Can Gamma of negative integers be proven using a Laurent expansion?

[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 doesn't seem to make much sense being divergent for x->-k ...
Any tips?

-Pere

 

[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.

 

[Pere Callahan]

Thanks for your reply. Yes, I mean the limit, of course

How does the existence of this limit imply analyticity of Gamma over the real numbers?

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

I am not yet convinced ..

-Pere