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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

Ángel Mario Gallegos
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Felipe Evaristo
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    "I've been working on this problem for days now." - well, can you show us some of your work? – The Chaz 2.0 Nov 05 '15 at 03:27
  • for an elementary way to compute the integral, change variable to $v = \left(\frac{1}{x}-1\right)^{1/3}$, the integral becomes $\int_0^\infty \frac{3v dv}{v^3 + 1}$. – achille hui Nov 05 '15 at 04:07
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    This is a possible duplicate of THIS ANSWER in which I used two different methodologies to evaluate the closed-contour integral with the same integrand. In the process, one can easily obtain the integral of interest here. – Mark Viola Nov 05 '15 at 05:11
  • In my work I have for the most part tried to correctly parametrize z=x on the different segments of the dog-bone-shaped contour attached to the question. I didn't post it here because it's all scratchwork. My difficulty lies exactly in determining what the correct parameterizations on the different segments will be since both 0 and 1 are branch points for the integrand. I basically have a hard time determining on which Riemann sheet I am as I go along each segment. Can anyone help me with that? – Felipe Evaristo Nov 05 '15 at 19:57
  • @Dr.MV In the case of the answer you referred to in your comment, how did you determine the arguments for zˆ(−2/3) and (1−z)ˆ(−1/3) on the different segments of the slit? Why are the arguments for each of them different and how will the arguments on the extremes of the dog bone look like if you go from one Riemann surface to another as you go over them? Finally, how many Riemann surfaces do you go through as you go over each dog-bone extreme? I apologize if these questions are too many. Arfken doesn't bring too deep explanations on this topic and I'm having a hard time finding material on it. – Felipe Evaristo Nov 05 '15 at 20:11
  • The short answer is that there is not a unique set of branch cuts. And the arguments of the components will depend on the choice of those cuts. If you look through the answer, you will find a description. – Mark Viola Nov 05 '15 at 20:23
  • @Dr.MV I think I finally understood your solution! Thanks a lot for referring to it! – Felipe Evaristo Dec 16 '15 at 18:28
  • You're welcome! My pleasure. - Mark – Mark Viola Dec 16 '15 at 18:35
  • And feel free to give that answer an up vote as you see fit of course. - Mark – Mark Viola Dec 16 '15 at 18:35
  • Just did, however it said I need at least 15 rep points for it to show. Not exactly sure if my up vote would still count in that case... – Felipe Evaristo Dec 16 '15 at 18:38

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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.

Claude Leibovici
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