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

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Discussion Overview

The discussion revolves around evaluating the integral of the function (1-x^3)^(-1/3) from 0 to 1, exploring the implications of branch cuts, contour integration, and the behavior of the function in the complex plane. Participants delve into the mathematical intricacies of contour integration and the choices involved in defining branch cuts for the integral.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the rationale behind dividing the integral into three sheets and seeks clarification on the behavior of branch points, particularly regarding real cuts.
  • Another participant suggests an alternative method for computing the integral using a combination of the Mercedes Benz contour and a circular contour, emphasizing the importance of branch cut placement.
  • Some participants discuss the necessity of restricting the angle theta for discontinuity along the cut and express confusion about calculating limits above and below the cut when it is not aligned with the real axis.
  • There is a proposal to define polar coordinates in a way that relates the integral along the real axis to the desired integral, leading to a discussion about the contributions from different contour segments.
  • One participant inquires about how to determine the phase of each segment and seeks guidance on the intervals for theta in local polar coordinates.
  • Another participant elaborates on defining the function f(z) = (1-z^3)^(-1/3) and the implications of choosing branch cuts and polar angles for different contours to ensure consistency in the evaluation of the integral.

Areas of Agreement / Disagreement

Participants express various viewpoints on the placement of branch cuts and the implications for contour integration. There is no consensus on the best approach, and multiple competing views remain regarding the handling of the integral and the definition of the function across different contours.

Contextual Notes

Participants highlight the importance of correctly defining branch cuts and polar angles, noting that arbitrary choices can lead to different interpretations of the integral. The discussion reveals a dependence on the specific definitions and assumptions made during the evaluation process.

  • #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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