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I'm trying to integrate a complex valued function over the extended real domain $[0,+\infty]$, for which I'm attempting to use the residue calculus. My function has a number of poles above the real axis, so I'm computing the residues for some of these to see what I get. All residues work out to nice values, except for one at $z=2\pi i$ which I was surprised to see has a value of $\infty$. My question is this: is it possible for a residue at a pole to take on an infinite value?

As example, say I try to compute the residues of $f(z)=\frac{\arctan(z/(2\pi))}{e^z-1}.$

If my understanding is correct I can compute the residue using $\text{Res}(f;2\pi i)=\lim_{z\rightarrow 2\pi i}\frac{\arctan(z/(2\pi))}{e^z}=i\infty,$

where I have just differentiated the denominator.

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    What is residue value at point $z=\infty$? Sum of all residue values including residue at infinity is equal $0$.2012-10-03

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I think the problem is that the numerator in the example is not analytic in a neighbourhood of $z=2\pi i$. Hence, any contour I use should in fact avoid this point.

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    yes, $\arctan$ around $z=i$ is a lot like $\log$ around $z=0$, which is trouble. Eliminating the trouble means pulling back by doing a change of variable like $z = 2 \pi \tan u$, though now you will see that you're not really integrating on a contour anymore.2012-10-03