (1.5) by similar -factors. The function satisfies the reflection identity
[TABLE]
Therefore
[TABLE]
and we come to the integrand in (1.2).
1.4. Further structure of the paper. In Section 2 we derive our integral (1.2). For a calculation we use a unitary integral transform defined in [16], see below Subsect. 2. This transform is an analog of a classical integral transform known under the names generalized Mehler–Fock transform, Olevskii transform, Jacobi transform, see [14], [15], [17].
We write an appropriate family of functions , and our integral (1.2) arises as the identity .
Section 3 contains a further discussion of Theorem 1.1.
2 Calculation
2.1. Convergence of the integral.
Lemma 2.1
Let , , i.e., . Then
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In particular,
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Remark. Our expression is single valued. Indeed,
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But , and therefore the result does not depend on a choice of a branch of .
Proof. We use two expressions for , namely,
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If , then we apply the Stirling formula (see e.g., [2]) to the first expression. If , we apply it to the second expression.
Corollary 2.2
If the parameters satisfy (1.4), then the integral (1.3) absolutely converges.
2.2. The Gauss hypergeometric function of the complex field. For we denote
[TABLE]
Following [8] (see also [9], Subsect. II.3.7) we define the beta function and the Gauss hypergeometric function of the complex field. Let , . Then
[TABLE]
The integral absolutely converges iff
[TABLE]
The right hand side gives a meromorphic continuation of to the whole .
For , , we define the hypergeometric function
[TABLE]
The integral has an open domain of convergence on any connected component of the set of parameters , it admits a meromorphic continuation to the whole set , see [16], Section 3.
The functions admit explicit expressions in terms of sums of products of Gauss hypergeometric functions . The standard properties of Gauss hypergeometric functions can be transformed to similar properties of functions , see [16], Section 3.
Below we need the following analog of the Gauss formula for :
[TABLE]
which is valid if
[TABLE]
see [16], Proposition 3.2, the last condition coincides with a condition of continuity of at .
2.3. The index hypergeometric transform. Fix real , such that
[TABLE]
Consider the measure on given by
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Let
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Consider the following function on :
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and consider the space of even functions on with inner product
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Next, define the kernel on by
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In [16] there was obtained the following statement:
The operator defined by
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is a unitary operator
[TABLE]
2.4. Application of the Mellin transform. We define a Mellin transform on as the Fourier transform on the multiplicative group of . Since , the Mellin transform is reduced to the usual Fourier transform and Fourier series. We have
[TABLE]
where . In the cases discussed below a function on is differentiable except the point , where a singularity has a form , . Also in our cases the integral (2.6) absolutely converges for being in a certain strip , therefore the Mellin transform is holomorphic in the strip. The inversion formula is given by
[TABLE]
the integration is taken over arbitrary line , where . We understand the integral (which can be non absolutely convergent) as
[TABLE]
The inversion formula holds at points of differentiability of , also it holds at points of singularities of the form with . In this case we can repeat the standard proof of pointwise inversion formula for the one-dimensional Fourier transform and pointwise convergence of Fourier series, see, e. g., [12], Sect. VIII.1, VIII.3; for advanced multi-dimensional versions of the Dini condition, see, e.g., [22], Sect. 9.
Convolution on is defined by
[TABLE]
As usual, we have
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this identity holds in intersection of strips of holomorphy and . We also define a function . Then
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So we have the following corollary of the convolution formula:
[TABLE]
where the integration contour is contained in the intersection of domains of holomorphy of and .
Lemma 2.3
a)* Let . Then the Mellin transform sends a function*
[TABLE]
to
[TABLE]
it is holomorphic in the strip
[TABLE]
b)* Assume that*
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Then the Mellin transform sends a function
[TABLE]
to the function
[TABLE]
defined in the strip
[TABLE]
(if it is non-empty)**
Proof. The statement a) follows from the definition of -function (2.1).
b) The function has singularities at , , with asymptotics of the form
[TABLE]
see [16], Theorem 3.9. This gives us a strip of convergence of the Mellin transform.
We must evaluate the integral
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We change an order of the integration and integrate in :
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Integrating in we again met a -function and after simple cancellations and an application of the reflection formula (1.7) we come to (2.8). The successive integration is valid under conditions (2.9).
We must justify the change of order of integrations. In fact, (2.13) is absolutely convergent as a double integral, i.e.,
[TABLE]
It is a special case of integral (2.13), we integrate it successively in and in under the same condition as for successive integration in (2.13).
