Multivalued functions and branch points: Evaluate the following integral I have this integral :
$$ {J}=\int_{-1}^{+1}\frac{(1-x)^{-\frac{2}{3}}(1+x)^{-\frac{1}{3}}}{4+x^2}\,dx$$
There are three branch points for this integral : $1,-i,+i$ and two pole points $-2i$ and $+2i$ which lie outside. So can we say that there are no pole points for this integral and how to solve it.
 A: OK, here is a solution that works with the integral directly, branch points and all.  First, consider the complex integral
$$\oint_C dz \frac{\left ( z-1 \right )^{-2/3} \left ( z+1 \right )^{-1/3}}{z^2+4} $$ 
where $C$ is the following contour

The outer circle has radius $R$ and the inner arcs have radius $\epsilon$, and there are branch points at $z=\pm 1$.  Note that because $C$ is a closed contour, the poles at $z=\pm i 2$ are inside the contour and we compute their residues.
(As an aside, I find the way people perform these integrations, using a "dogbone" contour and considering the residues at infinity, to be utterly confusing.)
Note that the contour $C$ is detoured around the branch points.  The argument of $z$ just above the cut is $\pi$  while below the cut it is $-\pi$.  Thus, the integrand is restricted to have arguments between $[-\pi,\pi]$.  This is significant.
The complex integral above may now be written as follows:
$$e^{i \pi} \int_R^{1+\epsilon} dx \frac{(x+1)^{-2/3} e^{-i 2 \pi/3} (x-1)^{-1/3} e^{-i \pi/3}}{x^2+4} + \int_{-1+\epsilon}^{1-\epsilon} dx \frac{(1-x)^{-2/3} e^{-i 2 \pi/3} (1+x)^{-1/3}}{x^2+4} \\ + \int_{1-\epsilon}^{-1+\epsilon} dx \frac{(1-x)^{-2/3} e^{+i 2 \pi/3} (1+x)^{-1/3}}{x^2+4} + e^{-i \pi} \int_{1+\epsilon}^R dx \frac{(x+1)^{-2/3} e^{i 2 \pi/3} (x-1)^{-1/3} e^{i \pi/3}}{x^2+4} \\ + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{\left ( R e^{i \theta}-1 \right )^{-2/3} \left ( R e^{i \theta}+1 \right )^{-1/3}}{R^2 e^{i 2 \theta} + 4}  $$
A few notes: the factors of $e^{\pm i 2 \pi/3}$, etc. in the integrals come from conversion of minus signs to their proper arguments, and then taken to the given powers. Also note that I left out the integrals over the small arcs about the branch points, as they go to zero as $\epsilon \to 0$.
That said, it should be clear that the fifth integral over $\theta$ goes to zero as $R \to \infty$.  Further, the first and fourth integrals cancel, as the exponents sum to $-1$.  Thus, we are left with, as the contour integral:
$$\left (e^{-i 2 \pi/3} - e^{i 2 \pi/3} \right ) \int_{-1}^{1} dx \frac{(1-x)^{-2/3} (1+x)^{-1/3}}{x^2+4} $$
The residue theorem states that the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$, i.e., $z=\pm i 2$.  Thus, we have
$$-i 2 \sin{\frac{2 \pi}{3}} \int_{-1}^{1} dx \frac{(1-x)^{-2/3} (1+x)^{-1/3}}{x^2+4} = i 2 \pi \frac1{i 4} \left [(-1+i 2)^{-2/3} (1+i 2)^{-1/3} - (-1-i 2)^{-2/3} (1-i 2)^{-1/3} \right ] $$
Are we done?  No!  We cannot leave the above in this form because it is not clear how to compute it.  What are the arguments of each term?  Well, as I stated above, the choice of branch we used is significant: all arguments must be between $-\pi$ and $\pi$.  Under this constraint, we may write the RHS in polar form where it simplifies amazingly:
$$\frac{\pi}{2 \sqrt{5}} \left (e^{-i 2 \pi/3} e^{i \frac13 \arctan{2}} - e^{i 2 \pi/3} e^{-i \frac13 \arctan{2}} \right ) $$
Simplifying everything, we get our final result:

$$\int_{-1}^{1} dx \frac{(1-x)^{-2/3} (1+x)^{-1/3}}{x^2+4} =  \frac{\pi}{\sqrt{15}} \sin{\left ( \frac{2 \pi}{3} - \frac13 \arctan{2} \right ) }$$

Numerically, this is about $0.801488$, which agrees with Dan's answer and Mathematica.  Mathematica's analytical answer, however, is a mess of hypergeometrics, etc., so pen and paper still have the advantage.
