Differentiation under an improper integral can fail without uniform domination

Given: For $x\ge 0$ and $t\in \mathbb{R}$, let $f(x,t):=t^3{e}^{-t^{2}x}$ and $F(t):={\int }_0^{\mathrm{\infty }}f(x,t)dx$.

On an open domain and open parameter interval, if $f$ and ${\mathrm{\partial }}_{t}f$ are continuous, one slice is absolutely improperly integrable, and $\mathrm{\mid }{\mathrm{\partial }}_{t}f\mathrm{\mid }$ has an integrable bound uniform on each compact parameter interval, then $F^{\mathrm{\prime }}(t)={\int }_D{\mathrm{\partial }}_{t}f(x,t)dx$ (Differentiation under an improper multiple integral under an integrable derivative bound).

A monotone differentiable substitution preserves a convergent improper integral under the stated compact-truncation hypotheses (Change of variable in an improper integral).

The derivative of the exponential is the exponential (The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$).

The improper integral ${\int }_0^{\mathrm{\infty }}h$ is the finite limit of ${\int }_0^{R}h$ as $R\to \mathrm{\infty }$, when that limit exists (Improper integrals over unbounded intervals).

If $G^{\mathrm{\prime }}=h$ on a compact interval and $h$ is integrable, then $\int h$ is the endpoint difference of $G$ (The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G^{\mathrm{\prime }}=f$ and $f$ is integrable, then ${\int }_a^{b}f=G(b)-G(a)$).

One has $\exp (x)\to 0$ as $x\to -\mathrm{\infty }$, and the exponential is normalized by $\exp (0)=1$ (The exponential tends to $+\mathrm{\infty }$ at $+\mathrm{\infty }$ and to $0$ at $-\mathrm{\infty }$, The power-series, product-limit, IVP, functional-equation, and Picard definitions agree).

The one-variable chain rule gives $(g\circ h{)}^{\mathrm{\prime }}=(g^{\mathrm{\prime }}\circ h)h^{\mathrm{\prime }}$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$).