4

I'm currently working from Zwiebach's A First Course in String Theory, Second Edition, and I am wondering about the following question (which is paraphrased):

"Use the equation $$ \Gamma(z) = \int_0^\infty dte^{-t}t^{z-1},~Re(z)>0 $$ to show that for $Re(z)>0,$

$$\Gamma(z)=\int^1_0dtt^{z-1}\left(e^{-t}+\sum\limits_{n=0}^{N}\frac{(-t)^n}{n!}\right)+\sum\limits_{n=0}^{N}\frac{(-1)^n}{n!}\frac{1}{z+n}+\int^\infty_1dte^{-t}t^{z-1},$$ and explain why the above right hand side is well defined for $Re(z)>-\left(N+1\right).$ It follows that the right hand side provides the analytic continuation of $\Gamma(z)$ for $Re(z)>-\left(N+1\right).$ Conclude that $\Gamma(z)$ has poles at $z=0, -1, -2,\cdots,$ and give the value of the residue at $z=-n,$ where $n\in\mathbb{Z}_{\geq 0}.$"

I understand why the gamma function has poles whenever $z$ is $0$ or a negative integer, because of the recursive formula for the gamma function, $\Gamma(z+n+1)=(z+n)\Gamma(z)$ and I calculated the residue by this method instead, ie. $$\left(z+n\right)\Gamma(z)=\frac{\Gamma(z+n+1)}{\left(z+n-1\right)!}\to \frac{(-1)^n}{n!},$$ and in the series above I see the issue with $z+n=0$ but I do not see how to derive this above formula for the gamma function. If a hint could be posted that would be preferable, since this is for my own interest.

user131051
  • 41

2 Answers2

0

In your series expansion, fix some n for which you want the residue, and multiply both sides by $z+n$, you'll get the same result.

Arkya
  • 618
0

If the problem is to derive the above formula then note that for ${\rm Re \ } z>0$: $$ \int_0^1 t^{z-1} t^n \; dt = \frac{1}{n+z}$$ This shows that the two sums cancel out completely. The rest then combines to the usual Gamma function integral. Now, you should argue that the first integral extends to any ${\rm Re\ } z>-(N+1)$, simply because you have increased the order of the zero at $t=0$ and that the other sum also extends provided $-z$ is not an integer. After that the residues pop out easily from the second sum.

H. H. Rugh
  • 35,236