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I just wanted to verify something, since I saw some notes online that confused me. They were doing some contour integral and $0$ was the only branch point of the function, and they chose the branch cut connecting $0$ to ${+\infty}$ on the real axis. The contour they chose was a semi-circular contour. The part that confused me, was that at $0$ rather than just "go through" $0$ - they made some small $\epsilon$ semi-circle around $0$, but then allowed themselves to integrate along the branch cut. I thought that as long as the function is analytic at $0$ (which it was) - it's fine to integrate over a contour containing branch points/cuts on the contour? So long as you don't actually "cross" that branch cut or encircle any problematic branch points. Am I correct? Or is there a valid reason they did this? Thank you!

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One might say you shouldn't integrate along a branch cut since (depending on exactly how things are defined) the value of the function on the branch cut is ambiguous (it's one thing on one side of the cut, another thing on the other). So integrate on a path that's very close to the branch cut (and on the side you want), and take the limit as your path approaches the branch cut. The result should be the same as integrating along the branch cut (using the values that come from the side you want).

As for a path that goes through a branch point, as long as the function has a finite limit as you approach the branch point, there should be no problem. Again you could avoid the branch point by taking a small detour around it, but in the limit as the size of the detour goes to $0$ you will get the same result.

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  • $\begingroup$ I see that makes sense to me. So integrating along a branch cut isn't a problem, but because of ambiguity in which "side" you are talking about it helps to instead take a contour "hovering" just above or below it and then take a limit? And with the branch point part - with the whole "detour" part, this is exactly what I was thinking - as long as it's nicely behaved at the branch point, any detour will give $0$ in the limit so I thought it seemed pointless. Thank you so much for taking the time to respond :) $\endgroup$ – Riemann'sPointyNose Feb 23 at 18:23

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