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How to prove: $$\frac{1}{2\pi i}\int_{-i\infty}^{i\infty} \frac{\Gamma(\alpha_1+x)}{\beta_1^{\alpha_1+x}}\, \frac{\Gamma(\alpha_2-x)}{\beta_2^{\alpha_2-x}}\, dx=\frac{\Gamma(\alpha_1+\alpha_2)}{(\beta_1+\beta_2)^{\alpha_1+\alpha_2}} \qquad \text{Re}(\alpha_1),\text{Re}(\alpha_2),\text{Re}(\beta_1),\text{Re}(\beta_2)>0$$ Does anyone can make the proof easier in the link provide below in the comment?

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I don't know. Is this an exercise in some book? If so, is there anything else in the book that looks at all like it? – Gerry Myerson Oct 5 '11 at 3:01
Where did you find this identity? Can you provide some motivation? If we let $\alpha_1 = z - \alpha_2$, drop the subscript off $\alpha_2$ and $B = \beta_2/\beta_1$ then the required identity is $$ \Gamma(z) = \frac{(1+B)^z}{2\pi i B^{\alpha}} \int^{i\infty}_{-i\infty} B^x \Gamma(z-\alpha+x) \Gamma( \alpha -x) dx. $$ My next move would be to try showing the right hand side satisfies the conditions of the Bohr-Mollerup theorem: – Ragib Zaman Oct 5 '11 at 3:06
@Victor: There is no harm in posting that link. – AD. Oct 5 '11 at 4:12
here it… – Victor Oct 5 '11 at 21:23
There is a proof in German right below the formula if you click on "Ausklappen". They define two functions, calculate a Fourier Transform, apply the Convolution formula and then substitute. I'm too lazy to retype it all in latex here, but looking at the formulas and replacing the text above for the German text should do. – user12014 Oct 5 '11 at 21:54

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