# Gamma of negative integers

1. Jan 21, 2008

### Pere Callahan

\Hi,

I know the Gamma function has a first order pole at the negative integers, and the residue at -k is (-1)^k/k!

So now I want to compute Gamma(-k)/Gamma(-l) , k and l positive integers, and I have a feeling it should be [ (-1)^k/k! ] / [ (-1)^l/l! ] =(-1)^{k+l} l!/k!

What would be an easy way to prove that...?

I thought about considering Gamma(x)/Gamma(x-l+k) and then doing a Laurent expansion of the numerator and denominator around x=-k

This would be

$$\frac{\frac{(-1)^k}{(x+k)k!}+reg.}{\frac{(-1)^l}{(x-l+2k)l!}+reg.}$$

However this doesnt seem to make much sense being divergent for x->-k ...
Any tips?

-Pere

2. Jan 22, 2008

### HallsofIvy

Having said that neither $\Gamma(-l)$, not $\Gamma(-k)$ exist, you have already said that $\Gamma(-l)/\Gamma(-k)$ does not exist. Apparently you really mean $\lim_{x\rightarrow -l}\Gamma(x)/\Gamma(x+l-k)$. If that limit existed, then Gamma would be analytic over all real numbers, and it is not.

3. Jan 22, 2008

### Pere Callahan

Isn't it true that if g und f have first order poles at k, then $$\lim_{x \to k}\frac{f(x)}{g(x)}= \frac{Res_k(f)}{Res_k(g)}$$?