Evaluate the integral $ \int _{\vert z \vert =1} \frac{1}{\sin(e^{-1/z})}\,dz$

Question

Evaluate the integral $$\int _{\vert z \vert =1} \frac{1}{\sin(e^{-1/z})}\,dz.$$

I tried using Laurent series but not getting result. Please help.

Answer 1

After letting $w=-1/z$, by the Residue Theorem, $$\begin{align} \int _{|z| =1} \frac{1}{\sin(e^{-1/z})}\,dz &=-\int _{|w |=1} \frac{1}{\sin(e^{w})}\cdot \frac{dw}{w^2}\\ &= -2\pi i\,\operatorname{Res}\left(\frac{1}{w^2 \sin(e^{w})},0\right)\\ &=-2\pi i \frac{d}{dw}\left(\frac{1}{\sin(e^{w})}\right)_{w=0}\\ &= \frac{2\pi i\cos(1)}{\sin^2(1)}. \end{align}$$ P.S. Note that $\sin(e^{w})=0$ iff $e^w=n\pi$, but $0<|e^w|\leq e<\pi$ for $|w|\leq 1$.