<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] Here is an approach that makes use of Euler's Beta function. Enforcing a substitution of $x \mapsto \sqrt[3]{x}$ one has
\begin{align*}
\int_0^\infty \frac{dx}{x^3 + 1} &= \frac{1}{3} \int_0^\infty \frac{x^{-2/3}}{x + 1} \, dx\\
&= \frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3} - 1}}{(x + 1)^{\frac{1}{3} + \frac{2}{3}}}\\
&= \frac{1}{3} \text{B} \left (\frac{1}{3}, \frac{2}{3} \right ) \tag1\\
&= \frac{1}{3} \frac{\Gamma \left (\frac{1}{3} \right ) \Gamma \left (\frac{2}{3} \right )}{\Gamma (1)} \tag2\\
&= \frac{1}{3} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\
&= \frac{1}{3} \cdot \frac{\pi}{\sin (\pi/3)} \tag3\\
&= \frac{1}{3} \cdot \frac{2\pi}{\sqrt{3}}\\
&= \frac{2\pi}{3 \sqrt{3}}.
\end{align*}
Explanation
(1) Since $\displaystyle{\text{B}(x,y) = \int_0^\infty \frac{t^{x - 1}}{(1 + t)^{x + y}} \, dt}$.
(2) Since $\text{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$. 
(3) From Euler's reflection formula.
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