I have to resolve this exercise with complex integration:
Show that $$\int_0^\infty\frac {x^\frac 1 3 } {1+x^2}dx=\frac \pi {\sqrt{3}}$$
I integrated along the path consisting of the segment $[-R,R]\subset\mathbb R$ and the semicircle $\{Re^{i\theta}:0\le\theta\le \pi\}\subset\mathbb C $. Inside this region, that I'll call $A$, the function has only a pole, in $i$: then, by Cauchy's formula $$\oint _A\frac {z^\frac 1 3 } {(z+i)(z-i)}dz=2\pi i \cdot\frac {i^\frac 1 3} {2i} =\pi i^\frac 1 3 $$ At this point I'm concerned with two things: first, I don't know if there is some agreement in $\mathbb C$ to choose the cubic root of $i$ (like in the real numbers whith the square root for example). Then, it seems to me that the integral along the semicircle tends to zero if $R$ goes to infinity ($\int |f(z)dz|\to \pi R^{-\frac2 3}\to 0$) and that $-\int ^{\infty}_0f(x)dx=\int _{-\infty}^0f(x)dx$. However according to this the integral along $A$ should be zero, in contrast with Cauchy's formula. Where am I wrong? Thank you in advance