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

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.
 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?

 Quote by 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?
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.
 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

 Quote by 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
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.
 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?
 ????