TITLE: How can I evaluate this contour integral?
QUESTION [5 upvotes]: Suppose we have the following contour integral, in the complex plane:
$$
\int_{\gamma} \frac{e^{\frac{1}{z}}}{z^{2}} \; dz
$$
where $\gamma (t) = e^{it}$ for $0 \leq t \leq 2 \pi$.
To solve this, I would express the integral in the form
$$
\int_{\gamma} \frac{f(z)}{z-z_{0}} \; dz
$$
Where we would have
$$
f(z) = \frac{e^{\frac{1}{z}}}{z} \\
z_{0} = 0
$$
Then, by Cauchy's integral formula, we have that
$$
\int_{\gamma} \frac{f(z)}{z-z_{0}} \; dz = 2 \pi i \cdot n( \gamma , z_{0} ) \cdot f(z_{0})
$$
However, $f(z_{0})$ is undefined for $z_{0} = 0$. What can I do to avoid this problem?

REPLY [5 votes]: You can use residue theorem. 
$$e^{1/z}=1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+\dots$$
$$\frac{e^{1/z}}{z^2}=\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{2z^4}+\frac{1}{6z^5}+\dots$$
So the residue of $\frac{e^{1/z}}{z^2}$ in $z=0$ is $0$.
So $\int_{\gamma} \frac{e^{\frac{1}{z}}}{z^{2}} \; dz =0$.