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?
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: en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem –  Ragib Zaman Oct 5 '11 at 3:06