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I found the integral $$ \int_C e^{-1/z}\sin(1/z)dz $$ over the circle $|z|=1$ while doing some problems in Schaum's Outline for Complex Variables.

This integral has me stumped. The answer is $2\pi i$, but I can't see why. There are no poles, so I don't think I can apply the residue theorem. Reparametrizing with $z=e^{it}$ for $t\in(0,2\pi)$ looks very messy.

How can this integral be approached? I tried to find the Taylor series, and the first few terms are $$ -\frac{1}{2z^2}-\frac{2}{3!z^3}-\cdots $$ so $0$ looks like an essential singularity, but I don't know if that's useful.

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See also the very similar… – imallett May 6 '13 at 16:29
up vote 4 down vote accepted

With a change of variables $z\mapsto\frac1z$ $$ \begin{align} \int_Ce^{-1/z}\sin(1/z)\,\mathrm{d}z &=\int_Ce^{-z}\sin(z)\frac{1}{z^2}\,\mathrm{d}z\\ &=\int_C\left(\frac1z-1+\frac{z}{3}+\dots\right)\,\mathrm{d}z\\ &=2\pi i \end{align} $$ Note that the change of variables reverses the direction of the path.

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That's very clever, thanks robjohn. – Hana Bailey Mar 28 '12 at 6:24

My Laurent series centered at $z=0$ of $ e^{-1/z}\sin(1/z)$ disagrees with yours: I have it as the product $\displaystyle\sum_{i=0}^{\infty} \dfrac{(-1)^i}{i!z^{i}}$ $\times$ $\displaystyle\sum_{j=0}^{\infty} \dfrac{(-1)^j}{(2j+1)!z^{2j+1}}$. The only $z^{-1}$ term of my product emerges when $j=0$ and $i=0$, and is equal to $\dfrac{1}{z}$, where it seems you have no nontrivial term of that order. I think that mine is correct, but perhaps we should both check our calculations?

In fact, the residue theorem is strong enough to apply in this case; $ e^{-1/z}\sin(1/z)$ has no poles and only one essential singularity at $z=0$, so our function is holomorphic on the complement of finitely many points in C, the unit circle centered at the origin. As a result, we can compute a residue at $z=0$ as above and apply the residue theorem, which tells us that our integral is $2 \pi i$.

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Thanks, I think I did make a mistake computing the series. – Hana Bailey Mar 28 '12 at 6:25

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