How do I prove that $\int_0^1 \frac{1}{(x^2-x^3)^{1/3}} =\frac{2\pi}{\sqrt{3}}$?

Question

This is a problem from Mathematical Methods for Physicists, by Arfken, 7th edition (Problem 11.8.27). I know the integrals in the circular paths around 0 and 1 will vanish, but am completely lost on why and how that happens because there are 2 branch points and 3 Riemann surfaces for the function in the integrand in this case (all examples in the book work with functions with 1 branch point and 2 Riemann surfaces only in the integrand, plus he never really explains why the integrals around branch points disappear). Also, do I need to go down all 3 Riemann surfaces to calculate the integral over the bottom line segment path or going down to the second one only will be enough? I've been working on this problem for days now. Thanks a lot for any help! Problem 11.8.27 Contour for problem 11.8.27

Answer 1

What achille hui suggested is, for sure, a very efficient way for the solution. Using $v = \left(\frac{1}{x}-1\right)^{1/3}$ that is to say $x=\frac{1}{1+v^3}$, $dx=-\frac{3 v^2}{\left(v^3+1\right)^2}dv$ we then have $$I=\int \frac{dx}{(x^2-x^3)^{1/3}} =-\int \frac{3 v}{v^3+1}dv=-3\int \frac{ v}{(v+1)(v^2-v+1)}dv$$ Using partial fraction decomposition, we then have $$\frac{ v}{(v+1)(v^2-v+1)}=\frac{1}{v+1}-\frac{v+1}{v^2-v+1}=\frac{1}{v+1}-\frac 12 \frac{2v-1}{v^2-v+1}-\frac 32 \frac{1}{v^2-v+1}$$ The first two terms do not present any problem; for the last one, completing the square $v^2-v+1=(v-\frac 12)^2+\frac 34$ takes you to a very classical integral.

I am sure that you can take from here.