= Exponential integral with a square-root endpoint singularity
{title2=$F(t)=\int_0^1\frac{e^{tx}-1}{x\sqrt{1-x}}\,dx$}
For every real $t$, the <mean value theorem> bounds $|e^{tx}-1|\leq |t|xe^{|t|}$ 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 $|t|\leq M$, 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>, with
$$
F^{(k)}(t)=\int_0^1\frac{x^{k-1}e^{tx}}{\sqrt{1-x}}\,dx\quad(k\geq1).
$$
In particular, $F(0)=0$, $F'(0)=2$, and $F''(t)>0$. Thus $F$ is <strictly convex>, lies above its tangent $2t$ at zero, and satisfies $F'(t)>2$ for $t>0$ and $F'(t)<2$ 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:
$$
F(t)=2\int_0^1\frac{e^{t(1-s^2)}-1}{1-s^2}\,ds,\qquad
F'(t)=2\int_0^1e^{t(1-s^2)}\,ds.
$$
\Image[/past-exam-of-the-integration-course-of-the-university-of-oxford/2017-exponential-endpoint-integral.png]
{title=Convexity and derivative bounds for the exponential endpoint integral}
{description=The derivative is below two for negative parameters and above two for positive parameters. The integral lies above its tangent at zero. Substitution removes the endpoint singularity before numerical quadrature.}
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