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$$\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) dz.$$

Is there an alternative to the residue theorem if we want to calculate the above integral?

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Change variables $w=\frac{1}{z}$ and see things simplify. –  Sasha Jul 23 '12 at 21:17

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Sasha posted a terse comment suggesting the same substitution I was about to suggest. Here's why that's a good idea: the exponential function behaves in a messy way at $\infty$, and $1/z=\infty$ at a point that's inside the circle $|z|=1$. "Messy" means an essential singularity rather than a pole. After this substitution, the zero in the denominator is no longer in the exponential function. The singularity at $z=0$ is at $w=\infty$, and that point is not inside the curve you'll integrate along.

So $z=1/w$, $dz=\text{what?}$.

Then as $z$ moves along the curve $|z|=1$, what path does $w$ follow?

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