# Reference request: proofs of integrals presented in Erdélyi's *Table of Integral Transforms*

I have asked related questions on MSE and received no answer, so: I am looking for proofs of the integrals presented in Erdélyi's Table of Integral Transforms vol. i-ii, specifically proofs of the various gamma integrals presented on page 297-299 of volume 2, for instance:

$$\int_{\mathbb{R}} \frac{dx}{\Gamma(\alpha + x)\Gamma(\beta - x)}= \frac{2^{\alpha + \beta -2}}{\Gamma(\alpha + \beta - 1)},~\Re(\alpha + \beta)>1.$$

That the proofs for the identities presented in the work were reviewed closely is mentioned in the preface, though I cannot find any separate references containing the proofs.

• If $\alpha$ and $\beta$ are positive integers, and the integral over real $x$ is replaced by a sun over integer $x$, then this becomes the identity: $$\sum_k \binom{\alpha+\beta-2}k = 2^{\alpha+\beta-2}$$Can a proof of that identity be transformed to a proof of the integration? – Matt F. Mar 7 '20 at 4:33

$$\int_{\mathbb{R}}\frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{\Gamma(\alpha+\beta-1)dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{(\alpha+\beta-2)!dx}{(\alpha+x-1)!(\beta-x-1)!}\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\binom{\alpha+\beta-2}{\alpha+x-1}dx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{1}{2\pi i}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-2}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-2}}{z^{\alpha}}\frac{1}{2\pi i}\int_{\mathbb{R}}z^{-x}dxdz\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\frac{1}{2\pi }\int_{\mathbb{R}}e^{-i\theta x}dxd\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\delta(-\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-2}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-2}}{e^{i(\alpha -1) 0}}\\ =\frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)}$$