The integral I want to evaluate is the integral of $\displaystyle\frac{z^n}{e^z-1}$ over the circle $|z| = 3 \pi $. I was thinking of using Cauchy's Residue Theorem, but then I got stuck on evaluating the residues at $z=0$, because I couldn't figure out how to change the form into one where I could find a Laurent series.
How do I evaluate the Complex Integral $(z^n)/(e^z - 1)$ using residue theory?
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2One formula you should be remembering for your test (it's probably the most useful one for calculating residues) is that $\operatorname{Res}_{z_{0}} \frac{f}{g} = \frac{f(z_0)}{g'(z_{0})}$ if $f(z_{0}) \neq 0$ and $g(z_{0}) = 0$ and $g'(z_{0}) \neq 0$. (i.e. $z_{0}$ is a simple zero of $g$). You can find other useful ones on the Wikipedia page on the [residue theorem](http://en.wikipedia.org/wiki/Residue_(complex_analysis)). – 2011-05-13
1 Answers
OK, this is an attempt to rewrite my answer without mentioning L'Hôpital or Riemann, as requested.
The integrand $f(z) = \frac{z^n}{\mathrm e^z-1}$ is the quotient of two holomorphic functions, each of which we can write as a power series at any point $z_0$. For $z_0=0$, this yields
$f(z)=\frac{z^n}{\left(\sum_{k=0}^\infty z^k/k!\right)-1}=\frac{z^n}{\sum_{k=1}^\infty z^k/k!}=\frac{z^n}{z\sum_{k=1}^\infty z^{k-1}/k!}=\frac{z^n}{z\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}\;.$
If $n$ is a positive integer, we can cancel one factor $z$ to obtain
$f(z)=\frac{z^{n-1}}{1+\sum_{k=2}^\infty z^{k-1}/k!}\;.$
The denominator doesn't go to zero as $z\to0$, and thus there is no pole at $z=0$.
On the other hand, if $n=0$, we have
$f(z)=\frac{1}{z\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}\;.$
Thus we can determine
$\lim_{z\to0}zf(z) = \lim_{z\to0}\frac{1}{\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}=1\;,$
and this is the residue of $f$ at $z=0$.
The residues at $z=\pm2\pi\mathrm i$ can be determined similarly using $\mathrm e^{z}=\mathrm e^{z\mp2\pi\mathrm i}$.
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0@joriki: Yes, this is exactly what I was trying to point out (not very clearly, as I see now): The existence of the limit can be established exactly as you outlined (and is$a$simple and weaker version of de l'Hôpital). Note also that you're using crucially that the singularity at $0$ can be removed because this is one of the hypotheses of the residue theorem. Anyway, I think we agree now. – 2011-05-13