# Gamma function (complex and negative)

1. Feb 26, 2007

### touqra

Has anyone seen this representation for gamma function before?

$$\Gamma(z) = \int_0^1\ dt\,\, t^{z-1}(e^{-t}\, -\, \sum_{n=0}^N\frac{(-t)^{n}}{n!})\,\, +\,\, \sum_{n=0}^N\frac{(-1)^{n}}{n!}\frac{1}{z+n}\,\, +\,\, \int_1^\infty dt\,\, e^{-t}\,t^{z-1}$$

for Re(z) > -N-1

I can't figure out how to prove this...

Last edited: Feb 27, 2007
2. Feb 27, 2007

### mjsd

ok, this is just the normal definition in disguise ie.
$$\Gamma(z)=\int_{0}^{\infty} dt\; t^{z-1} e^{-t}$$
the summation bit cancels out exactly. try it!
expand the first integral and do some simplifications (incl. doing one of the 0 to 1 integral for t)

3. Feb 27, 2007

### touqra

Thank you!
What a stupid head I have !!! Bump!

But why go into such pain as to disguise that gamma, having the summation bits cancelling exactly?
Prof says that the representation is an analytic continuation of gamma function to Re(z) > -N-1 . I don't understand. And the representation is suppose to lead us to conclude that the gamma function has poles at 0, -1, -2,... and give the value of the residue at z = -n with n positive.
I don't understand how can you conclude these if those summations are exactly cancelling...

4. Feb 28, 2007

### mjsd

ok, firstly accept the fact that z =-n are poles. you can immediate see that because you run into problems when trying to do that integral (that standard definition, for cases z = 0, -1... etc.) now suppose you don't know that and proceed to re-writing the definition in the form suggested by your prof. then you will see that the $$\int_0^1$$ and $$\int_1^\infty$$ integrals are all ok (for all z in the range your want), but the middle summation term tells you that you have problem at z = 0, -1, -2... etc. so by re-writing you have sifted out the poles effectively.