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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.

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One 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. – t.b. May 13 '11 at 7:29
up vote 7 down vote accepted

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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I wouldn't call this l'Hôpital but rather Riemann's theorem on removable singularities (Riemannscher Hebbarkeitssatz), but the conclusion remains the same, of course. – t.b. May 12 '11 at 23:34
@joriki Two other poles enter the picture, at $\pm2\pi\mathrm{i}$. Their contributions to the residue formula add up when $n$ is even and cancel each other when $n$ is odd. – Did May 12 '11 at 23:40
@I Love Cake: Do you know about Laurent series? If $f$ only has a simple pole at $z=a$, then the only negative power in the Laurent series centered at $a$ is the -1 power. This means that we can write $f(z) = c_{-1} \frac{1}{(z-a)} + \sum_{n=0}^\infty c_n (z-a)^n$. The residue of $f$ at $a$ is going to be the coefficient $c_{-1}$, so one way of recovering this is by multiplying $f(z)$ by $z-a$ and taking the limit as $z \to a$, since the nonnegative powers in the sum will vanish. Does that make sense? – user1736 May 13 '11 at 0:36
@Theo: I think you can actually derive L'hopital's rule for analytic functions. If $f$ has a zero of order $m$ and $g$ has a zero of order $n$, one way of finding the limit of the quotient is to factor out the zeros and see what cancels out. Alternatively, you can also differentiate $f$ and $g$ that many times directly, so that you eventually end up with the same expression. I guess you can make the argument that this means that you don't really ever need l'hopital's rule, but it should still be valid. – user1736 May 13 '11 at 0:43
@user: Ok, very good. Actually, I think you should elaborate your comments into an answer because this very much looks like what I Love Cake is looking for (I'm too tired now and I've written too many answers, recently) – t.b. May 13 '11 at 1:18

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