Complex Integration (Residue Theorem) 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.
 A: Hint:
$$\frac{e^{1/z}}{1+z^2}=\left(1+\frac1z+\frac1{2z^2}+\frac1{6z^3}+\ldots\right)\left(1-z^2+z^4-z^6+\ldots(\right)=$$
$$=\sum_{k=0}^\infty\frac{z^{-n}}{n!}\cdot\sum_{k=0}^\infty(-1)^nz^{2n}=\sum_{n=0}^\infty\sum_{k=0}^n\left(\frac{z^{-n}}{n!}(-1)^{n-k}z^{2(n-k)}\right)=\sum_{n=0}^\infty\sum_{k=0}^n\left(\frac{(-1)^{n-k}z^{n-2k}}{n!}\right)$$
We're interested in all the products above of the form $\;z^{-(2m+1)}z^{2m}=z^{-1}\;$ (first sum above), or
of the form $\;z^{n-2\left(\frac{n+1}2\right)}=z^{-1}\;$ (second-third sum). Observe then that for $\;n=1,3,5,7,9,11,...\;$ we get $\;k=1,2,3,4,5,6,...\;$, so the parity of $\;n-k\;$ flips between even and odd (beginning with the former):
$$\left.\sum_{n=0}^\infty\left(\frac{(-1)^{n}}{(2n+1)!}\right)z^{-1}\implies\text{Res}\,(f(z))\right|_{z=0}=\sum_{n=0}^\infty\left(\frac{(-1)^{n}}{(2n+1)!}\right)=\sin1$$
A: $I=\oint_{C:\left | z \right |= R}^{}\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz$, 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
$\oint_{C:\left | z \right |= R}^{}\frac{e^{^{\frac{1}{z}}}}{z^{2}+1}dz=2\pi i \sin 1$
