Evaluate (1-x^3)^-1/3 from 0 to 1

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SUMMARY

The discussion focuses on evaluating the integral of the function (1-x^3)^(-1/3) from 0 to 1 using contour integration techniques. Participants explore the necessity of branch cuts and the implications of choosing different contours, specifically the Mercedes Benz contour. The conversation highlights the importance of defining polar angles and branch cuts to avoid discontinuities and ensure the correct evaluation of the integral. The final result is confirmed to be 2 pi sqrt(3)/9, with detailed explanations on how to manage phase factors and analytical continuations.

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  • Complex analysis, specifically contour integration techniques
  • Understanding of branch cuts in complex functions
  • Familiarity with polar coordinates in complex analysis
  • Knowledge of the residue theorem and its applications
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  • #31
I still don't see why the residue are negatives for the example with the 2pi on top. I get negative 2I = 2pi*i residues, where the residues are from the positive sqrt z/z^2+1.
 
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  • #32
why is it that we choose the negative sqrt if the 2pi is on top?
 
  • #33
jack5322 said:
why is it that we choose the negative sqrt if the 2pi is on top?

You don't have to. Do it differently and see what happens.
 
  • #34
I know I don't have to, but I need to see that It's possible that we can do this. It will make sense If I can do this, because If the choice from 2pi can be shifted to the bottom then Ill see why we can manipulate how the cuts phases can be changed. Also If we shift the 2pi to the top, the zero is on the bottom right?
 
  • #35
I think I know now, we take the residues from the integral in the positive orientation, i.e. the one going from zero to infinity! That makes sense now.
 
  • #36
Actually that doesn't work. My question is this: Why is it that we calculate the residues from the negative square root if the 2pi is on top?
 
  • #37
jack5322 said:
Actually that doesn't work. My question is this: Why is it that we calculate the residues from the negative square root if the 2pi is on top?

In that case the polar angle goes from 2pi to 4 pi. When comnputing the residue, you have to evaluate a square root of some complex number. You write it in polar form, but then the angle will be between 2 pi and 4 pi.
 
  • #38
ok, why is it that the 2pi is on the top and the zero on the bottom for the example of something like (1-z)^-1/2 for -1<x<1 but the 2pi is on the bottom for the sqrtz with 0<x<infinity and zeroon the top?
 
  • #39
jack5322 said:
ok, why is it that the 2pi is on the top and the zero on the bottom for the example of something like (1-z)^-1/2 for -1<x<1 but the 2pi is on the bottom for the sqrtz with 0<x<infinity and zeroon the top?

If you have 1-z, then the line theta = 0 moves from z = 1 in te direction of z = 0. This is because you have to use the polar representation of 1-z. This is the only relevant difference between the two cases.
 
  • #40
following this logic, wouldn't the 8pi/3 be on the left of the cut for the MB contour, the one going out to the left? If not, why?
 

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