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

inserted missing factor from substitution z to exp (i theta) and also corrected final result from 2^{%-1} to 2^{%-2} and previous steps

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edited Mar 7, 2020 at 15:07

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Another wayapproach 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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{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-1}}{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-1}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i(\alpha -1) 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $$ \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)} $

Another way 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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{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-1}}{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-1}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i(\alpha -1) 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $

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)} $

inserted missing factor from substitution z to exp (i theta)

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edit approved Mar 7, 2020 at 15:07

Michael Engelhardt

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Another way 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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{e^{i\alpha \theta}}\frac{1}{2\pi i}\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-1}}{e^{i\alpha \theta}}\delta(-\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha\theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i\alpha 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $$ \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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{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-1}}{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-1}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i(\alpha -1) 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $

Another way 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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{e^{i\alpha \theta}}\frac{1}{2\pi i}\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-1}}{e^{i\alpha \theta}}\delta(-\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha\theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i\alpha 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $

Another way 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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{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-1}}{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-1}}{e^{i(\alpha -1) \theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i(\alpha -1) 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $

added 48 characters in body

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edited Mar 7, 2020 at 8:18

Dabed

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

Another way appears on a comment on your question, so this is just a rip-off trying to make things tidier but surely there are other ways 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}}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{z^{\alpha}}\int_{\mathbb{R}}z^{-x}dxdz\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha \theta}}\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-1}}{e^{i\alpha \theta}}\delta(-\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha\theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i\alpha 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $

Another way 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-1}}{z^{\alpha+x}}dzdx\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\oint_{|z|=1}\frac{(1+z)^{\alpha+\beta-1}}{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-1}}{e^{i\alpha \theta}}\frac{1}{2\pi i}\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-1}}{e^{i\alpha \theta}}\delta(-\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\int_{-\pi}^{\pi}\frac{(1+e^{i\theta})^{\alpha+\beta-1}}{e^{i\alpha\theta}}\delta(\theta)d\theta\\ =\frac{1}{\Gamma(\alpha+\beta-1)}\frac{(1+e^{i0})^{\alpha+\beta-1}}{e^{i\alpha 0}}\\ =\frac{2^{\alpha+\beta-1}}{\Gamma(\alpha+\beta-1)} $

Inserted missing factors in exponents in 7th adn 8th lines

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edit approved Mar 7, 2020 at 8:14

Michael Engelhardt

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deleted 4 characters in body

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edited Mar 7, 2020 at 6:59

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created Mar 7, 2020 at 6:51

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