Help calculating $\int_C e^{-1/z}\sin(1/z)dz$ over the unit circle? I found the integral 
$$
\int_C e^{-1/z}\sin(1/z)dz
$$
over the circle $|z|=1$ while doing some problems in Schaum's Outline for Complex Variables.
This integral has me stumped. The answer is $2\pi i$, but I can't see why. There are no poles, so I don't think I can apply the residue theorem. Reparametrizing with $z=e^{it}$ for $t\in(0,2\pi)$ looks very messy. 
How can this integral be approached? I tried to find the Taylor series, and the first few terms are 
$$
-\frac{1}{2z^2}-\frac{2}{3!z^3}-\cdots
$$
so $0$ looks like an essential singularity, but I don't know if that's useful.
 A: With a change of variables $z\mapsto\frac1z$
$$
\begin{align}
\int_Ce^{-1/z}\sin(1/z)\,\mathrm{d}z
&=\int_Ce^{-z}\sin(z)\frac{1}{z^2}\,\mathrm{d}z\\
&=\int_C\left(\frac1z-1+\frac{z}{3}+\dots\right)\,\mathrm{d}z\\
&=2\pi i
\end{align}
$$
Note that the change of variables reverses the direction of the path.
A: My Laurent series centered at $z=0$ of $ e^{-1/z}\sin(1/z)$ disagrees with yours: I have it as the product $\displaystyle\sum_{i=0}^{\infty} \dfrac{(-1)^i}{i!z^{i}}$ $\times$ $\displaystyle\sum_{j=0}^{\infty} \dfrac{(-1)^j}{(2j+1)!z^{2j+1}}$. The only $z^{-1}$ term of my product emerges when $j=0$ and $i=0$, and is equal to  $\dfrac{1}{z}$, where it seems you have no nontrivial term of that order.  I think that mine is correct, but perhaps we should both check our calculations?
In fact, the residue theorem is strong enough to apply in this case; $ e^{-1/z}\sin(1/z)$ has no poles and only one essential singularity at $z=0$, so our function is holomorphic on the complement of finitely many points in C, the unit circle centered at the origin.  As a result, we can compute a residue at $z=0$ as above and apply the residue theorem, which tells us that our integral is $2 \pi i$.
