Gamma function, residue

[sophia782]

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.

 

[galactus]

Wow, no one jumped on this one?.

As you probably know, Gamma is defined as \(\displaystyle {\Gamma}(z)=\int_{0}^{\infty}e^{-t}t^{z-1}dt\)

Which we can rewrite as:

\(\displaystyle \int_{0}^{1}e^{-t}t^{z-1}dt+\int_{1}^{\infty}e^{-t}t^{z-1}dt\)

Now, expand e^(-t) into a power series and then integrate term by term.

This gives the closed form:

\(\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!(z+n)}\)

Now, let z= -n and we have that aforementioned simple pole.

That is, \(\displaystyle n \in \mathbb{N}\)