<s>[INST] How to show $\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}$ I am trying to show $\displaystyle{\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}}.$
Any help?
(I am having troubles using the half circle infinite contour)
Or more specifically, what is the residue $\text{res} \left(\frac{1}{z^3+1},z_0=e^\frac{\pi i}{3} \right )$ 
Thanks!
 [/INST] $$\int_{0}^{+\infty}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{dx}{1+x^3}+\int_{1}^{+\infty}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{1+x}{1+x^3}\,dx$$
clearly equals $\int_{0}^{1}\frac{dx}{1-x+x^2}$, which is an elementary integral. We may also state
$$\int_{0}^{1}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{1+x-x^3-x^4}{1-x^6}\,dx = \sum_{n\geq 0}\left[\frac{1}{6n+1}+\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+5}\right]$$
and by denoting as $\chi_3$ and $\chi_6$ the non-principal Dirichlet's characters $\!\!\pmod{3}$ and $\!\!\pmod{6}$ the RHS can be written as $L(\chi_6,1)+\frac{1}{2}\,L(\chi_3,1)$, which is related to $\sqrt{3}$ and $\pi$ via the class number formula.
Another approach is given by the reflection formula for the $\psi$ (digamma) function, since
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(x)-\psi(1-x)=-\pi\cot(\pi x)$$
ensure
$$ \sum_{n\geq 0}\left[\frac{1}{6n+1}-\frac{1}{6n+5}\right]=\tfrac{\pi}{6}\cot\left(\tfrac{\pi}{6}\right),\qquad \sum_{n\geq 0}\left[\frac{1}{3n+1}-\frac{1}{3n+2}\right]=\tfrac{\pi}{3}\cot\left(\tfrac{\pi}{3}\right). $$
</s>