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

[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?