TITLE: Complex Integration (Residue Theorem)
QUESTION [2 upvotes]: How do I integrate $ \oint_{C:\left | z \right |= R}^{}\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz $ ,
with $ 0< R< 1 $ ?
I am supposed to use the residue theorem but there's no Laurent series around z=0 only a Taylor series. So I guess the Residue is just equal to 0, but there's an essential singularity at z=0. I'm confused.

REPLY [0 votes]: $I=\oint_{C:\left | z \right |= R}^{}\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz$, so one can go to infinity where $e^{1/z}$ behaves like $1$ and use the residue theorem at $\pm i$ where it is much easier to compute the residues than at $0$
Note that $z^2+1=(z+i)(z-i)$ so the sum of residues at $\pm i$ of the function is $\frac{e^{-i}}{2i}-\frac{e^{i}}{2i}=-\sin 1$
Also for $|z|=M$ large $|\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz|=O(M/(M^2-1)=O(1/M) \to 0, M \to \infty$
since $|dz|=M, |e^{1/z}|=1+O(1/M), |1/(z^2+1)| \le 1/(M^2-1)$
Applying Cauchy for large $M$:
$I-2\pi i \sin 1=\oint_{C:\left | z \right |= M}^{}\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz \to 0, M \to \infty$, hence
$\oint_{C:\left | z \right |= R}^{}\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz=2\pi i \sin 1$
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