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

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.