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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.

Link to e-PDFs of the work in question.

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    $\begingroup$ 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? $\endgroup$ – Matt F. Mar 7 at 4:33
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Titchmarsh’s Fourier integrals (1937, 7.6.4) has proof and attribution to Ramanujan.

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Another approach appears as a comment on your question, so this is just a rip-off trying to make things tidier but surely there are other ways to Titchmarsh and this to prove it

$ \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)} $

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  • $\begingroup$ Thanks Michael Engelhardt for edits, for the second one I selected improve instead of approve because I had another error not sure if that means the second edit won't appear as accepted $\endgroup$ – Daniel D. Mar 7 at 15:16
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    $\begingroup$ You're welcome - everything's fine, this is a cute derivation. $\endgroup$ – Michael Engelhardt Mar 8 at 3:57

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