I need to evaluate the following integral $$ \int^{1}_{0} \Gamma(\alpha+x) \Gamma(1-\alpha-x) \Gamma(\beta+x) \Gamma(1-\beta-x) \, dx $$ where $\Im(\alpha)>0$ and $\Im(\beta)<0$ are some arbitrary numbers. The above equation can be simplified to $$ \int^1_0 {\pi^2 \over \sin[\pi(\alpha+x)]\sin[\pi(\beta+x)]} dx $$ which would result in $0$ if $\alpha$ and $\beta$ didn't have any imaginary parts. However, since they're complex numbers, the integral should develop an imaginary part as well, I suppose.

Thanks for any insight!

EDIT: Corrected a sign in one of the arguments of a $\sin$ function and two Gamma functions.


The integral is $0$, because: fix $\alpha$. Then $$\int_0^1 \frac{\pi^2}{\sin(\pi (\alpha - x)) \sin(\pi (-\alpha + x))} \, \mathrm{d}x = 0$$ as one can see by directly integrating: $$\int \frac{\pi^2}{\sin(\pi (\alpha - x))^2} \, \mathrm{d}x = \pi \cot(\pi (\alpha - x)).$$ This is a periodic integral and is the same as $$\int_{x_0}^{x_0 + 1} \frac{\pi^2}{\sin(\pi (\alpha - x))\sin(\pi (-\alpha + x))} \, \mathrm{d}x$$ for any $x_0 \in \mathbb{R}$, from which we see that $$\int_0^1 \frac{\pi^2}{\sin(\pi (\alpha - x)) \sin(\pi (\beta + x))} \, \mathrm{d}x = 0$$ for any $\beta \in -\alpha + \mathbb{R}.$ Finally, this is a holomorphic function of $\beta$ (as long as $\mathrm{Im}(\beta) < 0$), so the identity theorem implies it equals $0$ everywhere.

Note that this integral is not defined when $\alpha$ or $\beta$ is real (so I don't understand the comment) and may be nonzero if $\alpha$ and $\beta$ lie on the same side of the real axis.

  • $\begingroup$ Thank you very much for your answer. As it happens I have mistaken one of the signs: the sin with $\alpha$ should have argument $\alpha+x$ and not $\alpha-x$. I have edited the answer. Does this change anything? This problem comes from a physical setting where it is expected to give a pure imaginary number so if it still gives $0$ it's because I haven't given the right set up for $\alpha$ and $\beta$. $\endgroup$ – ABarr Jul 13 '17 at 9:47

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