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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.2012-10-03
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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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