Gamma function, residue

• May 4th 2010, 11:34 AM
pascal4542
Gamma function, residue
Show that for $\displaystyle m \geq 0$, the residue of $\displaystyle \Gamma(z)$ at $\displaystyle z = -m$ is $\displaystyle \frac{(-1)^m}{m!}$.

$\displaystyle \Gamma(z)$ is the gamma function. The gamma function is meromorphic. It is defined in the right half-plane by $\displaystyle \Gamma(z)= \int_0^{\infty} e^{-t}t^{z-1}dt$ for $\displaystyle \text{Re}(z)>0$. There is also another representation of $\displaystyle \Gamma(z)=\frac{\Gamma(z+m)}{(z+m-1) \cdots (z+1)z}$ where the right-hand side is defined and meromorphic for $\displaystyle \text{Re}(z)>-m$ with simple poles at $\displaystyle z=0, -1, \ldots, -m+1$. However, I still do not see how to prove this. I need help with this. Thank you.
• May 4th 2010, 11:44 AM
Bruno J.
In general, the residue of $\displaystyle f(z)$ at $\displaystyle z=z_0$ is the residue of $\displaystyle f(z+z_0)$ at $\displaystyle z=0$. Now we have $\displaystyle \Gamma(z)=(z-1)\dots(z-m)\Gamma(z-m)$. Therefore what you are looking for is the residue of $\displaystyle \frac{\Gamma(z)}{(z-1)\dots(z-m)}$ at $\displaystyle z=0$. The denominator is analytic at $\displaystyle z=0$ and its value is $\displaystyle (-1)^mm!$; so the solution will be $\displaystyle \frac{(-1)^mR}{m!}$ where $\displaystyle R$ is the residue of $\displaystyle \Gamma(z)$ at $\displaystyle z=0$... can you finish now?