If $h(x,t)=\int_{\mathbb R}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy$ why can I permute $\partial _t$, $\partial _x$ and integral? Let $$h(x,t)=\int_{\mathbb R}\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy.$$ 
Why $$\partial _th(x,t)=\int_{\mathbb R}\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy$$
and $$\partial _xh(x,t)=\int_{\mathbb R}\partial _x\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}dy\ \ ?$$
I know that if for a.e. $y$, $x\mapsto p(x,y)$ is differentiable and if there is $k\in L^1$ s.t. $|\partial _xp(x,y)|\leq k(y),$
then $$\partial _x\int_{\mathbb R}p(x,y)dy=\int_{\mathbb R}\partial _x p(x,y)dy.$$ 
Unfortunately, if $p(x,y,t)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}$, I can't bound uniformly in $t$ the function $|\partial _t p(x,y,t)|$ neither bound $|\partial _x p(x,y,t)|$ uniformly in $x$. Any idea ? 

In the book PDE of Evans they say : since $\Phi:(x,t)\mapsto \frac{1}{\sqrt{t2\pi}}e^{-\frac{x^2}{2t}}$ is infinitely differentiable on $\mathbb R\times [\delta ,\infty )$ for all $\delta >0$, we have that $$h_t=\int_{\mathbb R}\Phi_t\quad \text{and}\quad h_{xx}=\int_{\mathbb R}\Phi_{xx},$$
but I don't really understand the argument. 
 A: For $0 < \delta_1 \leqslant t  \leqslant \delta_2$ we have
$$\left|\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\right| = \left|\frac{-1}{2\sqrt{2\pi}t^{3/2}}e^{-\frac{(x-y)^2}{2t}} + \frac{(x-y)^2}{2\sqrt{2\pi}t^{5/2}}e^{-\frac{(x-y)^2}{2t}}\right| \\\leqslant \frac{1}{2\sqrt{2\pi}\delta_1^{3/2}}e^{-\frac{(x-y)^2}{2\delta_2}} + \frac{(x-y)^2}{2\sqrt{2\pi}\delta_1^{5/2}}e^{-\frac{(x-y)^2}{2\delta_2}}$$
The RHS is integrable with respect to $y$ over $\mathbb{R}$ and by the Weierstrass M-test we have uniform convergence for $t \in [\delta_1, \delta_2]$ of
$$\int_{\mathbb R}\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\,dy$$
Clearly, the uniform convergence holds for $t$ in any compact interval contained in $(0,\infty)$ and that is enough to prove that for all $t >0 $,
$$\tag{*}\partial _th(x,t)=\int_{\mathbb R}\partial _t\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}} \, dy$$
The switching of the partial $x$-derivative and the integral can be justified in a similar way.
We also have
$$\left|\partial _x\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\right| = \left|\frac{-2(x-y)}{\sqrt{2\pi}t^{3/2}}e^{-\frac{(x-y)^2}{2t}} \right| = \frac{2|x-y|}{\sqrt{2\pi}t^{3/2}}e^{-\frac{|x-y|^2}{2t}}$$
Note that for $\delta_1 \leqslant |x| \leqslant \delta_2$ we have $|x-y| \leqslant |x| + |y| \leqslant \delta_2 +|y|$ and $-|x- y|^2 \leqslant -||x| - |y||^2 = -|y|^2 + 2|x||y| - |x|^2$, and, thus, 
$$\exp\left(-\frac{|x-y|^2}{2t}\right) \leqslant \exp\left(-\frac{|y|^2}{2t}\right)\exp\left(\frac{|x||y|}{t}\right)\exp\left(-\frac{|x|^2}{2t}\right) \\ \leqslant \exp\left(-\frac{|y|^2}{2t}\right)\exp\left(\frac{\delta_2|y|^2}{t}\right)\exp\left(-\frac{\delta_1^2}{2t}\right)$$ 
Again, we can find an integrable upper bound independent of $x$ is a compact interval and can apply the Weierestrass M-test to obtain uniform convergence of the integral of the partial $x$-derivative.  This justifies the integral-derivative switch for all $x \in \mathbb{R}$. 
A: Fix $\delta>0$. Observe
\begin{align}
\left|\frac{\partial}{\partial x}\left(\frac{1}{\sqrt{2\pi t}}\exp\left( -\frac{(x-y)^2}{2t}\right) \right)\right|  = \frac{1}{\sqrt{2\pi t}}\exp\left( -\frac{(x-y)^2}{2t}\right)\frac{|x-y|}{t}.
\end{align}
Since
\begin{align}
1+\frac{(x-y)^2}{2t}+\frac{(x-y)^4}{8t^2}\leq \exp\left( \frac{(x-y)^2}{2t}\right)
\end{align}
then it follows
\begin{align}
\exp\left( -\frac{(x-y)^2}{2t}\right) \leq& \frac{8t^2}{8t^2+4t(x-y)^2+(x-y)^4}\\
 \leq&\ \frac{8t^2}{t^2+2t(x-y)^2+(x-y)^4} = \frac{8t^2}{(t+(x-y)^2)^2}.
\end{align}
Hence it follows
\begin{align}
 \frac{1}{\sqrt{2\pi t}}\exp\left( -\frac{(x-y)^2}{2t}\right)\frac{|x-y|}{t} \leq&\ \frac{8}{\sqrt{2\pi}}\frac{|x-y||t|^{\frac{1}{2}}}{(t+(x-y)^2)^2} \leq \frac{8}{\sqrt{2\pi}} \frac{1}{t+(x-y)^2}\\
 \leq&\  \frac{C}{\delta+(x-y)^2} =K_\delta(x, y)
\end{align}
for all $x, y \in \mathbb{R}$ and $t \in [\delta, \infty)$. In particular, we see that
\begin{align}
\int^\infty_{-\infty} K_\delta(x, y)\ dy = \int^\infty_{-\infty}\frac{dy}{\delta+y^2} = \frac{C}{\sqrt{\delta}}<\infty. 
\end{align}
Similar type of estimates holds for $\partial_t\Phi$.  
A: If you are allowed to use Measure Theory, the most efficient tool to use, in cases you want to show that integration and differentiation commute, is Lebesgue Dominated Convergence Theorem. In could be applied in the following version:
If $f=f(x,t)$, where $(x,t)\in\mathbb R\times (0,\infty)$  is differentiable in $t$ and $\int_{\mathbb R}|f(x,t)|\,dx<\infty$ for all $t$, and there exists a $g=g(x,t)$, such that
$$
\Big|\frac{f(x,t+h)-f(x,t)}{h}\Big|\le g(x,t)
$$
and $\int_{\mathbb R} g(x,t)\,dx\le G(t)<\infty$, then $\int_{\mathbb R}|f_t(x,t)|\,dx<\infty$ and
$$
\frac{d}{dt}\int_{\mathbb R}f(x,t)\,dx=\int_{\mathbb R}f_t(x,t)\,dx.
$$ 
