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How to integrate

\begin{equation} \int_0^{2 \pi } \frac{e^{e^{-i \theta }}}{e^{4 i \theta }} \, d\theta \end{equation}

I tried to expand the $e^{e^{-i\theta}}$ but that involves terms like $\sin(\sin\theta)$, which I have no idea how to deal with.

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  • $\begingroup$ Do you know a little complex analysis? $\endgroup$ Commented Sep 16, 2013 at 22:05
  • $\begingroup$ yes just a little bit, this is actually from a question of my complex analysis course. The initial question is to do a contour integral of $\oint_C \frac{e^{1/z}}{z^5}dz$ around the unit circle in the positive direction. And I don't know how to deal with the singularity at z=0, so I though it may be easier to do as a integral of $\theta$. $\endgroup$ Commented Sep 16, 2013 at 22:11
  • $\begingroup$ Have you already heard of residues? $\endgroup$ Commented Sep 16, 2013 at 22:13

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If you already know about residues, you can see that

$$\int_{\lvert z\rvert = 1} \frac{e^{1/z}}{z^5}\,dz = 2\pi i \operatorname{Res}_0 \left(\frac{e^{1/z}}{z^5}\right).$$

The residue is determined to be $0$ by the Laurent expansion.

If you don't know about residues, you can use the fact that the exponential series

$$e^{1/z} = \sum_{k=0}^\infty \frac{1}{k!}z^{-k}$$

converges uniformly on the unit circle, and thus you can interchange summation and integration,

$$\int_{\lvert z\rvert = 1} \frac{e^{1/z}}{z^5}\,dz = \sum_{k=0}^\infty \frac{1}{k!} \int_{\lvert z\rvert = 1} z^{-k-5}\,dz = 0,$$

since $\int_{\lvert z\rvert = 1} z^m\,dz = 0$ for $m \neq -1$.

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  • $\begingroup$ Thanks very much, but I also see the formula $\frac{d^nf(z)}{dz^n}=\frac{n!}{2 \pi i}\oint \frac{f(z')}{\left(z'-z\right)^{n+1}} \, dz'$, but if we set f(z)=e^(1/z), and take f''''(z) at z=0, it gives $\infty$, where have did I wrong? $\endgroup$ Commented Sep 16, 2013 at 22:27
  • $\begingroup$ $e^{1/z}$ has a singularity in $0$. That formula applies only to functions that are holomorphic in the domain over whose boundary you integrate. $\endgroup$ Commented Sep 16, 2013 at 22:30
  • $\begingroup$ OK, I see, thanks a lot! $\endgroup$ Commented Sep 16, 2013 at 22:33

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