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Why do the singularities in Cauchy's residue theorem have to be within the contour, and why do they still count if they're not on the path of integration, like I'd suspect for real integrals? Sorry if this is basic but I haven't found it anywhere else.

Pineapple Fish
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    Are you familiar with Cauchy's Integral Theorem? If $f$ is analytic in and on a rectifiable curve, then the contour integral on that path is $0$. Now, deform the contour to exclude any singularities and proceed. – Mark Viola Jan 24 '19 at 21:24

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I find homotopy to the most intuitive explanation: suppose you have two closed contours $\gamma_1$ and $\gamma_2$ and a function $f(z).$ If you can smoothly deform $\gamma_1$ into $\gamma_2$ without "crossing over" a singularity of $f,$ then $$\oint_{\gamma_1}f(z)\,dz=\oint_{\gamma_2}f(z)\,dz.$$ So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.

Adrian Keister
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  • Nice, why does homotopy apply? I've thought about it before. – Pineapple Fish Jan 24 '19 at 21:42
  • Well, within the limits of the assumptions of homotopy, it always applies! Homotopy doesn't get you everywhere you need to go, but it is quite intuitive. – Adrian Keister Jan 25 '19 at 18:24
  • I think I get it, it works because the integral is linear, you can write it as a sum of the integral along other paths, then each of those be be changed thus you can change the whole closed contour... I’ll think more about the gritty details of singularities. – Pineapple Fish Jan 25 '19 at 18:44