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

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 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?

Count Iblis

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

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, 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?

jack5322

????