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I am trying to compute $$\text{Res}\: (e^{z+\frac1z}, 0)$$ and can't get a solvable integral using the definition of a residue. I already know other ways to compute residues of poles of arbitrary order, but nothing for essential singularities- and I believe this is the case here.

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    $\begingroup$ $\exp(x)*\exp(1/x)$ Now use taylor expansion on second exponential. $\endgroup$ – MrYouMath Jul 25 '16 at 20:41
  • $\begingroup$ @MrYouMath $exp(1/z) = \sum_{k=0}^\infty \frac{1}{k!z^k} = \sum_{k=-\infty}^0 \frac{z^k}{k!}$, and thus it I only dealt with that it would have been $Res = a_{-1} = 1$, but I have to deal with the multiplication with the infinite sum of the other part. $\endgroup$ – mankala Jul 25 '16 at 21:01
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    $\begingroup$ hint for $0 < |z| < \infty$ : $(\sum_{k=0}^\infty \frac{z^k}{k!})(\sum_{m=0}^\infty \frac{z^{-m}}{m!}) = \sum_{n=-\infty}^\infty z^n \sum_{m \ge 0,k \ge 0, k-m = n} \frac{1}{m!}\frac{1}{k!}$ (both series converge absolutely so you can change the order of summation) $\endgroup$ – reuns Jul 25 '16 at 21:12
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$f(z)=e^{z+\frac{1}{z}}=[1+z+\frac{z^2}{2!}+......][1+\frac{1}{z}+\frac{1}{z^2.2!}+....]$.

$Res[f(z);z=0]=$coefficient of $1/z$ in $f(z)=\sum_{n=0}^{\infty}$$\frac{1}{n!.(n+1)!}$

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