Let
$\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) dz.$
Is there an alternative to the residue theorem if we want to calculate the above integral?
Let
$\int_{|z|=1} \sin\left(e^{\frac{1}{z}}\right) dz.$
Is there an alternative to the residue theorem if we want to calculate the above integral?
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?