I want to know if there can be a general statement about the poles (Laurent expansion) of such products of Gamma functions as a function of $p \in \mathbb{R}$ in the limit $\epsilon \rightarrow 0$,

$\Gamma[ \frac{(-1-2p)}{2} - \epsilon ]\Gamma[\frac{-1+p}{2} + \epsilon]\Gamma[\frac{(5+p)}{2} + \epsilon]$

For any given $p$ such that the $p$ dependent part of the argument is a negative integer one can do the usual Laurent expansion of the Gamma function in $\epsilon$ for each of the 3 factors and then multiply. But what can be said about the Laurent expansion in general as a function of $p$?

I wish one could write down the Laurent expansion in $\epsilon$ as a function of $p$!

One sees that there are these special cases like if $p$ is such that there are two integers $N$ and $M$ satisfying, $p= 1-2N^2 = -5 -2M^2$ then the later two Gamma function can have poles simultaneously. (like $N=2, M =1, p= -7$) Existence of such special $p$ naively seems to make things more tricky.

  • $\begingroup$ Are you missing a square root in the definition of your product? Otherwise I can't imagine why it should matter that the argument of $\Gamma$ is near a square. $\endgroup$ – Greg Martin Apr 10 '13 at 21:38
  • $\begingroup$ @Greg Martin Are you referring to the fact that I have written $N^2$ and $M^2$? That I did so that, $(p-1)/2 = - N^2$ is a negative definite integer and hence the second Gamma function has a pole. Similarly, $(5+p)/2 = -M^2$ is again a negative definite integer and hence the 3rd Gamma function has a pole. It would be great if you can help with the question. $\endgroup$ – Anirbit Apr 10 '13 at 23:11
  • $\begingroup$ Anirbit: the Gamma function has poles for every non-positive integer argument. Squares do not seem to enter the picture anywhere. $\endgroup$ – S. Carnahan Apr 16 '13 at 1:03
  • $\begingroup$ @S.Carnahan As you can see in my comment above to Greg, I have explained why I parametrized using squares - to ensure negative definiteness. $\endgroup$ – Anirbit Apr 16 '13 at 21:35
  • $\begingroup$ This has now found an answer on PhysicsOverflow: physicsoverflow.org/22983/… $\endgroup$ – Dilaton Aug 29 '14 at 14:03

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