Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / D / D4 / 2 / Solution

A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.