Residue Theorem for infinite oscillation

Question

Compute: $\int_{\lvert z \rvert = \frac{1}{2}} \frac{dz}{z \sin\left( \frac{1}{z} \right)}$

So far I have done the following: $\int_{\lvert z \rvert = \frac{1}{2}} \frac{dz}{z \sin\left( \frac{1}{z} \right)} = 2\pi i \sum_{n \in \mathbb{Z} \setminus{\{ 0 \}}}\text{Res}_{\frac{1}{n \pi}} \left( {\frac{1}{z \sin\left( \frac{1}{z} \right)}} \right) + \text{Res}_{0} \left( {\frac{1}{z \sin\left( \frac{1}{z} \right)}} \right) $

Now these are all simple poles and we get:

$\text{Res}_{\frac{1}{n \pi}} \left( {\frac{1}{z \sin\left( \frac{1}{z} \right)}} \right) = \lim_{z\rightarrow \frac{1}{n\pi}}{\frac{{z-\frac{1}{n \pi}}}{z \sin\left( \frac{1}{z} \right)}} = (-1)^{n+1} \frac{1}{n\pi}$

$\text{Res}_{0} \left( {\frac{1}{z \sin\left( \frac{1}{z} \right)}} \right) = \lim_{z\rightarrow 0}{\frac{z}{z \sin\left( \frac{1}{z} \right)}} = \lim_{z\rightarrow 0}{\frac{1}{ \sin\left( \frac{1}{z} \right)}}$

And here I am stuck on how to evaluate this residue. Any ideas on how to continue?

Answer 1

Let us make the computations inside out. Consider $w=1/z$. So $dw/w=-dz/z$, and $$ \int_{|z|=1/2}\frac 1{\sin(1/z)}\;\frac{dz}z = -\int_{|w|=2}\frac 1{\sin w}\frac {dw}w=0\ . $$ We integrate an even function, so the expansion starts with $w^{-2}$ and continues with good even powers of $w$, no $w^{-1}$, zero residue in zero. No other poles inside the disk of radius two centered in the origin.

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Numerical check, pari/gp:

? f(z) = 1/z/sin(1/z)
%1 = (z)->1/z/sin(1/z)
? intnum(t=0, 2*Pi, f( exp(I*t)/2 ) * I*exp(I*t)/2 )
%2 = 7.7685678008533630521918953008669306988 E-44 - 1.3236791693963119234612282128495142633 E-6*I