Lemma 2.4
Let and
[TABLE]
Then
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Proof. We apply formula (2.7) assuming
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We evaluate Mellin transforms of , applying Lemma 2.3. In the integrand in the right hand side of (2.7) we get a product of two factors. The first factor is
[TABLE]
it is holomorphic un the strip
[TABLE]
The second factor
[TABLE]
is holomorphic in the strip
[TABLE]
It , are sufficiently small, then strips (2.16) and (2.17) have a non-empty intersection and we can apply formula (2.7). Two factors cancel and we get a factor independent on and the integral
[TABLE]
The integrand up to a constant factor is a Mellin transform of a function . By the inversion formula, integral (2.15) converts to
[TABLE]
For sufficiently small we can apply the Gauss identity (2.3). Thus we get (2.15) for sufficiently small , .
Keeping in mind (2.10)–(2.12), we can easily verify that the integral in the left-hand side of (2.15) converges for
[TABLE]
Thus, under these conditions the left hand side is holomorphic. The right hand side also is holomorphic in this domain. Therefore, they coincide.
2.5. Proof of Theorem 1.1. Let . Consider a function on given by
[TABLE]
Lemma 2.5
a)* iff*
[TABLE]
b)**
[TABLE]
The statement a) is trivial, b) is reduced to -function.
The -image of is done by Lemma 2.4. Since is unitary, we have
[TABLE]
This is Theorem 1.1, where the parameters , , , are , , , . Our calculation is valid for positive reals , , , satisfying conditions , . For extending the identity to the domain (1.4) we refer to Corollary 2.2, the integral (1.3) is holomorphic in the domain (1.4), the right hand side also is holomorphic.
3 Final remarks
3.1. Barnes–Ismagilov integrals. Let . Let , . Following Ismagilov [11], we define integrals of the form
[TABLE]
By [11], Lemma 2, such integrals admit a representation of the form
[TABLE]
where are products of -factors and parameters of hypergeometric functions are linear expressions in , and , . It is reasonable to claim that integrals are hypergeometric functions of the complex field.
By Lemma 2.4.b, the functions defined by (2.2) are compatible with this definition. Ismagilov considered -expressions that are counterparts of the Racah coefficients for unitary representations of the Lorentz group . Our theorem is an example of a hypergeometric identity for .
Some integrals of type (1.2) with products of -functions were obtained by Kells [13] and Derkachov, Manashov, and Valinevich in [6].
3.2. A difference problem. The de Branges–Wilson integral, the Dougall formula, and our integral (1.2) are representatives of beta integrals in the sense of Askey [1]. Quite often integrands in beta integrals are weight functions for systems of hypergeometric orthogonal polynomials. In particular, orthogonal polynomials corresponding to the de Branges–Wilson integral are the Wilson polynomials, see [21], [2]. Recall that they are even eigenfunctions of the following difference operator
[TABLE]
where . If an integrand of a beta integral decreases as a power function, then only finite number of moments converge; however in this case a beta integral can be a weight for a finite system of hypergeometric orthogonal polynomials (this phenomenon was firstly observed by Romanovski in [20]), the system of orthogonal polynomials related to the Dougall formula was obtained in [17]. On the other hand, such finite systems are discrete parts of spectra of explicitly solvable Sturm–Liouville problems (see, e.g., [10], [18]).
In the case of our integral (1.2) the integrand decreases as , we have no orthogonal polynomials. However a difference Sturm–Liouville problem can be formulated. We consider a space of meromorphic even functions on , a weight on defined by the integrand (1.2), and the following commuting difference operators:
[TABLE]
[TABLE]
See a simpler pair of difference operators of this kind in [16]. On the other hand, see a one-dimensional operator with continuous spectrum similar to in Groenevelt [10].
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1] Askey R., Beta integrals and the associated orthogonal polynomials , Lecture Notes in Math., 1395, Springer-Verlag, Berlin, 1989, 84-121.
- 2[2] Andrews G. E., Askey R., Roy R., Special functions , Cambridge University Press, Cambridge, 1999.
- 3[3] Bailey, W. N., Series of hypergeometric type which are infinite in both directions , Quart. J. Math., Oxford Ser., 7 (1936), 105-115.
- 4[4] de Branges, L. Gauss spaces of entire functions , J. Math. Anal. Appl., 37 (1972), 1-41.
- 5[5] de Branges L., Tensor product spaces , J. Math. Anal. Appl., 38 (1972), 109-148.
- 6[6] Derkachov, S. E.; Manashov, A. N.; Valinevich, P. A. SL ( 2 , ℂ ) SL 2 ℂ \mathrm{SL}(2,{\mathbb{C}}) Gustafson integrals . SIGMA 14 (2018), Paper No. 030, 16 pp.
- 7[7] Dougall, J., On Vandermonde’s theorem and more general expansions , Proc. Edinburgh Math. Soc., 25 (1906), 114-132.
- 8[8] Gelfand, I. M., Graev, M. I.; Retakh, V. S. Hypergeometric functions over an arbitrary field. Russian Math. Surveys 59 (2004), no. 5, 831-905.
