2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.