integration - How to show $int_{0}^{infty} frac{dx}{x^3+1} = frac{2pi}{3sqrt{3}}$

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!

Answer

There are already two answers showing how to find the integral using just calculus. It can also be done by the Residue Theorem:

It sounds like you're trying to apply RT to the closed curve defined by a straight line from $0$ to $A$ followed by a circular arc from $A$ back to $0$. That's not going to work, because there's no reason the intergal over the semicircle should tend to $0$ as $A\to\infty$.

How would you use RT to find $\int_0^\infty dt/(1+t^2)$? You'd start by noting that $$\int_0^\infty\frac{dt}{1+t^2}=\frac12\int_{-\infty}^\infty\frac{dt}{1+t^2},$$and apply RT to the second integral.

You can't do exactly that here, because the function $1/(1+t^3)$ is not even. But there's an analogous trick available.

Hint: Let $$f(z)=\frac1{1+z^3}.$$If $\omega=e^{2\pi i/3}$ then $$f(\omega z)=f(z).$$(Now you're going to apply RT to the boundary of a certain sector of opening $2\pi/3$... be careful about the "$dz"$...)