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I have a problem that involves integrating a complicated Schwarz-Christoffel integrand. In order to test out my results so far, I've tried to give simpler integrands of the same type to Mathematica, and it spits out... oddities.

Here's what I believe is a minimal working example:

f[w]:=1/(w(1-w^2))^(1/2)
F[ζ_?NumericQ]:=NIntegrate[f[w],{w,0,ζ}]
ParametricPlot[With[{ζ=x + I y},{{ReIm[F[ζ]]}}],{x,-20,20},{y,0,20}]

This produces for me an image with a screwed up upper left corner (it is supposed to be a square, for those not familiar with these types of integrands).

I hesitate to give it any more complicated integrands until I can get this one to display correctly. Can anyone tell me what's wrong with this code, or what I might have to add in to make it show a nice pristine square? Furthermore, is this code going to be OK for a polygon with many sides? For my problem, there are as many as 24 sides, leading to 23 terms in the integrand of the form $(w-a_i)^{\theta/\pi - 1}$.

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  • $\begingroup$ The integral depends on the path of integration, doesn't it? What should the path be? $\endgroup$
    – Michael E2
    Jan 14 '20 at 4:41
  • $\begingroup$ I don't think it matters, but for the sake of nailing one down, let's say the straight line path from 0 to the point. $\endgroup$ Jan 14 '20 at 4:52
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    $\begingroup$ What I mean is that when I specify different integration paths, I get different results. None look like what you describe, btw, and I'm not familiar with this type of plot. Different plots for different paths means you've got to pick the right path, assuming nothing else is wrong with the code. $\endgroup$
    – Michael E2
    Jan 14 '20 at 5:07
  • $\begingroup$ @MichaelE2 That's very interesting that the same code produces different things for different people. How can I change the path myself then, to play with it? $\endgroup$ Jan 14 '20 at 7:02
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    $\begingroup$ You can specify a polygonal path in the complex plane by listing the vertices: {w, 0, 1+I, zeta} -- Otherwise you can manually parametrize the path. $\endgroup$
    – Michael E2
    Jan 14 '20 at 13:10

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