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.