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Is it possible to apply Cauchy's residue theorem to a function which has an infinite number of isolated singularities within the contour of integration (say a semicircle whose radius goes to infinity), with known residues $r_n$, assuming that $$\sum_{n=1}^\infty r_n$$ converges?

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In a strict sense, the residue theorem only applies to bounded closed contours. The hypotheses of the residue theorem cannot be fulfilled if the contour contains infinitely many singularities, since the union of the contour and its interior is compact, so the singularities must have an accumulation point, which would be a non-isolated singularity for which no residue can be defined.

In a looser sense, the residue theorem is sometimes applied to unbounded "contours", e.g. the real axis closed by a semi-circle "at infinity". To justify this rigorously, one defines a sequence of closed contours such that one part of the contour "goes to infinity" and its contribution to the integral can be shown to go to $0$, while the remainder of the contour either remains finite or grows such that the limit of its contribution is an improper integral, e.g. over the entire real axis.

Thus we are never actually dealing with infinite contours. If we have a sequence of contours that "goes to infinity" and the number of singularities it encloses also goes to infinity in the process, we can apply the finite version of the residue theorem to each contour in the sequence, and then the convergence of the integral to $\sum_n r_n$ enters into the convergence argument that has to be made anyway, and no special argument for applying the residue theorem to infinitely many singularities is required.

To summarize, you can apply the residue theorem to an "infinite contour" enclosing infinitely many singularities, as long as any of the actual contours in the sequence justifying this argument enclose only finitely many singularities; otherwise you have a non-isolated singularity to which the residue theorem doesn't apply.

Alternatively, you can think of contours that extend to infinity as closed contours on the Riemann sphere. Since the Riemann sphere is compact, infinitely many singularities necessarily have an accumulation point on it (which may be the point at infinity), so in this case there can only be finitely many singularities for the function to be meromorphic on and inside the contour.

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@pbs: I don't understand. What part of the argument is that meant to be an objection against? The positive integers aren't enclosed by any bounded closed contour, and for the case of contours "going to infinity" and eventually enclosing infinitely many singularities, e.g. at the positive integers, I explained why we only need the finite version of the theorem to deal with them. – joriki Oct 1 '12 at 7:02
Sorry, I was thinking of a contour, say some hemisphere, whose radius goes to infinity (oriented appropriately), and I was just thinking of isolated singularities within the contour. Talk of non-isolated singularities just threw me a bit. – poirot Oct 1 '12 at 7:05
@pbs: You mean a semicircle, not a hemisphere. I'm not sure about the status of your objection now -- did that resolve it? – joriki Oct 1 '12 at 7:12
yes, I think this answers my question now (paragraph 3 of your answer deals with my question). Thanks. – poirot Oct 1 '12 at 7:34

Yes, the residue theorem does not presuppose that the number of poles is finite. It does, however, suppose that the number of poles is isolated (and hence countable); if this is not the case, then the function in question is not meromorphic, in which case the residue theorem cannot be applied.

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so if there are a countable number of isolated singularities within the contour, say at the positive integers only, then we can apply the residue theorem so long as the infinite sum of their residues converges? – poirot Oct 1 '12 at 7:02
As mentioned significantly more rigorously by @joriki, the genuine, classical residue theorem is only on bounded, closed contours. In this case, it is easy to verify that if we want the poles to be isolated, then there must in fact be only finitely many. The more "general" result, also as explained nicely by joriki, is that you can then take a limit by considering a sequence of increasing large contours, if you wish to consider infinitely many poles. Everything should work out if the resulting limit of partial sums of residues converges. – Christopher A. Wong Oct 1 '12 at 7:36

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