Dominated differentiation under the integral sign ($F^{\prime }(t)={\int }_X{\partial }_{t}h(t,x)d\mu (x)$)

Let $t_0$ be interior to a parameter interval. Suppose $h(t,\cdot )$ is a measurable function for each $t$, $h(t_0,\cdot )$ is a Lebesgue integrable function, and $h(\cdot ,x)$ is differentiable near $t_0$ for $x$ almost everywhere. If one integrable nonnegative function $D$ satisfies $|{\partial }_{t}h(t,x)|\le D(x)$ throughout that neighbourhood off one fixed null set, then $F(t)={\int }_{X}h(t,x)d\mu (x)$ is defined there and satisfies the displayed derivative formula at $t_0$.

The mean value theorem gives $|h(t,x)-h(t_0,x)|\le |t-t_0|D(x)$, proving local integrability. It also bounds the difference quotient by $D$. That quotient converges almost everywhere to ${\partial }_{t}h(t_0,x)$, so the dominated convergence theorem permits passing the limit through the Lebesgue integral.

Suppose $f$ and its partial derivative ${\partial }_{t}f$ are jointly continuous on a compact rectangle $J\times [a,b]$. Then $F(t)={\int }_a^{b}f(t,x)dx$ is continuously differentiable in the interior of $J$, and the displayed formula holds, with one-sided derivatives at parameter endpoints where the corresponding partial derivative is given.

To prove it, the extreme value theorem bounds $|{\partial }_{t}f|$ by $M_1$. For each fixed $x$, the mean value theorem bounds the difference quotient $[f(t+h,x)-f(t,x)]/h$ by $M_1$. It converges pointwise to ${\partial }_{t}f(t,x)$, and the constant $M_1$ is integrable over $[a,b]$. Dominated convergence gives the derivative formula. Apply continuity of parameter integrals on compact rectangles to ${\partial }_{t}f$ to obtain continuity of $F^{\prime }$. The proof works locally on unbounded parameter intervals as well.

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: