I want to calculate the integral:
$$I \equiv \int_0^{\infty} \frac{dx}{1+x^3}$$
using residue calculus. I'm having trouble coming up with a suitable contour. I tried to take a contour in the shape of a quarter of a circle of radius $R$, then take the limit. The circular arc tends to zero, but the vertical fragment is problematic. I'm getting:
$$\int_0^{\infty} \frac{dx}{1+x^3} + \int_0^{\infty} \frac{idy}{1-iy^3} = w$$
where $w$ is the residue from the singularity at $e^{i \frac{\pi}{3}}$. If the second integral was purely imaginary, then it would be no problem, but it has a real part:
$$\int_0^{\infty} \frac{dx}{1+x^3} – \int_0^{\infty} \frac{x^3dx}{1+x^6} + i\int_0^{\infty} \frac{dx}{1+x^6} =w$$
So to get the answer I would have to know the value of the second integral, which doesn't look any easier than the first, and indeed I am unable to relate the two. Perhaps this approach is doomed, and there is a simpler way?
Best Answer
Since $(e^{2\pi i/3}z)^3 = z^3$, a suitable contour is a third of a circle, with the rays on the positive real axis and $[0,\infty)\cdot e^{2\pi i/3}$. That gives you
$$2\pi i \operatorname{Res}\left(\frac{1}{1+z^3}; e^{\pi i/3}\right) = \int_0^\infty \frac{dx}{1+x^3} - e^{2\pi i/3}\int_0^\infty \frac{dx}{1+(e^{2\pi i/3}x)^3}.$$
From then on, it's a simple manipulation.