Applying Fourier transforms in two variables separately Let $G(x, t) := (4\pi t)^{-d / 2}e^{-|x|^2 / 4t}$ for all $x \in \mathbb{R}^d$ and $t \in (0, \infty)$. The function $G$ is not integrable on $\mathbb{R}^d \times (0, \infty)$ since
$$
\int_0^\infty \int_{\mathbb{R}^d} G(x, t)\,dx\,dt
= \int_0^\infty \,dt
= \infty,
$$
so the Fourier transform of $G$ is not well-defined. However, $G$ is integrable when viewed as functions of the variables $x$ and $t$ separately. In particular, if we let $\mathcal{F}_x$ and $\mathcal{F}_t$ denote the Fourier transforms in $x$ and $t$, respectively, then we can compute
$$
\mathcal{F}_x G(\xi, t)
= e^{-4\pi^2t|\xi|^2},
$$
and
$$
\mathcal{F}_t \mathcal{F}_x G(\xi, \tau)
= \int_0^\infty e^{-4\pi^2t|\xi|^2} e^{-2\pi i \tau t}\,dt
= \frac{1}{4\pi^2|\xi|^2 + 2\pi i\tau}.
$$
But it seems like we have computed the Fourier transform of $G$! What is going on here? I am currently learning (the very basics) of tempered distributions, and it seems like this computation is justified in a certain sense (because $\mathcal{F} = \mathcal{F}_t \mathcal{F}_x$ is in fact true for Schwartz functions). On the other hand, the computation above doesn't even mention distribution theory—is there any other rigorous way of understanding this computation without requiring such machinery?
 A: *

*What is going on here?

Indeed, in the $L^1$ theory of the Fourier transform, you need Fubini-type theorems to exchange the order of Fourier transforms and to write $\mathcal F_{t,x} = \mathcal F_t \mathcal F_x$. Here indeed, $G$ is not integrable and as you can see in your last computation, the Fourier transform is actually still not defined when $\xi = 0$, and the result is not bounded, while it would be continuous if $G$ was $L^1$.
Your computation is rigorous and makes sense without distribution theory but is saying that for all $\xi\neq 0$, $\mathcal F_t \mathcal F_x G = (|2\pi\,\xi|^2 + 2i\pi\,\tau)^{-1}$. It does not tell anything about $\mathcal F_{t,x}G$ and about $\mathcal F_x\mathcal F_tG$ and about what happens when $\xi=0$. To investigate these questions, distribution theory is indeed a good framework.


*Distribution theory

For tempered distributions, the Fourier transform is defined by $\langle \mathcal F f,\varphi\rangle = \langle f,\mathcal F \varphi\rangle$ for any Schwarz class test function $\varphi$, and coincides with the classical integral definition if $f\in L^1$, so indeed the relation $\mathcal F_{t,x} = \mathcal F_t \mathcal F_x = \mathcal F_x \mathcal F_t$ is valid, as well as your first classical computation of the Fourier transform of a Gaussian with $t>0$.
Warning however, distribution theory is not just a way to justify formal computations. For example $\Delta(1/|x|) = 0$ when $x\neq 0$ in dimension $3$ but $\Delta(1/|x|) = 4\pi\,\delta_0$ in distribution theory. Here, in your last computation, there is a singularity when $\xi = 0$ and so one should use test functions or approximations to justify it.
A: You can explain the integral in the sense of distribution, even the integral does not converge in the Lebsgue sense.
Let me take a simple example. Formally, We can abuse notation and write
$$
\delta(\xi) = \hat{1}(\xi) = \int_{\mathbb{R}^n} e^{-2\pi i\xi\cdot x}\,dx.
$$
The above integral does not converge, but you can explain it by applying it to a test function $\phi\in\mathcal{S}(\mathbb{R}^n)$:
$$
\langle\int_{\mathbb{R}^n} e^{-2\pi i\xi\cdot x}\,dx, \phi(\xi)\rangle := \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\phi(\xi)e^{-2\pi i\xi\cdot x}\,d\xi\,dx = \int_{\mathbb{R}^n}\hat{\phi}(x)\,dx = \phi(0) = \langle\delta, \phi\rangle.
$$
Your problem can be explained in the same way. Such notations are convenient for computation, because they enable us to pretend 'generalized functions' as functions.
