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Fix $x\in {\mathbb{R}}^n$ and a coordinate direction $e_i$. The fundamental theorem of calculus givesFor sufficiently small $h$, the points $(t,x+sh{e}_i)$ lie in a fixed compact set. The continuity of $D_{i}g$ makes it uniformly continuous there, so the right-hand side converges to $D_{i}g(t,x)$ uniformly in $t$. We may consequently pass the limit through the finite integral:

To prove continuity, if $x_k\to x$, joint continuity makes $D_{i}g(t,x_k)\to D_{i}g(t,x)$ uniformly for $t\in [0,1]$ once the $x_k$ lie in a compact neighbourhood of $x$. Hence $D_{i}G(x_k)\to D_{i}G(x)$. This proves the stated differentiation under the integral sign result and the continuity of every partial derivative.

For fixed $(x_1,x_2)$, apply the given one-variable identity to the smooth functionThe chain rule givesThereforewhereEvery derivative of the integrand is continuous. Repeated differentiation under the integral sign therefore shows that $h$ is smooth. This is the Second-order Hadamard lemma.

Differentiation under the integral sign first givesOn the other hand,Its integral is zero because $\sin \theta$ vanishes at both endpoints. Hence

Differentiating once more,For $x\ne 0$, the preceding identity and ${\sin }^2\theta +{\cos }^2\theta =1$ giveThe equation extends through $x=0$ in its regular limiting form.

Let $p_t(z)=(2\pi t{)}^{-1/2}{e}^{-z^2/(2t)}$ be the heat kernel for $u_t=\frac{1}{2}{u}_{xx}$. The symmetry of the Gaussian distribution gives the method-of-images formulaThis is the Neumann heat kernel on a half-line. Differentiation under the integral sign shows that $v\in C^{1,2}$ for $t,x>0$ and that $v_t=\frac{1}{2}{v}_{xx}$. At $x=0$, the two differentiated kernel terms cancel, so $v_x(t,0)=0$. The Gaussian approximate identity gives $v(t,x)\to f(x)$ as $t\downarrow 0$, while the dominated convergence theorem gives continuity up to $x=0$. Finally $|v(t,x)|\le \Vert f{\Vert }_{\infty }$, which is stronger than the required exponential bound. Thus $v$ satisfies every condition in the displayed boundary-value problem.

Because $x^{-2}$ is integrable on $[1,\infty )$, the Dominated convergence theorem applied to the defining integral proves that $v$ is continuous on $\mathbb{R}$.

For $\lambda >0$, the change of variables formula $t=\lambda x$ givesand for $\lambda <0$ the analogous formula is obtained with $e^{it}$. On either open half-line the lower endpoint stays away from zero locally, so repeated differentiation under the integral sign proves smoothness. Thus

A reparameterization gradient differentiates an expected value by representing the simulated random variable as a differentiable transformation of parameter-independent noise. For $X_{\eta }=g_{\eta }(\epsilon )$, the identity ${\nabla }_{\eta }\mathbb{E}[F_{\eta }(X_{\eta })]=\mathbb{E}[{\nabla }_{\eta }{F}_{\eta }(g_{\eta }(\epsilon ))]$ requires justified differentiation under the integral sign. For a normal distribution, use $X=m+e^{\rho }\epsilon$ with $\epsilon$ drawn from the standard normal distribution.

For real $x$, the integral representation isUsing differentiation under the integral sign twice and then integration by parts shows ${\mathrm{Hi}}^{\prime \prime }-x\mathrm{Hi}=1/\pi$. The cubic decay makes the integral converge for every real $x$.