Past exam of the integration course of the University of Oxford / 2014 / 2 / b / ii

We prove the required differentiation under the integral sign directly. Fix $t_0>0$, and take a compact parameter interval $J\subseteq [0,\infty )$ containing $t_0$ in its interior. The jointly continuous partial derivative ${\partial }_{t}f$ is bounded on the compact rectangle $J\times [0,1]$ by some $M_1$.

For nonzero $h$ sufficiently close to zero, defineFor each $x$, the mean value theorem in the parameter gives $|D_h(x)|\le M_1$. The definition of the partial derivative also gives $D_h(x)\to {\partial }_{t}f(t_0,x)$. Since $M_1$ is integrable on $[0,1]$, dominated convergence applied along any sequence $h_n\to 0$ yieldsThusAt $t=0$, repeat the argument on $J=[0,\delta ]$ with $h>0$. It proves that the right derivative exists and satisfies the same formula at zero. This is the appropriate meaning of differentiability on the closed half-line. In fact $F^{\prime }$ is continuous there: apply part (i) to the continuous function ${\partial }_{t}f$.

The proof establishes differentiation of parameter integrals on compact rectangles under the exact regularity assumptions used here. In particular, it supplies the theorem's proof instead of assuming an interchange of a limit and an integral.