I want to evaluate this contour integral:
$\int_{\gamma}e^{\frac{-1}{z}}dz$ where $\gamma(t)=e^{it}$ on $[0,2\pi]$
I have tried several methods. The first one I tried is direct parameterization:
First, I write $z$ with respect to $\gamma(z)$:
$\int_{\gamma}e^{\frac{-1}{e^{it}}}\frac{dz}{dt}dt$
I then calculate $\frac{dz}{dt}$:
$\frac{d(e^{it})}{dt}=ie^{it}$
Replacing and rearranging I get:
$\int e^{\frac{-1}{e^{it}}}ie^{it}dt$
and from here... I am stuck.
I tried to apply the generalization of the fundamental theorem of calculus, but in order to find a "clean" antiderivative I need to use the exponential integral Ei and it seems awfully complicated and messy.
Any ideas or suggestions ? Are my steps right?
$\exp(-1/z)$ has the Laurent series $$\exp(-1/z)=\sum_{n=0}^\infty\frac{(-1)^n}{n!z^n}.$$ The residue at the origin is the coefficient of $1/z$, that is $-1$. Thus $$\int_\gamma \exp(-1/z)\,dx=-2\pi i.$$