# Thread:

1. ## Gamma function, residue

Show that for $m \geq 0$, the residue of $\Gamma(z)$ at $z = -m$ is $\frac{(-1)^m}{m!}$.

$\Gamma(z)$ is the gamma function. The gamma function is meromorphic. It is defined in the right half-plane by $\Gamma(z)= \int_0^{\infty} e^{-t}t^{z-1}dt$ for $\text{Re}(z)>0$. There is also another representation of $\Gamma(z)=\frac{\Gamma(z+m)}{(z+m-1) \cdots (z+1)z}$ where the right-hand side is defined and meromorphic for $\text{Re}(z)>-m$ with simple poles at $z=0, -1, \ldots, -m+1$. However, I still do not see how to prove this. I need help with this. Thank you.

2. In general, the residue of $f(z)$ at $z=z_0$ is the residue of $f(z+z_0)$ at $z=0$. Now we have $\Gamma(z)=(z-1)\dots(z-m)\Gamma(z-m)$. Therefore what you are looking for is the residue of $\frac{\Gamma(z)}{(z-1)\dots(z-m)}$ at $z=0$. The denominator is analytic at $z=0$ and its value is $(-1)^mm!$; so the solution will be $\frac{(-1)^mR}{m!}$ where $R$ is the residue of $\Gamma(z)$ at $z=0$... can you finish now?
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