The theorem (as I know it) only allows for a finite set of isolated singularities.

I integrated, along a square box, a function that has simple poles at all the non-zero integers -- and a triple pole at zero. Then I let the box get infinitely large to help prove that the sum of $1/n^2$ is $\pi^2 / 6$.

But why can the Residue Theorem still be applied, even though the box is getting infinitely large, and the poles will eventually become an infinite set?

...I know that a box is compact, and poles at the integers means this set of poles is a discrete set, hence the set of poles in this compact box ...is finite. But something about taking the limit is bugging me.


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    $\begingroup$ As the sides of the box grow in size, they are passing through singularities of your function. So it is dubious that you can just take a limit. However, what you are really arguing is that partial sums of the series correspond to certain contour integrals. And these contour integrals can be made as close as you like to $\pi^2/6$. $\endgroup$ – Philip Hoskins May 9 '15 at 7:57
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    $\begingroup$ You cannot make the contour infinitely large. Instead, you choose a family of contours, deriving an equation from each contour, then take limit. $\endgroup$ – Yai0Phah May 9 '15 at 7:58
  • $\begingroup$ If the contour you're talking about passes through the real axis, then as Philip mentioned it is very dubious you can take the limit of the contour integrals. It should be justified thay you can do that in each case, perhaps inductively. The actual case and showing your work could probably help. $\endgroup$ – Timbuc May 9 '15 at 8:01
  • $\begingroup$ the square box I integrated along intersects the real axis at N+1/2, N an integer. So, I believe the box will safely enclose the poles, with no poles on the box itself. Then I take a limit as N goes to infinity...this part I am not too sure about. Can I specify that N is an integer, and let N go to infinity? I don't think I've ever done that in any of my coursework, but perhaps it's ok to do? Thanks, $\endgroup$ – user239123 May 9 '15 at 8:09
  • $\begingroup$ and this coincides with your suggestion, I think, @philipHoskins. $\endgroup$ – user239123 May 9 '15 at 8:15

To reiterate and amplify the comments: You are not actually making a contour infinitely large. Instead, you're considering a sequence of contours $(\gamma_{N})$, each enclosing finitely many poles. For each $N$, you use the residue theorem to deduce an equation of the form $$ a_{N} = \frac{1}{2\pi i} \oint_{\gamma_{N}} f(z)\, dz = b_{N} + c_{N} + \cdots, $$ in which $a_{N}$ is the sum of the residues of the enclosed poles, and the terms on the right are contributions to the integral from smooth portions of the contour.

Next you focus on the termwise equality of sequences $$ a_{N} = b_{N} + c_{N} + \cdots, $$ taking the limit as $N \to \infty$. Chances are you've arranged that $a_{N}$ converges to the sum of some series (such as $\zeta(2)$) or can be evaluated explicitly (e.g., because there are only finitely many poles), and that the limits of the expressions on the right can be evaluated (or are themselves integrals of interst).

In the end, you have a useful evaluation of some quantity, often an improper integral or the sum of an infinite series.

For brevity, one often speaks of the process as if there's a "limiting contour" enclosing all the poles, but technically that's not the underlying reasoning.


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