Integrate: $\int_0^1 \frac{1}{\sqrt[3]{x^2 - x^3}}dx$

How to integrate using Residue theorem. $$\int_0^1 \frac{1}{\sqrt[3]{x^2 - x^3}}dx$$

How do I choose my branch-cut particularly? I was reading this article on wikiepdia and I think it is related. What I don't understand is

1. "The cut of $z^{3/4}$ is therefore $(−∞, 0]$ and the cut of $(3−z)^{1/4}$ is $(−∞, 3]$. It is easy to see that the cut of the product of the two, i.e. $f(z)$, is $[0, 3]$". What is this product? Is it $(-\infty, 3]\setminus (-\infty, 0] \cup {0}$ ?
2. Why is $0\le\arg((3-z)^{(1/4)}) \le 2\pi$? the start of angle is taken at negative $x$ axis (counterclockwise) while $0\le\arg(z^{(3/4)}) \le 2\pi$ is taken along positive axis (counterclockwise)?

Also in my integral, I think I don't have residue outside the contour, do I? Thanks for your help in advance!!

My given hint says to use this contour but this quite different from the one in Wikipedia?

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In a case like this, you have to contend with a residue at infinity. You can see this from the contour in your hint: as the radius of the circular contour $R \to \infty$, the integral about that contour approaches

$$i R \int_0^{2 \pi} d\phi \, \frac{e^{i \phi}}{\left(R^2 e^{i 2 \phi}-R^3 e^{i 3 \phi}\right)^{1/3}} \sim i 2 \pi (-1)^{-1/3}$$

However, you have to subtract out that dumbbell piece which excludes the branch points from the interior. This is where things get tricky. To this effect, we define

$$z^{-2/3} = e^{-(2/3) \log{z}}$$

such that $\arg{z} \in [-\pi,\pi)$. This definition is a result of the original branch cut of this factor being $(-\infty,0]$. The reaosn the branch is defined this way is because the argument of the log is negative real along the branch cut. Further define

$$(1-z)^{-1/3} = e^{-(1/3) \log{(1-z)}}$$

such that $\arg{(1-z)} \in [0,2\pi)$. This definition is a result of the original branch cut of this factor being $(-\infty,1]$. The reason the branch is defined like this is because, along the branch, $1-z$ is positive real along the branch cut.

To summarize, on the lines above and below the real axis, $z=x \in [0,1]$ and therefore $\arg{z} = 0$. On the line above the real axis, however, $\arg{(1-z)} = 2 \pi$. Therefore above the real axis, $z^{-2/3} (1-z)^{-1/3} = x^{-2/3} (1-x)^{-1/3} e^{-i 2 \pi/3}$ Below the real axis, $z^{-2/3} (1-z)^{-1/3} = x^{-2/3} (1-x)^{-1/3}$ because there, $\arg{(1-z)} = 0$.

Further, it should be clear that the integrals about the small circular arcs of radius $\epsilon$ around the branch points vanish as $\epsilon^{1/3}$.

Therefore, we may write

$$\left ( 1-e^{-i 2 \pi/3}\right) \int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi (-1)^{-1/3}$$

Because that residue was calculated from the $1-z$ term, then $-1=e^{i \pi}$ and we have

$$\int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi \frac{e^{-i \pi/3}}{1-e^{-i 2 \pi/3}} = \frac{\pi}{\sin{(\pi/3)}} = \frac{2 \pi}{\sqrt{3}}$$

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@RandomVariable: not ignoring you, just under a massive workload and cannot give the question the time it deserves right away. Please give me a little bit and I will answer. – Ron Gordon Jun 24 '13 at 18:41