How can I easily determine whether we can interchange integrals over a infinite region? In his Partial Differential Equations, p187, Evans adopts the following method to calculate the inverse Fourier transform of $\frac1{1+|y|^2}$, where $y\in \mathbb{R}^n$: he uses $\frac1{1+|y|^2}=\int_0^\infty e^{-t(1+|y|^2)}\,dt$, and claims that
$$
\left(\frac1{1+|y|^2}\right)^\vee=\frac1{(2\pi)^{n/2}} \int_0^\infty e^{-t}
\left(\int_{\mathbb{R}^n}e^{ix\cdot y-t|y|^2}\,dy\right) dt,\quad (*)
$$
thus shifts attention to the evaluation of $\int_{\mathbb{R}^n}e^{ix\cdot y-t|y|^2}\,dy$.
However I can't see why (*) is correct on the spot. After thinking for a while I managed to justify it.
My attempt: I tried to simplify the notation to $\int \left(\int_0^\infty f(t,y) \,dt\right) e^{ixy} dy$, and let $I_b^a(y) =\int_b^a f(t,y) \,dt$. Then we have to show
$$
\int\lim_{a\to +\infty} I_0^a(y)e^{ixy} \,dy=\int_0^{+\infty}\int f(t,y)e^{ixy} \,dydt.
$$
Maybe we should first admit that (Well after some thought I realized that this also requires a justification)
$$
\int I_0^b(y)e^{ixy} \,dy=\int_0^{b}\int f(t,y)e^{ixy} \,dydt
$$
and then prove that
$$
\int\lim_{a\to +\infty} I_0^a(y)e^{ixy}\,dy=\lim_{a\to +\infty} \int I_0^a(y)e^{ixy} \,dy
$$
which is the same as
$$
\left|\int I_b^\infty e^{ixy} dy\right| \text{ can be arbitrarily small as long as } b \text{ is sufficiently large,}
$$
which in turn is the same as (if we work out $I_b^\infty$ explicitly)
$$
\int\frac{e^{-b(1+|y|^2)+ixy}}{1+|y|^2}\,dy \to 0.\quad (**)
$$
So I think it remains to show
$
\int\frac{e^{-b(1+|y|^2)}}{1+|y|^2}\,dy\to 0
$.
So I rewrite the integral as
$$
\int \frac{e^{-b(1+|y|^2)}}{1+|y|^2} \,dy=\int_0^\infty
 \int_{|y|=R} \frac{e^{-b(1+R^2)}}{1+|R|^2} dydR=\int_0^\infty \frac{e^{-b(1+R^2)}\omega_n R^{n-1}}{1+R^2} dR.
$$
Then we can use Gamma function to prove that it is so.
However, I deem my arguments as involved and not general. At (**), I can't imagine how harder it would be if $I_b^\infty$ could not be calculated out directly. Yet Evans takes (*)  for granted, not even bothering the interchange problem at all. I think there must be some easier and more general way to justify that but, what is it?
 A: $\newcommand{\d}{\,\mathrm{d}}$Forgive me if I am making a trivial oversight, but I think Fubini’s theorem doesn’t apply unless $n=1$ (and the Fourier transform doesn’t exist at $x=0$ if $n\gt1$, and it may also fail to exist for other $x$). I think you need to instead consider powers $(1+|y|^k)^{-1}$ where $k\ge n+1$, to guarantee existence of the transform and interchangeability of the integrals with Fubini. It may be the case that the integrals are interchangeable anyway, but one must use a different device than Fubini.
The Fourier Transform has the lucky property that $|e^{ixy}|\equiv1$, which simplifies convergence computations. To use Fubini’s theorem, in the forms given in the Wiki article, the standard hypotheses are equivalent to the claim that $(y,t)\mapsto e^{ixy}e^{-t(1+|y|^2)}$ is absolutely integrable through $\int_{\Bbb R^n}\int_{[0,\infty)}$, regardless of the value of $x$.
We can compute:

$$\begin{align}\int_{\Bbb R^n}\int_0^\infty|e^{ixy}e^{-t(1+|y|^2)}|\d t\d y&=\int_{\Bbb R^n}\int_0^\infty e^{-t(1+|y|^2)}\d t\d y\\&=\int_{\Bbb R^n}\frac{1}{1+|y|^2}\d y\end{align}$$

One way to identify the last integral diverges ($n\gt1$) is through the coarea formula. In the notation of the article, if $u:\Bbb R^n\to\Bbb R$ maps $y\mapsto\|y\|$ then $u$ is Lipschitz with gradient a unit vector, and we can set $g(y)=(1+|y|^2)^{-1}$. Then: $$\begin{align}\int_{\Bbb R^n}\frac{1}{1+|y|^2}\d y&=\int_{\Bbb R^n}g(y)|\nabla u(y)|\d y\\&=\int_{\Bbb R}\int_{u^{-1}(t)}g(x)\d H_{n-1}(x)\d t\\&=\int_0^\infty\int_{S(t)}\frac{1}{1+t^2}\d H_{n-1}(x)\d t\\&=\int_0^\infty\frac{1}{1+t^2}H_{n-1}(S(t))\d t\\&=\sigma_n\int_0^\infty\frac{t^{n-1}}{1+t^2}\d t\\&=\begin{cases}\pi&n=1\\\infty&n\gt1\end{cases}\end{align}$$
Where the $\sigma_n$ are some known constants for the surface area of an $n$-sphere.
This doesn’t exclude the integrals from being interchangeable. However, we can’t use vanilla Fubini to justify the interchange here.
However, the same proof shows that we can use Fubini’s theorem to interchange the integrals if you instead consider the Fourier transform of $\frac{1}{1+|y|^{n+1}}$.
