Evaluating Contour Integral How do I go about evaluating the following by contour integration?
$$ \int^1_0 \frac{dx}{(x^{2} - x^{3})^{1/3}} $$
The question does not fit in the standard form of : $\int^{2\pi}_0$ or $\int^\infty_{-\infty}$ 
etc.
I tried substituting $ x = cos(\theta) $ but got stuck.
 A: You have branch points at $x=0$ and $x=1$, so one would consider the following integral:
$$\oint_C dz \, z^{-2/3} (1-z)^{-1/3}$$
where $C$ is a dumbbell-shaped contour that encircles the branch points in a counterclockwise manner.  Around each of the branch points, the integral consists of a circular segment of radius $\epsilon$.  As $\epsilon \to 0$, the integrals around each of the circular segments vanishes.  (About $z=0$ as $\epsilon^{1/3}$, about $z=1$ as $\epsilon^{2/3}$.  Can you see why?)  
This leaves a top segment and a bottom segment.  On the top segment, $\arg{z} = 0$, while on the bottom, $\arg{z}=2 \pi$ for only the $z^{-2/3}$ piece.  Why?  This is a consequence of how we defined the respective  branch cuts of $\log{z}$ and $\log{(1-z)}$.  That is, for $\log{z}$, the cut is applied to the positive real axis, while the cut for $\log{(1-z)}$ is applied to the real axis where $\Re{z} \ge 1$.  Thus, $\log{z}$ experiences a $2 \pi$ jump in its argument by crossing its branch cut, while $\log{(1-z)}$ does not.  Then the contour integer is equal to
$$\int_0^1 dx \, x^{-2/3} (1-x)^{-1/3} + e^{-i 2 \pi/3} \int_1^0 dx \, x^{-2/3} (1-x)^{-1/3}$$
By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue at infinity.  The residue at infinity of a function $f(z)$ is computed by computing the residue at $z=0$ of $f(1/z)/z^2$.  In this case, this is
$$\frac1{z^2} \left ( \frac1{z^2}-\frac1{z^3}\right )^{-1/3} = \frac1{z} (z-1)^{-1/3}$$
Noting that, because of how we set up the branch cuts, $-1=e^{i \pi}$, we have that
$$\left ( 1-e^{-i 2 \pi/3}\right ) \int_0^1 dx \, x^{-2/3} (1-x)^{-1/3} = i 2 \pi \, e^{-i \pi/3}$$
which means that
$$\int_0^1 dx \, x^{-2/3} (1-x)^{-1/3} = \frac{\pi}{\sin{(\pi/3)}} = \frac{2 \pi}{\sqrt{3}}$$
which is what you get from seeing that the integral is a beta function.
