Past exam of the integration course of the University of Oxford / 2017 / 3 / a

A sufficient theorem is dominated differentiation under the integral sign. Let $t_0$ be interior to a parameter interval, suppose $u(\cdot ,t)$ is measurable for each $t$, and suppose $u(\cdot ,t_0)$ is integrable. Assume $u(x,\cdot )$ is differentiable near $t_0$ for $x$ outside one fixed null set, and that throughout this neighborhoodfor one nonnegative Lebesgue integrable function $H$. Then $F(t)={\int }_{E}u(x,t)dx$ is defined near $t_0$, is differentiable at $t_0$, andIndeed the mean value theorem bounds both $|u(x,t)-u(x,t_0)|$ by $|t-t_0|H(x)$ and the difference quotient by $H(x)$. The first bound proves local integrability; the second lets the dominated convergence theorem pass the limit of difference quotients through the integral.

Fix $t\in \mathbb{R}$. The mean value theorem for the exponential givesConsequentlyThe function is continuous on $(0,1)$ and hence measurable, so $u(\cdot ,t)$ is Lebesgue integrable for every real $t$. In particular, the numerator cancels the apparent $1/x$ singularity at zero: $u(x,t)\to t$ as $x\downarrow 0$. Only the integrable square-root singularity at one remains. At $t=0$ the function is identically zero.

Fix any real $t_0$ and restrict $t$ to $|t-t_0|<1$. The parameter derivative isThis bound is integrable on $(0,1)$ and is uniform throughout the parameter neighborhood. Part (ii) supplied integrability at $t_0$, so dominated differentiation under the integral sign proves differentiability there. As $t_0$ was arbitrary,For $t>0$, $e^{tx}\ge 1$; for $t<0$, $e^{tx}\le 1$. Since ${\int }_0^1(1-x{)}^{-1/2}dx=2$, these comparisons giveThe first two inequalities are actually strict, because the exponential differs strictly from one throughout $(0,1)$. Differentiating once more under the same type of bound gives $F^{\prime \prime }(t)={\int }_0^{1}x{e}^{tx}/\sqrt{1-x}dx>0$. Hence $F$ is strictly convex and lies above its tangent $2t$ at zero. These properties are collected in exponential integral with a square-root endpoint singularity.