Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D / 3 / c

Substitute $u=\log x$, giving $dx=e^{u}du$. The nonnegative integral becomesFor $\alpha >1$ an integration by parts gives $1/(\alpha -1{)}^2$; its boundary term vanishes because exponential decay dominates $u$. Hence it is finite, proving integrability and

For $x>1$ the geometric expansion givesEvery summand is nonnegative, so monotone convergence permits integration term by term without a prior integrability assumption. The preceding part applies with $\alpha =p+k+1>1$, yieldingThe series is finite for every fixed $p>0$, by comparison of its tail with $\sum k^{-2}$. This also proves integrability of the original function.