I have recently encountered a truely terrible integral which I need to compute. I am not sure it's doable but before throwing the whole project in the bin I thought I would ask here. At the moment, a step I require is evaluating the integral of $f(z)$ along a closed coutour $C$ containing zero where $f(z)$ is something like


$f(z)=e^{(\frac{a}{z}+\frac{b}{z-c})}g(z)$


$c$ is located outside the contour $g(z)$ is holomorphic inside the disk enclosed by $C$ which has a very long but finite taylor expansion. The reason no traditional tricks work (using the coordinate change $z=1/w$, looking at the series and trying to collect together all the $\frac{1}{z}$ terms) is of course this $\exp((\frac{b}{z-c}))$ term, for which the series expansion has an infinite number of terms, so trying to sum up all those containing $1/z$ is doomed from the start... Has anybody ever encountered something similar? I tried reading about Bessel functions, but they didn't quite fix the problem.



The integrals of things looking like $e^{(\frac{a}{z}+\frac{b}{z-c})}$ on closed contours
You can use the residue theorem, given the series expansion of
$$h(z)=e^{b/(z-c)}g(z)=\sum\_{n=0}^\infty h\_n z^n,$$
the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates to
$$\oint\_C e^{a/z}e^{b/(z-c)}g(z)\,dz=2\pi i\sum\_{n=1}^\infty \frac{ h\_{n-1}a^n}{n!}=2\pi i\sum\_{n=1}^\infty \frac{ a^n}{n!(n-1)!}h^{(n-1)}(0),$$
with $h^{(n)}(0)$ the $n$-fold derivative of $h(z)$ evaluated at $z=0$. 


*I would love to be shown wrong, but I'm pretty certain this is the best one can do in the general case $-$ there is no short-cut to the residue at an essential singularity.*


