Evalute the contour Integration

Question

Evaluate by contour integration $$\int_{0}^1\frac{dx}{(x^2-x^3)^{1/3}}$$

I am not getting any idea as to how to solve this using contour integration in complex analysis. Can some one please help me?

Answer 1

Of course we can evaluate this as a Beta function, but that is not the question. Points $0$ and $1$ are singularities of the integrand, which is a hint that contour integration may work on it.

The hint is this. The integrand $$ \frac{1}{(z^2-z^3)^{1/3}} $$ has a single-valued branch in the complement of the interval $[0,1]$. Let $f(z)$ be the branch that looks like $$ \frac{1-i\sqrt{3}}{2}\;\frac{1}{z} + O\left(\frac{1}{z^2}\right) \tag{1}$$ near $z=\infty$. There are no poles in the finite plane outside $[0,1]$.

Integration of $f(z)$ around a circle of large radius $R$ gives us the same answer as integration around a contour $\Gamma_\varepsilon$ described like this:

Go just below the real axis from $0-i\varepsilon$ to $1-i\varepsilon$, go around the point $1$ in a counterclockwise semicircle, go just above the real axis from $1+i\varepsilon$ to $0+i\varepsilon$, go around the point $0$ counterclockwise in a semicircle.

The limit (as $R \to +\infty$) of the integral around the large circle is found from (1). The limit (as $\varepsilon \to 0$) of the integral around the contour $\Gamma_\varepsilon$ is found in terms of the real integral you want to compute. You get a linear equation to solve for your integral.

Answer 2

Such integral can be evaluated through Euler's Beta function:

$$ \int_{0}^{1}x^{-2/3}(1-x)^{-1/3}\,dx = B\left(\tfrac{1}{3},\tfrac{2}{3}\right) = \Gamma\left(\tfrac{1}{3}\right)\,\Gamma\left(\tfrac{2}{3}\right)\stackrel{\text{reflection f.}}{=}\frac{\pi}{\sin\frac{\pi}{3}} =\color{red}{\frac{2\pi}{\sqrt{3}}}.\tag{1} $$ As an alternative, by letting $x=\sin^2\theta$ the LHS is converted into

$$ 2\int_{0}^{\pi/2}\left(\cot\theta\right)^{1/3}\,d\theta=2\int_{0}^{+\infty}\frac{dt}{t^{1/3}(1+t^2)}=6\int_{0}^{+\infty}\frac{t}{1+t^6}\,dt=3\int_{0}^{+\infty}\frac{dt}{1+t^3}.\tag{2}$$ The last integral can be evaluated through standard contours, or just via $$ \int_{0}^{+\infty}\frac{dt}{(1+t)(1-t+t^2)}=\int_{1}^{+\infty}\frac{dt}{t(t^2-t+1)}=\int_{0}^{1}\frac{t\,dt}{t^2-t+1}=\left[\frac{1}{\sqrt{3}}\arctan\frac{2t-1}{\sqrt{3}}+\frac{1}{2}\log(t^2-t+1)\right]_{0}^{1}=\frac{2\pi}{\sqrt{3}}.\tag{3}$$