Complex analysis: contour integration

Question

Evaluate by contour integral:

$$\int_0^1{ dx\over (x^2-x^3)^\frac 13}$$

Should I go for some kind of substitution so that the limit changes to $0$ to $\pi/2$?

Answer 1

Draw a dumbbell contour $C$ about the branch cut $z \in [0,1]$. That is, two small circular segments about the branch points, and an upper and lower path connecting them above and below the real axis, respectively, as illustrated below:

We consider the integral

$$\oint_C dz \, f(z) = i 2 \pi \text{Res}_{z=0} \left [\frac{1}{z^2} f\left(\frac{1}{z}\right) \right ]$$

where

$$f(z) = z^{-2/3} (1-z)^{-1/3}$$

and the term on the right in the former equation is the residue at infinity. This residue may be computed by seeing that

$$\frac{1}{z^2} f\left(\frac{1}{z}\right) = \frac{(z-1)^{-1/3}}{z}$$

so that

$$\text{Res}_{z=0} \left [\frac{1}{z^2} f\left(\frac{1}{z}\right) \right ] = (-1)^{-1/3}$$

I will elaborate on this in a bit.

Now, 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 $(-\infty,0]$. 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 $[1,\infty)$.

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

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

Answer 2

Here is a way without contour integration. $$I = \int_0^1 \dfrac{dx}{(x^2-x^3)^{1/3}} = \int_0^1 \dfrac{dx}{x^{2/3}(1-x)^{1/3}}$$ Let $x=\sin^2(a)$. We then get that $$I = \int_0^{\pi/2} \dfrac{2 \sin(a) \cos(a) da}{\sin^{4/3}(a) \cos^{2/3}(a)} = 2 \int_0^{\pi/2} \sin^{-1/3}(a) \cos^{1/3}(a) da = \beta(1/3,2/3) = \dfrac{2 \pi}{\sqrt3}$$