Past exam of the options I course of the University of Oxford / 2005 / D / 2

Since $\log x\le 0$ on $(0,1]$, substitute $x=e^{-s}$ to obtainThe functions are measurable, so this proves Lebesgue integrability. Also $0<e^{-x}\le 1$ gives $|h_t(x)|\le |g_t(x)|$. Hence $h_t$ is integrable on $(0,1]$ as well.

A useful differentiation theorem says: if $f(t,x)$ is integrable in $x$, differentiable in $t$ for almost every $x$, and on a neighborhood of each parameter value the derivative is bounded in absolute value by one integrable function of $x$, then the integral is differentiable and its derivative is the integral of ${\partial }_{t}f$. The proof applies dominated convergence to difference quotients, bounded by that same majorant by the mean value theorem.

Fix $t_0>0$ and choose $0<a<t_0<b$. For $t\in [a,b]$,The first part is integrable by part (i). The second is integrable by the permitted hypothesis for $h_b$; alternatively exponential decay dominates its polynomial and logarithmic factors. The theorem therefore applies locally at every $t_0$, giving