Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / D / D3 / 2 / Solution

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. All terms after multiplication by $(\log x{)}^2$ are nonnegative, so monotone convergence permits integration term by term. Substituting $x=e^{-t}$ and integrating by parts twice givesConsequently the integral is ${\sum }_{k\ge 1}2/k^2$, and the supplied sum gives $\boxed{{\pi }^2/3}$.