I want to solve this integral:
$\int_{0}^{2\pi}e^{\frac{-1}{e^{it}}}e^{it}$
When I plug it into Wolfram Alpha, the result is $-2\pi$.
However the indefinite integral is insane and it involves the exponential integral.
Is there any "simple" way to solve it?
I tried u-substitution and integrating by parts, but it's not making it any easier..
On
$\int_{0}^{2\pi}e^{\frac{-1}{e^{it}}}e^{it} dt=i \int_{0}^{2\pi}e^{\frac{-1}{e^{it}}}ie^{it} dt=i \int_{|z|=1}e^{\frac{-1}{z}} dz= i 2 \pi i Res(f; 0)=-2 \pi Res(f; 0)$,
where $f(z)=e^{\frac{-1}{z}}$.
Since $f(z)=1- \frac{1}{z}+\frac{1}{2z^2}-+...$, we have $Res(f;0)=-1$, hence
$$\int_{0}^{2\pi}e^{\frac{-1}{e^{it}}}e^{it} dt=2 \pi.$$
Here's a couple ways to see the answer: $$ \int_0^{2 \pi} e^{-1/e^{it}} e^{it} dt = \int_0^{2 \pi} e^{-e^{-it}} e^{it} dt = \int_0^{2\pi}\sum_{n=0}^{\infty}\frac{1}{n!} (-e^{-i t})^n e^{i t} dt $$ which is pretty straightforward to integrate term by term, which you can do because the sum is uniformly convergent on a bounded interval.
Or, with a $u$-sub by defining $u = e^{-it}$:
$$ \int_0^{2 \pi} e^{-1/e^{it}} e^{it} dt = i \int_0^{2 \pi} \frac{e^{-e^{-it}}}{e^{-2 i t}} (-i) e^{-it} dt = i \oint_\gamma \frac{e^u}{u^2} du $$
where $\gamma$ is a clockwise loop around the origin (in fact in this case it is the unit ring with clockwise orientation). This integral can be evaluated with Cauchy's integral formula.