MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand how to apply the residue theorem to solve

$\frac{1}{2\pi j}\int^{\gamma+j\infty}_{\gamma-j\infty}\Gamma(n-s)\Gamma(s)\Gamma(1-s) {}_1F_1(s;b;c) \left(\frac{b}{c}\right)^s\,\mathrm{d}s$

where $n\in\mathbb{N}$, $\,n\ge0$, $\,(b,c)\in\mathbb{R}$ and $\,(b,c)>0$. I understand that I need to sum the residues in points at which the above expression is not analytic. I also understood that the function ${}_1F_1$ is analytic over the s-plane, and hence I need to look for points for the Gamma functions.

Is the residue from $\Gamma(1-s)$ cancelling the residue from $\Gamma(s)$ and vise versa?

how to get the right residues for this combination?

Is any ${}_pF_q$ analtyic over the s plane?

share|cite|improve this question
Why don't you just expand your gammas in a power series at the relevant points, and see what happens? – Igor Rivin Dec 5 '11 at 13:40
@Igor -- The $\Gamma$ function has poles at every negative integer, so this would give him a sum over infinitely many terms. @Remmy -- Yes, there is cancellation. Rather than working out the residues manually, it would be easier to use Euler's Reflection formula to replace the two gamma's with a trig function and a polynomial. – David Speyer Dec 5 '11 at 14:04
@David: The contours will surround only finitely many integers, so if he actually computed things, he would see what was going on. I am pretty sure that's what Euler did :) – Igor Rivin Dec 5 '11 at 14:53
@Igor: @David: following your advices, the Taylor series expansion at s=n is given by $\frac{\pi}{\sin(n\pi)}-\frac{\pi^2}{\tan(n\pi)\sin(n\pi)}(s-n)+\left( \frac{\pi^3}{2\sin(n\pi)}+\frac{\pi^3}{\tan(n\pi)^2\sin(n\pi)}\right )(s-n)^2+\dots $ and hence the residue of $\Gamma(1-s)\Gamma(s)$ at $s=n$ is given by \sum^\infty_{n=0}-\frac{\pi^2}{\tan(n\pi)\sin(n\pi)} now, I wasn't able to get beyond this point... – Remy Dec 5 '11 at 15:48
@Remmy: do you know that the integral can actually has a simple form? – Igor Rivin Dec 5 '11 at 18:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.