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[SOLVED] Poles of \Gamma(z)

dwsmith

Well-known member
Feb 1, 2012
1,673
Compute the residue of $\Gamma(z)$ at each of its poles.

So the poles are at the negative integers and 0. I suspect there must be a formula than since this is an infinite set.

$\Gamma(z) = \dfrac{e^{-\gamma z}}{z}\prod\limits_{n=1}^{\infty}\left(1+\dfrac{z}{n}\right)^{-1}e^{z/n}$

Should I start by logarithmically differentiating?
 

Jose27

New member
Feb 1, 2012
15
Just a thought (I don't know if it'll work), but maybe using the $\Gamma$ function's integral representation (remember we have to split the integral several times in order to extend it past a certain integer) and using Fubini's theorem. Something along these lines.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Just a thought (I don't know if it'll work), but maybe using the $\Gamma$ function's integral representation (remember we have to split the integral several times in order to extend it past a certain integer) and using Fubini's theorem. Something along these lines.
$\displaystyle\Gamma(z) =\int_0^{\infty}e^{-t}t^z\dfrac{dt}{t}$

What should I do with this?

---------- Post added at 21:29 ---------- Previous post was at 20:37 ----------

The $\text{Res} = \lim\limits_{z\to -m}(z+m)\Gamma(z)$ and we can write $\Gamma(z) = \dfrac{\Gamma(z+m)}{(z+m-1)(z+m-2)\cdots (z+1)z}$ for the $\text{Re} z> -m$.

By substitution, we have
$$
\text{Res} = \lim_{z\to -m}(z+m)\Gamma(z) = \lim_{z\to -m}(z+m)\frac{\Gamma(z+m)}{(z+m-1)(z+m-2)\cdots (z+1)z}.
$$

By definition, we have $(z+m)\Gamma(z+m) = \Gamma(z+m+1)$.

So
$$
\text{Res} = \lim_{z\to -m}\frac{\Gamma(z+m+1)}{(z+m-1)(z+m-2)\cdots (z+1)z} = \frac{\Gamma(1) = 1}{(-1)(-2)\cdots (-m)} = \frac{(-1)^m}{m!}.
$$
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
$$
\text{Res} = \ldots = \frac{(-1)^m}{m!}.
$$
That is correct. The functional equation $\Gamma(z+1) = z\Gamma(z)$ is the key to the calculation.