I've come across this integral and I'm having some problems with it. I get to a solution, but looks a bit weird and I may be doing something wrong.
$\int_C\cos(e^{(1/z)})dz $
Being $C$ the unit circle.
I've tried to calculate the residue at z=0, which is the only singularity (and it's essential). I do the following:
$e^{1/z} = \sum_0^\infty \frac{1}{n!z^n}$
$\cos (x) = 1-x^2/2+x^4/4! + \ldots$
I've found that the term that goes with 1/z for every x^n in the cosine, after substituting x=exp(1/z), is n/z, that means, the term 1/z will be:
$\sum_1^\infty \frac{(-1)^n}{(2n-1)!z}=\sum_0^\infty \frac{(-1)^{n+1}}{(2n+1)!z}$
That means the residue is:
$\sum_0^\infty \frac{(-1)^{n+1}}{(2n+1)!}$
Which is $-sin(1)$, so the integral would be $-2\pi i\sin(1)$
Don't know why, but doesn't look good. Can someone confirm this is right or tell me what I'm doing wrong? Thank you.