\[ I\da \int_0^\infty {x^{1\over 3} \over 1 + x^2} \dx = {\pi \over \sqrt 3}. \]
Solution
Let \[ f(z)=\frac{z^{1/3}}{1+z^2}, \] using the branch \(0<\Arg z<\pi\) on the upper half-plane. Integrate over the upper semicircle of radius \(R\), indented by a semicircle of radius \(\varepsilon\) around the branch point \(0\).
On the large arc, \[ \abs{\int_{C_R}f(z)\,dz} \le \pi R\frac{R^{1/3}}{R^2-1} =O(R^{-2/3})\to0, \] and similarly the small arc is \(O(\varepsilon^{4/3})\to0\).
Along the positive real axis, \(z^{1/3}=x^{1/3}\). Along the negative real axis approached from above, \((-x)^{1/3}=e^{i\pi/3}x^{1/3}\) for \(x>0\). Taking account of the orientation, the two straight pieces therefore tend to \[ \qty{1+e^{i\pi/3}}I. \]
The only enclosed pole is \(z=i\), and on the chosen branch \[ \Res_{z=i}f(z) =\frac{i^{1/3}}{2i} =\frac{e^{i\pi/6}}{2i}. \] Hence the residue theorem gives \[ \qty{1+e^{i\pi/3}}I =2\pi i\frac{e^{i\pi/6}}{2i} =\pi e^{i\pi/6}. \] Since \[ 1+e^{i\pi/3}=2e^{i\pi/6}\cos\qty{\pi\over6} =\sqrt3\,e^{i\pi/6}, \] we obtain \[ I={\pi\over\sqrt3}. \]