How can I calculate this complex integral?:
$$ \int_{\vert z\vert =0.2}\frac{dz}{\sin(1/z)} $$
The only idea I have is considering the Laurent series of the function and then look for the residues, but I think this is very much for a simple exercise. Anyone knows a better way to solve it?
On
By using the substitution $z=\frac{1}{w}$ the given integral turns into $$ I=-\oint_{|w|=5}\frac{dw}{w^2 \sin(w)} \tag{1}$$ and the meromorphic function $f(w)=\frac{1}{w^2\sin w}$ has a triple pole at the origin and simple poles at $\pi\mathbb{Z}\setminus\{0\}$. By the residue theorem $$ I = -2\pi i\left(\text{Res}_{w=-\pi}f(w)+\text{Res}_{w=0}f(w)+\text{Res}_{w=\pi}f(w)\right) \tag{2}$$ where $$\text{Res}_{w=-\pi}f(w)=\text{Res}_{w=\pi}f(w)=-\frac{1}{\pi^2}\tag{3}$$ and $$\text{Res}_{w=0}f(w) = [x^2]\frac{x}{\sin x} = \frac{1}{6}\tag{4} $$ such that $I=\color{red}{2\pi i\left(\frac{2}{\pi^2}-\frac{1}{6}\right)}.$
There are infinitely many poles, and a nasty accumulation point, inside the contour, but there are only finitely many outside. Try the change of variables $z = 1/w$.