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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)?

enter image description here

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? enter image description here

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up vote 6 down vote accepted

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

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