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I am a bit confused on an aspect of contour integrals. Wikipedia gives the following example of an integral where the residue theorem can be used (it also finds the integral in a couple of other ways, but I am mainly interested in doing it using residues):

We want to calculate

[itex]\int_{- \infty}^\infty \frac{1}{(x^2+1)^2}\,dx[/itex].

To this end, we take the integral in the complex plane, and instead consider

[itex]\oint_{- \infty}^\infty \frac{1}{(z^2+1)^2}\,dz[/itex].

The singularities of [itex]f(z)=\frac{1}{(z^2+1)^2}[/itex] are at [itex]i[/itex] and [itex]-i[/itex]. The Laurent series about [itex]i[/itex] has a residue of [itex]-i/4[/itex]. In the upper half of the complex plane, we can choose our contour to be the semicircle with endpoints at opposite values along the real axis enclosing the pole at [itex]i[/itex], which is a closed loop, so that the residue theorem gives that the value of the closed integral is [itex]\pi/2[/itex].

Up to there, I am perfectly fine. However, how does this tell us the value of the original (the [itex]dx[/itex]) integral? For that integral to be [itex]\pi/2[/itex], the integral over the arc part of the semicircle would have to be [itex]0[/itex]. How do we know that this is the case? Furthermore, since the original integral will be less than [itex]\pi/2[/itex] if the integration limits aren't [itex]\pm \infty[/itex], but since we can still enclose the same pole with finite integration limits, the arc part of the integral must be something other than [itex]0[/itex] except in the limit where the integration limits become [itex]\pm \infty[/itex].

So, I guess what I am saying is that I will sometimes see examples where the residue theorem is used in which something is assumed about part of the integral that doesn't seem obvious to me, but that is crucial toward getting the correct result for the original, real integral, as in the example above. Is there something that I am missing?

Thanks for helping in advance.

-HJ Farnsworth

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# Contour integration confusion

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