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

  • Thread starter jack5322
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  • #26
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If you have a function of the form f(z)z^(1/2) and f(z) has a pole somewhere at z = p, then the residue at z = p clearly depends on how you define z^(1/2).
 
  • #27
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so the specified value of the multifunction is the same value on the top of the cut?
 
  • #28
1,838
7
so the specified value of the multifunction is the same value on the top of the cut?
Make some arbitrary choice for that and then the values are fixed everywhere. Why not try out a simple example, like the integral of sqrt(x)/[x^2 + 1] from x = 0 to infinity? In this case you can choose a contour from epsilon to R, half a circle to minus R and then from minus R to minus epsilon and then half a circle back to epsilon.
 
  • #29
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how do we know which part of the multifunction to compute for the residues?
 
  • #30
1,838
7
how do we know which part of the multifunction to compute for the residues?
The choice is fixed when you define the function. If you define sqrt(z) and then want to evaluate the residue at z = i, there is no ambiguity, because you know what to choose for the polar angle for z = i.
 
  • #31
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I still dont 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.
 
  • #32
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why is it that we choose the negative sqrt if the 2pi is on top?
 
  • #33
1,838
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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
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I know I dont 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
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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
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Actually that doesnt 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
1,838
7
Actually that doesnt 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
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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
1,838
7
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
59
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following this logic, wouldnt 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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