Contour Integral with Many Singularities How would one compute?
$$ \oint_{|z|=1} \frac{dz}{\sin \frac{1}{z}} $$

Can one "generalize" the contour theorem and take the infinite series of the residues at each singularity?
Another method for this case: Think of the outside of the unit disk as your domain.  Go around the contour in the opposite direction.  Outside the unit disk, the function
$$
\frac{1}{\sin\frac{1}{z}}
$$
is meromorphic.  It has only one singularity in that region: a pole with residue $-1/6$ at $\infty$.  So the value of the integral is
$$
(-2\pi i)\left(\frac{-1}{6}\right) = \frac{i\pi}{3} .
$$
If you want to generalize the residue theorem for infinitely many singularities, understand how the residue theorem is proven first. One replaces the loop by a circuit that encloses the singularities plus some arcs that are integrated in both directions. 

In the case of infinitely many singularities you can do that for all singularities at once, but you can enclose more and more. 

The extra point that you need is to show is that integral over the loop enclosing the non-isolated singularity must tend to zero as you make the loop smaller. Or more general, show that the limit exist. Then the integral is equal to the series of the residues at the isolated singularities, plus the limit at the non-isolated one.



Now, for that specific integral one doesn't have to go that way. 

$$\oint_{|z|=1}\frac{dz}{\sin\left(\frac{1}{z}\right)}=\oint_{|z|=1}\frac{dz}{\sin{\overline{z}}}$$.

You can expand in series and compute $\oint_{|z|=1}\overline{z}^kdz$.