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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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If it is possible, I'd expect the value to be simply $\infty$ rather than $+\infty$. I.e. in this context, there's only one infinity rather than the $\pm\infty$ sometimes considered in real analysis, and you approach $\infty$ by going in any direction whatsoever in $\mathbb C$, not just in one of the two "real" directions. – Michael Hardy Oct 3 '12 at 16:42
What is residue value at point $z=\infty$? Sum of all residue values including residue at infinity is equal $0$. – M. Strochyk Oct 3 '12 at 16:51

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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Under "numerator in the example is not analytic at $z=2\pi{i}$" you mean "numerator in the example is not analytic in the punctured neighbourhood $z=2\pi{i}"$? – M. Strochyk Oct 3 '12 at 17:06
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. – mercio Oct 3 '12 at 18:08

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