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You integrate in a loop around a singularity $z$ and get $2\pi i \text{Res}(z)$. Is there a path of integration such that the result is $\gamma 2\pi i \text{Res}(z)$ with $\gamma\in (0,1)$?

If it helps, I know that if the singularity is simple, then it is easy: approach and exit the singularity in rays $\gamma/2\pi$ apart.

Meow
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  • Are you familiar with winding numbers? And of course there is such a path. The full loop gives you a multiple of $2\pi i$ and since the integral as a function of the length of the contour is continuous there must be a partial contour that gives you a fraction of the residue. – Alex R. Nov 29 '14 at 07:58
  • @AlexR. I am. Apologies. I meant a contour that does not stop at any finite point (that is, goes off to infinity in both directions). – Meow Nov 29 '14 at 08:18

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