Problem 3. Compute: \(\int _ { 0 } ^ { \infty } { \frac { d x } { 1 + x ^ { 3 } } } .\)
Solution
The value is \[ \boxed{\frac{2\pi}{3\sqrt3}}. \]
1. A sector of angle \(2\pi/3\) relates the two radial integrals.
Let \[ F(z)=\frac1{1+z^3} \] and use the positively oriented boundary of the sector \[ 0\le\arg z\le\frac{2\pi}{3},\qquad |z|\le R, \] with \(R>2\). Along the positive real ray the contribution is \[ \int_0^R\frac{dx}{1+x^3}. \] Along the upper ray, traversed from \(Re^{2\pi i/3}\) back to zero, put \(z=e^{2\pi i/3}x\). Since \(z^3=x^3\) and \(dz=e^{2\pi i/3}dx\), that contribution is \[ -e^{2\pi i/3}\int_0^R\frac{dx}{1+x^3}. \] Thus the two radial sides contribute \[ \left(1-e^{2\pi i/3}\right)I_R, \qquad I_R=\int_0^R\frac{dx}{1+x^3}. \]
2. The arc vanishes and exactly one pole lies in the sector.
On the circular arc \(|z|=R\), \[ |F(z)|\le\frac1{R^3-1}, \] so its integral has modulus at most \[ \frac{(2\pi/3)R}{R^3-1}\longrightarrow0. \] The poles are the roots of \(z^3=-1\), namely \(e^{i\pi/3}\), \(-1\), and \(e^{5i\pi/3}\). Only \[ \zeta=e^{i\pi/3} \] lies in the interior of the sector. It is simple, with residue \[ \operatorname{Res}_{z=\zeta}F =\frac1{3\zeta^2} =\frac{e^{-2\pi i/3}}3. \] Hence the residue theorem and the limit \(R\to\infty\) give \[ \left(1-e^{2\pi i/3}\right)I =\frac{2\pi i}{3}e^{-2\pi i/3}, \] where \(I=\int_0^\infty(1+x^3)^{-1}dx\). The improper integral converges because the integrand is bounded near zero and is \(O(x^{-3})\) at infinity.
3. Simplifying the complex quotient gives the real value.
Let \(\omega=e^{2\pi i/3}=-1/2+i\sqrt3/2\). Since \(e^{-2\pi i/3}=\overline\omega=-1/2-i\sqrt3/2\), \[ \frac{i\overline\omega}{1-\omega} =\frac1{\sqrt3}. \] Therefore step 2 yields \[ I=\frac{2\pi}{3}\frac1{\sqrt3} =\frac{2\pi}{3\sqrt3}. \]