Interchanging spatial Fourier transform and time derivative for heat kernel Let $K_t := (4\pi t)^{-n / 2}e^{|x|^2 / 4t}$ for $x \in \mathbb{R}^n$ and $t \in (0, \infty)$. I would like to show that
$$
\tag{1}
\partial_t \widehat{K_t} = \widehat{\partial_t K_t},
$$
(which makes it easy to show that $K_t$ is a solution to the heat equation). This essentially involves differentiating inside the integral sign; i.e., showing that
$$
\partial_t\int_{\mathbb{R}^n} K_t(x)e^{-2\pi i \xi \cdot x}\,dx
= \int_{\mathbb{R}^n} \partial_tK_t(x)e^{-2\pi i \xi \cdot x}\,dx.
$$
How can I justify this computation? (This post provides a nice counterexample for why this operation is not true in general, even for Schwartz functions.)
 A: Let's just calculate the partial derivative using the product rule:
\begin{align}
\frac{\partial}{\partial t}\left(K_t(x)e^{-2\pi i x\cdot \xi}\right)&= e^{-2\pi i x\cdot \xi}\left[(4\pi t)^{-n/2}\left[e^{-|x|^2/4t}\cdot \left(\frac{-|x|^2}{4}\right)\cdot \left(-\frac{1}{t^2}\right)\right] +
\left[-\frac{n}{2}(4\pi t)^{-\frac{n}{2}-1}\cdot 4\pi\right]e^{-|x|^2/4t}\right]
\end{align}
Therefore, by ignoring some constant factors (and recalling that the complex exponential has magnitude 1), we have the following type of estimate:
\begin{align}
\left|\frac{\partial}{\partial t}\left(K_t(x)e^{-2\pi i x\cdot \xi}\right)\right| &\leq C\left[t^{-\frac{n}{2}-2}|x|^2e^{-|x|^2/4t} + t^{-\frac{n}{2}-1}e^{-|x|^2/4t}\right]\\
&\leq 2C\cdot \left(t^{-\frac{n}{2}-2}+t^{-\frac{n}{2}-1}\right)\left(|x|^2+1\right)e^{-|x|^2/4t},
\end{align}
for some constant $C>0$, which depends only on things like $4,\pi,n$ etc.
Now, let us fix a $\delta>0$. Then, by the previous remarks,
\begin{align}
\sup_{t\in (\delta, \infty), \xi\in\Bbb{R}^n}
\left|\frac{\partial}{\partial t}\left(K_t(x)e^{-2\pi i x\cdot \xi}\right)\right|  &\leq 2C\cdot \left(\delta^{-\frac{n}{2}-2}+ \delta^{-\frac{n}{2}-1}\right)\left(|x|^2+1\right)e^{-|x|^2/4\delta}.
\end{align}
If we now call the RHS $f(x)$, then we see that due to the exponential damping, $f\in L^1(\Bbb{R}^n)$. Thus, the dominating condition for Leibniz's integral rule is satisfied, so for any $t\in (\delta,\infty)$ and $\xi\in\Bbb{R}^n$, we can indeed swap the derivatives:
\begin{align}
\frac{\partial}{\partial t}\bigg|_{(t,\xi)}\int_{\Bbb{R}^n}K_t(x)e^{-2\pi i x\cdot \xi}\,dx&=\int_{\Bbb{R}^n}\frac{\partial}{\partial t}\bigg|_{(t,\xi)}K_t(x)e^{-2\pi i x\cdot \xi}\,dx
\end{align}
Actually, I didn't need the uniform estimate in the $\xi$ variable, but here it just came out for free so why not? Finally, since $\delta>0$ was arbitrary, this completes the proof of the differentiability, and the validity of the equation on all of $(0,\infty)$.

The reason you may have been worried is because there is no integrable upper bound if we have taken the supremum over all $t\in (0,\infty)$; and this is exactly right. However, we do not need such a strong estimate. Differentiabiliy is a local property, so if you're trying to prove differentiability at a point $t_0\in (0,\infty)$, you only need to find some interval $I\subset (0,\infty)$ which contains $t_0$, such that by taking the supremum over $t\in I$, we have an integrable upper bound. In this case, we were fortunate that we could just take the interval to be $I=(\delta, \infty)$ for any fixed $\delta>0$.
A: The trick is to turn everything into integration because Fubini's Theorem for interchanging orders of integration is far more powerful than general techniques for interchanging integration and differentiation. For example, let $\epsilon > 0$ and consider
$$
       \int_{\mathbb{R}^n}K(x,t)e^{-2\pi i(\xi\cdot x)t}dx =\\
    =\int_{\mathbb{R}^n}\int_\epsilon^t\frac{\partial}{\partial u}(K(x,u)e^{-2\pi i(\xi\cdot x)u})dudx+\int_{\mathbb{R}^n}K(x,\epsilon)e^{-2\pi i(\xi\cdot x)\epsilon}dx
$$
The second term on the right does not depend on $t$ and, so, may be discarded for the discussion at hand because ultimately we want the derivative with respect to $t$ of the above. The first term on the right may be re-written by interchanging orders of integration in $u$ and $x$ to obtain
$$
   \int_{\mathbb{R}^n}\int_{\epsilon}^{t}K_{u}(x,u)e^{-2\pi i(\xi\cdot x)u}du dx 
  =\int_{\epsilon}^{t}\int_{\mathbb{R}^n}K_u(x,u)e^{-2\pi i(\xi\cdot x)u}dx du \tag{*}
$$
Now it may be argued using dominated convergence that the following function of $u$ is continuous in $u$:
$$
         u\mapsto \int_{\mathbb{R}^n}K_u(x,u)e^{-2\pi i(\xi\cdot x)u}dx
$$
Therefore, the right side of $(*)$ is continuously differentiable in $t$, and the derivative is the above function evaluated at $u=t$. And that gives you what you want:
$$
    \frac{\partial}{\partial t}\int_{\mathbb{R}^n}K(x,t)e^{-2\pi i(\xi\cdot x)t}dx = \int_{\mathbb{R}^n}\frac{\partial}{\partial t}\left(K(x,t)e^{-2\pi i(\xi\cdot x)t}\right)dx.
$$
