= Solution
Let $\omega=e^{2\pi i/3}$ and integrate $1/(1+z^3)$ around the positively oriented sector from argument <zero> to $2\pi/3$, with outer radius $R>1$. The single pole inside is $z_0=e^{i\pi/3}$, with <residue> $1/(3z_0^2)=1/(3\omega)$. The outward real radial segment contributes $I_R=\int_0^R dx/(1+x^3)$; the return radial segment contributes $-\omega I_R$ because $(\omega x)^3=x^3$.
On the circular arc, $|1+z^3|\geq R^3-1$, so its integral has modulus at most $(2\pi R/3)/(R^3-1)\to0$. Thus the <residue theorem> yields
$$
(1-\omega)I=\frac{2\pi i}{3\omega}.
$$
Since $\omega(1-\omega)=\omega-\omega^2=i\sqrt3$,
$$
\boxed{\int_0^\infty\frac{dx}{1+x^3}=\frac{2\pi}{3\sqrt3}.}
$$
The ordinary improper integral converges at both endpoints, so no principal-value interpretation is needed here.
Back to article page
