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  Poles of products of Gamma functions

+ 2 like - 0 dislike

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.

This post imported from StackExchange MathOverflow at 2014-08-29 08:16 (UCT), posted by SE-user Anirbit
asked Apr 10, 2013 in Mathematics by Anirbit (585 points) [ no revision ]
retagged Aug 29, 2014
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.

This post imported from StackExchange MathOverflow at 2014-08-29 08:16 (UCT), posted by SE-user Greg Martin
@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.

This post imported from StackExchange MathOverflow at 2014-08-29 08:16 (UCT), posted by SE-user Anirbit
Anirbit: the Gamma function has poles for every non-positive integer argument. Squares do not seem to enter the picture anywhere.

This post imported from StackExchange MathOverflow at 2014-08-29 08:16 (UCT), posted by SE-user S. Carnahan
@S.Carnahan As you can see in my comment above to Greg, I have explained why I parametrized using squares - to ensure negative definiteness.

This post imported from StackExchange MathOverflow at 2014-08-29 08:16 (UCT), posted by SE-user Anirbit

1 Answer

+ 2 like - 0 dislike

Forget the $p$'s and consider a product $\Gamma(x-\epsilon)\Gamma(y-\epsilon)\ldots\Gamma(z-\epsilon)$. If one of $x,y,\ldots,z$ is not a negative integer, it is regular, and you can put $\epsilon=0$ in the factor. So it is enough to consider the case where all of $x,y,\ldots,z$ are negative integers. In this case, if there are $k$ factors,  $\epsilon^k\Gamma(x-\epsilon)\Gamma(y-\epsilon)\ldots\Gamma(z-\epsilon)$ is analytic at $\epsilon=0$.

Therefore you can expand it into a power series using the power series for  $\epsilon\Gamma(x-\epsilon)$ etc. to order $k$ or higher. At the end substitute for $x,y,\ldots,z$ your expressions in $p$,

answered Aug 29, 2014 by Arnold Neumaier (15,787 points) [ revision history ]

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