This is a question on contour integration. The particular problem has a (simple) pole on the contour which prohibits a direct application of Cauchy's Residue Theorem.
Daniel Fischer commented as follows
Not really. […] if the contour is smooth at the pole, it's as if half of the pole lies inside the contour and half outside. If the contour has a corner at the pole, with (inner) angle $\alpha$, the fraction is $\alpha2\pi$, so you get $\alpha i$ times the residue of the pole instead of $2\pi i$ times as for singularities properly enclosed by the contour.
The same result is mentioned in this question.
Unfortunately, Daniel didn't know a reference for this (generalised) result. Can anyone point me to a book/paper/recourse which covers this result? I'd like to see a proof and some maths underlying this intuition.
Thank you very much!
Best Answer
Very reasonable question! I wondered about this for decades, myself! :)
The integral does not literally converge. It does converge in a "Cauchy principal value" sense, but this requires that we make a convention, or do something. It is not in any way automatic, any more than $\int_{\mathbb R} f(x)/x\;dx$ "automatically" takes the "Cauchy principal value" value.
The more bare, real fact is that that integral "through" the pole is not well-defined, since, after all, as a literal integral (as opposed to something with conventions imposed) it does not converge at all.
This explains why there's no "proof" that a contour integral "through" a pole picks up half the residue. Because the assertion is not literally true, as stated. Sure, we can say something about the related principal value integral, but that is a very different thing.
(And the possibilities of other "angles" of contour through poles likewise need principal value interpretations, otherwise are not well-defined. And, NB, there is no mandate to take the PV interpretation, so, in particular, the literal integrals do not magically/automatically take those values.)
EDIT: also, in case people might too glibly assume that there's not real issue about "regularizing" such integrals, please do consider the precise assertion of the Sokhotski-Plemelj theorem (eminently google-able). That is, it turns out that it is easy to imagine false things in terms of regularization.