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I'm a physics student and learning complex analysis (to do definite integrals). There is a contour integral problem but I cannot solve it. Fortunately, I found the same problem on this stack exchange but I cannot understand the logic. There is the problem in this link. Evaluate by contour integration $\int_0^1\frac{dx}{(x^2-x^3)^{1/3}}$ and I want to ask about the first answer.

What I cannot understand is how to get a Laurent expansion of given integrand. (to get a residue at infinity) He/She said what I should do but I cannot see how x^(2/3) vanishes and e^(2pi*i/3) appears.

And another point confusing me is that I've never seen doing contour integral using residue at infinity. I googled for a bit and found that there is a property that total residue sum (including at infinity) is zero. (which is not written on my textbook.) My question is, can we just find the residues at 0 and 1 instead of finding residue at infinity?

I added a another confusing point in reply.

Septacle
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  • Let $f(z) = (z^2-z^3)^{-1/3}$. Then choosing the correct branch $g(z) = f(1/z) = (z^{-2}-z^{-3})^{-1/3} = z (z-1)^{-1/3}=z e^{-i \pi /3} (1-z)^{-1/3}$ is analytic for $|z| < 1$ and we can expand using the binomial series $(1-z)^{-1/3}= \sum_{k=0}^\infty {-1/3 \choose k} (-z)^k$ – reuns Sep 17 '17 at 00:24
  • $f$ has a branch point at $z= 0$ and $z=1$ so $\text{Res}(f(z),1),\text{Res}(f(z),0)$ don't make any sense. – reuns Sep 17 '17 at 01:53
  • Thanks! I got it. But There arose another confusing point. I wanna use branches -pi<arg(z)<pi and -pi<arg(1-z)<pi instead of 0<arg(1-z)<2pi in link. But then contour integral itself comes out as pure imaginary number and i (sqrt(-1))cancel with 2pi * i in residue integral, and e^(2pi*i/3) term remains which is not possible as the original integral was real. I cannot find my fault (which seems related to branch cut) Can you help me? – Septacle Sep 18 '17 at 01:08
  • I don't seee what you mean. You need a branch of $f(z) = (z^2-z^3)^{-1/3}$ analytic away from $[0,1]$. See https://i.stack.imgur.com/BmIo9.png the inner bone contour is $\gamma$, the outer circle is ${|z|= 2}$. Then $(1-e^{-2 i\pi /3})\int_0^1 (x^2-x^3)^{-1/3} = \int_\gamma f(z)dz = \int_{|z|=2} f(z)dz = \int_{|s|=1/2} f(1/s)\frac{ds}{s^2}$ $ = 2i \pi \text{Res}(\frac{f(1/s)}{s^2},0) =2i \pi \text{Res}(\frac{1}{s^2} s e^{-i\pi /3} \sum_{k=0}^\infty {-1/3 \choose k} (-s)^{k},0) = -2i \pi {-1/3 \choose 1}$ – reuns Sep 18 '17 at 01:38
  • Thanks for your reply. OK I'll try to explain where I stuck. I wanna know if it's possible to use branch for arg(1-z), -pi<arg(1-z)<pi instead of 0<arg(1-z)<2pi given in the link. So the arg(1-z) I use for upper and lower line becomes pi and -pi, not 2pi and 0. But Then the first coefficient (1-e^(-2pi * i/3)) in your reply becomes (e^(-pi * i/3)-e^(pi * i/3)) which is pure imaginary. So not all the imaginary terms cancel and the final answer is not real. I hope that I made it clear (sorry for not using LATEX. I don't know how to use it) – Septacle Sep 18 '17 at 01:46

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