- (20 points) Evaluate the integral:
\[ \int _ { 0 } ^ { \infty } { \frac { u ^ { 2 } d u } { u ^ { 6 } + 1 } } \]
Solution
The integral equals \[ \boxed{\frac{\pi}{6}}. \]
1. The upper-half-plane residues sum to \(-i/6\).
Let \[ F(z)=\frac{z^2}{z^6+1}. \] The poles are the six simple roots of \(z^6=-1\). In the upper half-plane these are \[ \zeta_1=e^{i\pi/6},\qquad \zeta_2=e^{i\pi/2}=i, \qquad \zeta_3=e^{5i\pi/6}. \] Since \((z^6+1)'=6z^5\), the residue at any such root is \[ \operatorname{Res}_{z=\zeta}F =\frac{\zeta^2}{6\zeta^5} =\frac1{6\zeta^3}. \] Now \[ \zeta_1^3=i,\qquad \zeta_2^3=-i,\qquad \zeta_3^3=i, \] so \[ \sum_{j=1}^3\operatorname{Res}_{\zeta_j}F =-\frac{i}{6}+\frac{i}{6}-\frac{i}{6} =-\frac{i}{6}. \]
2. The upper semicircle gives the real-line integral.
Integrate \(F\) over the contour formed by \([-R,R]\) and the counterclockwise upper semicircle \(\Gamma_R\), with \(R>1\). On \(|z|=R\), \[ |F(z)|\le \frac{R^2}{R^6-1}, \] so \[ \left|\int_{\Gamma_R}F(z)\,dz\right| \le \frac{\pi R^3}{R^6-1}\longrightarrow0. \] The residue theorem therefore yields, after letting \(R\to\infty\), \[ \int_{-\infty}^{\infty}\frac{x^2}{x^6+1}\,dx =2\pi i\left(-\frac{i}{6}\right)=\frac{\pi}{3}. \] The integral converges absolutely because the integrand is bounded near zero and is \(O(x^{-4})\) at infinity.
3. Evenness gives the requested half-line value.
The real integrand is even. Hence \[ 2\int_0^\infty\frac{u^2}{u^6+1}\,du =\int_{-\infty}^{\infty}\frac{x^2}{x^6+1}\,dx =\frac{\pi}{3}, \] so the desired integral is \(\pi/6\).