$\int_C \frac{1}{z^2}\cot \frac{1}{z}\,dz = 2\pi i$

Question

Let $C$ be the unit circle $|z|=1$ oriented counterclockwise. Wolfram Alpha seems to suggest that $$\int_C \frac{1}{z^2}\cot \frac{1}{z}\,dz = 2\pi i$$ Since 0 is not an isolated singularity of the integrand, the usual calculus of residues does not work. Does anyone know how to prove this?

Answer 1

Define $z'=1/z$, then $$ \int_C \frac{1}{z^2}\cot\frac{1}{z}dz = -\int_{C'}\cot{z'}dz'=\int_C \cot{z'}dz'=2\pi i $$ where $C'$ is the circle $|z'|=1$ orieneted clockwise.

I will leave the detailed derivation to yourself.