Exponential integral with a square-root endpoint singularity ($F(t)={\int }_0^1\frac{e^{tx}-1}{x\sqrt{1-x}}dx$)

For every real $t$, the mean value theorem bounds $|e^{tx}-1|\le |t|x{e}^{|t|}$ on $0<x<1$. Hence the apparent singularity at zero cancels, and the remaining bound $|t|e^{|t|}/\sqrt{1-x}$ is Lebesgue integrable. On a bounded parameter interval $|t|\le M$, every positive-order parameter derivative of the integrand is bounded in absolute value by $e^M/\sqrt{1-x}$. Repeated dominated differentiation under the integral sign therefore proves that $F$ is a smooth function, withIn particular, $F(0)=0$, $F^{\prime }(0)=2$, and $F^{\prime \prime }(t)>0$. Thus $F$ is strictly convex, lies above its tangent $2t$ at zero, and satisfies $F^{\prime }(t)>2$ for $t>0$ and $F^{\prime }(t)<2$ for $t<0$. The inequalities follow directly as well by comparing $e^{tx}$ to one. The substitution $x=1-s^2$ removes the square-root endpoint singularity for numerical integration: