Compute the following integral carefully justifying each step:
\[ \int_{0}^{\infty} \frac{\log x}{1+x^{3}} \text {. } \]
Solution
For \(0<s<3\), set \[ I(s)=\int_0^\infty\frac{x^{s-1}}{1+x^3}\,dx. \] With \(u=x^3\), so that \(x=u^{1/3}\) and \(dx=\frac13u^{-2/3}\,du\), we obtain \[ I(s)=\frac13\int_0^\infty\frac{u^{s/3-1}}{1+u}\,du. \] The Euler beta integral gives, for \(0<s<3\), \[ I(s)=\frac\pi3\csc\left(\frac{\pi s}{3}\right). \]
On every compact subinterval of \((0,3)\), differentiation under the integral sign is justified by domination near \(0\) and \(\infty\). Therefore \[ I'(s)=\int_0^\infty\frac{x^{s-1}\log x}{1+x^3}\,dx. \] At \(s=1\) this is the desired integral. Differentiating the closed formula, \[ I'(s) =-\frac{\pi^2}{9} \csc\left(\frac{\pi s}{3}\right) \cot\left(\frac{\pi s}{3}\right). \] Hence \[ \int_0^\infty\frac{\log x}{1+x^3}\,dx =-\frac{\pi^2}{9} \cdot\frac{2}{\sqrt3}\cdot\frac1{\sqrt3} =\boxed{-\frac{2\pi^2}{27}}. \]