Past exam of the options II course of the University of Oxford / 2010 / D / D4 / 2

A naive absolute bound on the original derivative would be $1/x$, which is not integrable on $[1,\infty )$. Instead make the change of variable $s=tx$, valid for $t>0$:The integral is absolutely convergent for every positive lower endpoint. Its integrand is continuous there, so the fundamental theorem of calculus givesThe product rule now yieldsThis proves differentiability and the equation without an unjustified differentiation of a conditionally convergent derivative integral.