 Solution

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.