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!
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 $t\,.$ $$\tag*{$\Box$} \quad $$