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I have a relatively simple question but I don't know how to do it. Let $p(t,x) := \frac{1}{\sqrt t} e^{-\frac{x^2}{2t}}$, $t>0, x\in \mathbb R$. I would like to justify

$$ \frac{\partial}{\partial t}\int_\mathbb{R} p(t,x)dx = \int_\mathbb{R} \frac{\partial}{\partial t} p(t,x)dx.$$

using the standard differentiation theorem. However, I cannot find an integrable function $g(x)$ such that

$$|\frac{\partial}{\partial t}p(t,x)| = \Big|-\left(\frac{x^2}{2t^2\sqrt t} +\frac{1}{2t\sqrt t} \right) e^{-\frac{x^2}{2t}}\Big| \leq g(x), \quad \forall t>0 \quad \text{and} \quad a.e \quad x. $$

Any help for this case is highly appreciated. If possible, I would like a general statement for $p(t,x) = e^{\frac{-|x|^2}{2t}}$, $t>0$, $x \in \mathbb R^n$ as well. Thank you very much!

Jeffrey Jao
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    This question needs some motivation. $\int p(t,x),dx = \sqrt{2\pi}$ for all $t>0$. – Andrew Jun 22 '23 at 02:26
  • The conditions for swapping derivative and integral is a common question: for example check this wikipedia link: https://en.wikipedia.org/wiki/Leibniz_integral_rule

    Seems like a similar question was also asked before: https://math.stackexchange.com/questions/2530213/when-can-we-interchange-integration-and-differentiation

     – Alan Chung Jun 22 '23 at 03:17
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    You do NOT need a uniform bound over all $t>0$. Differentiability is a local property, so all you need to do is for each $t_0>0$, find some interval $I$ around $t_0$ such that the supremum over all $t\in I$ is an integrable function of $x$. I’ve commented on a similar issue Interchanging spatial Fourier transform and time derivative for heat kernel. – peek-a-boo Jun 22 '23 at 08:43
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    @AlanChung the link doesn’t answer OP’s question. They already seem to know the general measure-theoretic theorem for differentiation under the integral sign. It’s just a matter of applying the theorem correctly, by recognizing the local nature of differentiability. – peek-a-boo Jun 22 '23 at 08:45
  • @peek-a-boo That's quite exactly what I want, thanks! :) – Jeffrey Jao Jun 22 '23 at 14:44

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It is well-known that $$\tag{1} \int_{\mathbb R}p(t,x)\,dx=\sqrt{2\pi} $$ for all $t>0\,.$ The time derivative is therefore zero. The fact $$\tag{2}\boxed{\quad\phantom{\Big|} \int_{\mathbb R}\partial_t\,p(t,x)\,dx=0\,.\quad} $$ can be shown directly without seeking an integrable function $g(x)$ that majorizes $|\partial_t\,p(t,x)|\,:$

In slight contrast to you I get $$\tag{3} \partial_t\,p(t,x)=\frac{p(t,x)(x^2-t)}{2t^2\sqrt{t}}\,. $$ Clearly, \begin{align} \int_{\mathbb R}\partial_t\,p(t,x)\,dx&=\frac{1}{2t^2\sqrt{t}}\underbrace{\int_{\mathbb R}x^2p(t,x)\,dx}_{\sqrt{2\pi} t}- \frac{\sqrt{2\pi} t}{2t^2\sqrt{t}}=0\,.\tag{4} \end{align} because the variance of a normal random variable with density $p(t,x)/\sqrt{2\pi}$ is $t\,.$ $$\tag*{$\Box$} \quad $$

Kurt G.
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