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

  • Thread starter jack5322
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In summary: The phase of this function on the real axis from one contour to the next is exp(2 pi i/3). So, the phase you are looking for is the phase of the power 1/3 of the function (1-z^3) on the real axis. This phase is exp(-2/3 pi i), when you go from the contour that encircles the origin in the positive direction to the contour that encircles the origin in the negative direction. So, the phase of the function on
  • #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?
 
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  • #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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