An identity involving an infinite integral with a sinh in the denominator

Question

I recently encountered the rather appealing looking integral, which appears in the theory of random matrices :

$$\int_{-\infty}^{\infty}\prod_{j=1}^{p-1}(j^{2}+z^{2})\frac{zdz}{\mathrm{sinh}(2\pi z)} = \frac{\Gamma(p+1/2)^{2}}{2\pi p}$$

I wondered if anyone had seen it before in some table of integrals or such like, or had an idea how to prove it? Here are some things I noticed/tried :

1) The following identity is straightforward to prove,

$$\int_{0}^{\infty}z^{k}\frac{dz}{\mathrm{sinh}(2\pi z)} = \frac{2(1-2^{-k-1})\zeta(k+1)k!}{(2\pi)^{k+1}}$$

and has a well known closed form expression for odd k. It is proved basically by expanding the sinh in a geometric series and integrating term by term. Noting that the first integral I mentioned has all odd powers of z in the numerator, I tried to multiply out the polynomial and integrate term wise. The trouble is one obtains a quite complicated combinatorial sum (involving central factorial numbers and bernoulli numbers) which I didn't know how to evaluate.

2) Contour integration methods. The pole structure of the integrand is quite easy to write down, and I tried a few box type contours which basically involved shifting the contour up the imaginary axis, and joining at the sides (noting that the integrand decays exponentially as Re(z) goes to + or - infinity). The residues at the poles can be evaluated easily, the trouble is that the integral along the top edge of the box is not obviously related to the original one, along the real line.

3) Generating function methods. The generating function,

$$\sum_{p=0}^{\infty}\frac{\prod_{i=1}^{p}(i^{2}+z^{2})}{(p!)^{2}}y^{p}$$

appears to be expressible as a hypergeometric function. How this could be used constructively, I'm not sure.

p.s. I know the identity is true because it appears as the consequence of solving a certain random matrix problem in two completely different ways.

Answer 1

This integral is equivalent to the special case $\alpha=\gamma=p$, $\phantom.\beta=\delta=1/2$ of the known contour integral $$ \int_{-i\infty}^{i\infty} \Gamma(\alpha+s) \phantom. \Gamma(\beta+s) \phantom. \Gamma(\gamma-s) \phantom. \Gamma(\delta-s) \phantom. ds = 2\pi i \frac{\Gamma(\alpha+\gamma) \phantom. \Gamma(\alpha+\delta) \phantom. \Gamma(\beta+\gamma) \phantom. \Gamma(\beta+\delta)} {\Gamma(\alpha+\beta+\gamma+\delta)}, $$ valid for all $\alpha,\beta,\gamma,\delta$ of positive real part. This is formula 641.2 on page 655 of Gradshteyn and Ryzhik [GR], which in turn cites formula 302(32) in Erdélyi et al. [E]. [EDIT Thanks to Richard Borcherds for recognizing this integral as a result of Barnes 1908. See postscript below about references and proofs for this integral.]

Indeed, the factor $z \phantom./\sinh(2\pi z)$ of the integrand factors as $$ \frac{1}{2} \frac{iz}{\sin(\pi i z)} \frac1{\cos(\pi i z)} = \frac1{2\pi^2} \Gamma(1+iz) \phantom. \Gamma(1-iz) \phantom. \Gamma\bigl(\frac12+iz\bigr) \phantom.\Gamma\bigl(\frac12-iz\bigr), $$ while the factor $\prod_{j=1}^{p-1} (j^2+z^2)$ is $$ \prod_{j=1}^{p-1} (j+iz) \phantom. \prod_{j=1}^{p-1} (j-iz) = \frac{\Gamma(p+iz)}{\Gamma(1+iz)} \frac{\Gamma(p-iz)}{\Gamma(1-iz)}. $$ Hence the integral is $$ \frac1{2\pi^2} \int_{-\infty}^{\infty} \Gamma(p+iz) \phantom. \Gamma(p-iz) \phantom. \Gamma\bigl(\frac12+iz\bigr) \phantom.\Gamma\bigl(\frac12-iz\bigr) \phantom. dz. $$ Now take $z=is$ to obtain the known integral, multiplied by $1/2\pi^2 i$.

