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I have seen examples of complex contour integrals whose value depends on the choice of the contour of integration and some integrals where the value does not depend on the choice of the contour. Is there any indication by which we can know whether the choice of contour matters?

Just to give an example, the value of the improper integral $I_1=\int_{-\infty}^{\infty}\frac{\sin x}{x}$ does not depend on the choice of the contour but $I_2=\int_{-\infty}^{\infty}\frac{e^{-ix}}{x^2-4}$ does! When we try to evaluate these improper integrals by converting them to complex contour integrals and then using Cauchy's residue theorem, $I_1$ is found to be independent of the choice of contour while $I_2$ is not.

Solidification
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  • If one contour is a continuous deformation of the other, a homotopy, and if the integrand is analytic everywhere on the region between the two contours (the region traversed by the homotopy) then the integrals are guaranteed to be equal (which is remarkable!) – FShrike May 17 '22 at 14:31
  • If either the homotopy or the "analytic on the region between" condition fails, generally the integrals will no longer be the same – FShrike May 17 '22 at 14:31
  • I made a video on this topic. The contour independence comes at the same time when the fucntion is "complex differentiable" inside the domain. of integration. Actually the idea underlying this is heavily related to the idea underlying some general vector calculus theorems, so in the video I talk about that first before this particular case. – tryst with freedom May 17 '22 at 14:32
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    Hmm the integrals you've written aren't contour integrals but rather just real integrals over the real line. You integrate a function over a contour. You don't integrate an integral over a contour – tryst with freedom May 17 '22 at 14:35
  • Thanks, I have clarified that a bit. – Solidification May 17 '22 at 14:37
  • I mean the integral is zero for very specific choices of contour, if singularity points are contained inside the contour then it is not zero – tryst with freedom May 17 '22 at 14:38
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    Hint: by joining two contours with the same endpoints, this is equivalent to asking when an integral around a closed contour is nonzero. – J.G. May 17 '22 at 14:40
  • now that I look at it a bit more carefully, and it seems the first integral is a bit special. It seems so that even if you have a loop enclosing the origin it will still be zero, because the value of $\sin(z)$ is zero. By cauchy's theorem $\int \frac{f(z)}{z} dz =2\pi i f(0)$ for a loop enclosing the origin – tryst with freedom May 17 '22 at 14:45
  • The second integral, it doesn't seem to have the property I said of the above. Where did you get that it is loop independent? – tryst with freedom May 17 '22 at 14:47
  • I said $I_2$ is not independent of the choice of contour while $I_1$ is! – Solidification May 17 '22 at 14:50
  • Oh. Lol should've have read carefully. – tryst with freedom May 17 '22 at 14:59
  • @FShrike Sorry to comment on an older post, but could you elaborate on the condition for the two integrals not to agree? For example, in $I_2$ if we take a semicircle as a contour with two smaller semicircles around the poles and close the contour below the two poles why would it give a different answer than if we rotated this contour 180 degrees and closed it upwards? – CBBAM Feb 09 '24 at 03:59
  • @J.G. What do you mean by joining the two endpoints? – CBBAM Feb 09 '24 at 03:59
  • @Solidification Did you ever find a satisfactory answer to this question? I am studying the Klein-Gordon propagator and have run into the same question. – CBBAM Feb 09 '24 at 04:01
  • @CBBAM I said join contours with the same endpoints, not join endpoints. My point is $\int_A^Bfdx$ is path-independent iff $\oint fdx$ is the same on all closed contours from $A$ to $B$ back to $A$. – J.G. Feb 09 '24 at 08:04
  • @J.G. Thank you. – CBBAM Feb 09 '24 at 08:21
  • @CBBAM The two integrals may fail to agree, it depends on the function. If your function is everywhere analytic then any two paths $a\to b$ yield the same integral. Otherwise, say, consider that the loop which is a circle of radius $1$ centred at $2$ and the loop which is a circle of radius $1$ centred at zero both can be viewed as loops based at $1$; if we integrate $\frac{1}{z},\mathrm{d}z$ around the first loop, we get $0$, but if we do so around the second we get $\pm2\pi i$. These are not the same, and all that has changed is that I've taken a different path $1\to 1$ – FShrike Feb 09 '24 at 11:08
  • @FShrike Thank you for your answer, that was very helpful! – CBBAM Feb 09 '24 at 19:02

1 Answers1

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For the first integral, the residue at the only pole z=0 is zero. However such a scenario is not there in the second scenario.

Generally speaking, other than just being analytic on the interior of the domain we integrate over, another way for a function to have 0 integral over all loops is if the residues are all zero. It's easy to see this from Cauchys residue theorem.

tryst with freedom
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