4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.