The formula for that integral, multiplied by the same factor $1/2\pi^2 i$, then yields $$ \frac1\pi \frac{\Gamma(2p) \phantom. \Gamma(p+(1/2))^2 \phantom. \Gamma(1)}{\Gamma(2p+1)} , $$ which is equivalent to the desired answer because $\Gamma(1)=1$ and $\Gamma(2p+1) = 2p \phantom. \Gamma(2p)$.

Postscript about the contour integral formula: Gary couldn't locate the Erdélyi reference; I found it, but it turns out the book just displays the formula without proof or source. Fortunately Richard Borcherds (in a comment to this answer) recognized it as a result of Barnes [B]; see Lemma 15 on pages 154-155. There's even a Wikipedia page on Barnes' work in this area, including this result which is called the "first Barnes lemma". Barnes' proof uses a hypergeometric identity of Gauss, which is natural in the context of his paper. Here's an alternative approach using Fourier convolutions. For $\mu,\nu$ of positive real part, and real $t>1$, we have

$$ \int_{-i\infty}^{i\infty} \Gamma(\mu+s) \phantom. \Gamma(\nu-s) \phantom. t^s \phantom. ds = 2\pi i \phantom. \Gamma(\mu+\nu) \phantom. t^\nu / (1+t)^{\mu+\nu}. $$ [This is basically [GR, p.657, 6.422#3], and can be proved by shifting the contour integral to the left: the poles of $\Gamma(\mu+s)$ yield the terms in the Laurent expansion of the right-hand side about $t = \infty$.] Now take $(\mu,\nu) = (\alpha,\gamma)$ and $(\delta,\beta)$ and multiply, finding that $$ \int_{-i\infty}^{i\infty} F(s) \phantom. t^s \phantom. ds = (2\pi i)^2 \Gamma(\alpha+\gamma)\phantom.\Gamma(\beta+\delta)\phantom.t^{\beta+\gamma} / (1+t)^{\alpha+\beta+\gamma+\delta} $$ where $F(\cdot)$ is the convolution on the imaginary $s$-axis of $\Gamma(\alpha+s) \phantom. \Gamma(\gamma-s)$ with $\Gamma(\delta+s) \phantom. \Gamma(\beta-s)$. Since this integral uniquely determines $F$, it follows that $$ F(s) = 2\pi i \frac{\Gamma(\alpha+\gamma) \phantom. \Gamma(\beta+\delta)} {\Gamma(\alpha+\beta+\gamma+\delta)} \Gamma(\alpha+\delta+s) \phantom. \Gamma(\beta+\gamma-s). $$ Taking $s=0$ recovers the Barnes formula.

References

B] Barnes, E.W.: A new development of the theory of the hypergeometric functions. Proc. LMS (1908) s2-6(1): 141–177.

[E] Erdélyi, A., et al.: Table of Integral Transforms II. New York: McGraw Hill, 1954.

[GR] Gradshteyn, I.S., and Ryzhik, I.M.: Table of Integrals, Series, and Products (4th ed.). New York: Academic Press, 1980.

Answer 2

I will follow route #3, via the generating function. You are correct, this is indeed a hypergeometric function. Maple quickly returns $ { }_{2}F_{1}\left(\left[1-iz,1+iz\right], \left[1\right] | y\right) $. More useful than this though is the ODE which this satisfies. gfun quickly returns $$ (y^2-y)w''(y) + (3y-1)w'(y) + (1+z^2)w(y), w(0)=1 $$ Translating this to a recurrence on the coefficients (i.e. u(p)y^p), we get $$-(p^2+2p+1)u(p+1) + (p^2+2p+z^2+1)u(p), u(0)=1$$ which has $$\frac{\Gamma(p+1-iz)\Gamma(p+1+iz)}{\Gamma(1-iz)\Gamma(1+iz)\Gamma(p+1)^2}$$ as solution. Seems you should be able to go from there to obtain your answer.