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

[jack5322]

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.

 

[jack5322]

why is it that we choose the negative sqrt if the 2pi is on top?

 

[Count Iblis]

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.

 

[jack5322]

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?

 

[jack5322]

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.

 

[jack5322]

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?

 

[Count Iblis]

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.

 

[jack5322]

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?

 

[Count Iblis]

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.

 

[jack5322]

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?