Can we interchange the integral order of this integral to start integration on $x$ ? (Taking $g$ and $f$ two functions of rapid decrease which are $o(x^2)$ near zero)


$$A=\int\_{0}^\infty \int\_0^{\infty} \int\_0^{\infty} \frac{1}{x} f(t) g(u)\sin(x(t-u)) \frac{1}{u-t} dt du dx$$


So can we write:


$$A=\int\_{0}^\infty \int\_0^{\infty} \int\_0^{\infty} \frac{1}{x} f(t) g(u)\sin(x(t-u)) \frac{1}{u-t} dx dt du$$


Of course we cannot apply usual therorem as the integral is not absolutely integrable. 
See my previous post for an example of interchange without direct absolute convergence of integral:
[Changing the order of integration of double integral: references and theorems](https://mathoverflow.net/questions/295790/changing-the-order-of-integration-of-double-integral-references-and-theorems)


(The question seems tricky to me: if the two "-" in the integral are replaced by "+" then I think the equality holds) Any reference on integral order interchange going further then the usual theorem with positive function or absolutely convergent functions is welcome.



Interchange of integration order (of a not absolutely convergent integral with sinus)
$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$


Let us denote by $I$ and $J$ your first and second displayed triple integrals, respectively. The innermost integral in $J$ does not exist in the Lebesgue sense. Understood as the "improper" integral, $\int\_0^{\infty-}\cdots\,dx:=\lim\_{b\to\infty}\int\_0^b\cdots\,dx$, this innermost integral equals 
\begin{equation\*}
 -f(t) g(u)\,\frac{\pi/2}{|t-u|}
\end{equation\*}
for $t\ne u$. So, the value of $J$ would be 
\begin{equation\*}
 -\int\_{0}^\infty \int\_0^{\infty} f(t) g(u)\,\frac{\pi/2}{|t-u|}\, dt\, du, 
\end{equation\*}
which would be $-\infty$ if e.g. the functions $f$ and $g$ are strictly positive and continuous. If $f$ and $g$ continuous and $f(t)g(t)>0>f(u)g(u)$ for some $t$ and $u$ in $(0,\infty)$, then no reasonable value can be assigned to $J$. So, I think nothing reasonably good can be done for $I$ and $J$. 


However, things indeed work out well if, as you suggested, the minuses in the expressions of $I$ and $J$ are replaced by pluses. Then we can also relax your conditions on $f$ and $g$: instead of taking $f$ and $g$ to be "of rapid decrease which are $o(x^2)$ near zero", it will suffice to assume that $f$ and $g$ are in $L^1(0,\infty)$ and $|f(t)|+|g(t)|\ll t^p$ for some $p>-1/2$ and all $t\in(0,1)$. Instead of $I$ and $J$, here we are going to consider (with factors in the expression of the integrand conveniently rearranged) 
\begin{equation\*}
 I^+:=\int\_{0}^{\infty-} \int\_0^{\infty} \int\_0^{\infty} \frac{\sin(x(t+u))}{x(t+u)} f(t) g(u)\, dt\, du\, dx
=\lim\_{b\to\infty}I^+\_b \tag{1}
\end{equation\*}
and 
\begin{equation\*}
 J^+:=\int\_{0}^\infty \int\_0^{\infty} \int\_0^{\infty-} \frac{\sin(x(t+u))}{x(t+u)} f(t) g(u)\, dx\, dt\, du=\lim\_{b\to\infty}J^+\_b, \tag{2}
\end{equation\*}
where 
\begin{equation\*}
 I^+\_b:=\int\_{0}^b \int\_0^{\infty} \int\_0^{\infty} \frac{\sin(x(t+u))}{x(t+u)} f(t) g(u)\, dt\, du\, dx 
\end{equation\*}
and
\begin{equation\*}
 J^+\_b:=\int\_0^{\infty} \int\_0^{\infty} \int\_{0}^b \frac{\sin(x(t+u))}{x(t+u)} f(t) g(u)\,dx\, dt\, du. 
\end{equation\*}
Namely, we are going to show that the "triple integrals" $I^+$ and $J^+$ are well defined (in the sense that the limits $\lim\_{b\to\infty}I^+\_b$ and $\lim\_{b\to\infty}J^+\_b$ exist), and we shall also show that $I^+=J^+$. 


To begin doing this, recall that $f$ and $g$ are in $L^1(0,\infty)$ and $|\frac{\sin v}v|\le1$ for $v>0$. So, by Fubini's theorem, for all $b>0$ 
\begin{equation\*}
 I^+\_b=J^+\_b\in\R. \tag{3}
\end{equation\*}


Moreover, $\int\_0^b\frac{\sin xv}x\,dx=\int\_0^{bv}\frac{\sin w}w\,dw$ is bounded uniformly in $(b,v)\in(0,\infty)^2$ and $\int\_0^b\frac{\sin xv}x\,dx\to\frac\pi2$ as $b\to\infty$, for each real $v>0$. Furthermore, switching to the polar coordinates and using the that conditions $f$ and $g$ are in $L^1(0,\infty)$ and $|f(t)|+|g(t)|\ll t^p$ for some $p>-1/2$ and $t\in(0,1)$, we have 
\begin{multline\*}
 \int\_0^{\infty} \int\_0^{\infty}\frac{|f(t) g(u)|}{t+u}\,dt\, du \\ 
 \ll \int\_0^{\infty} \int\_0^{\infty}|f(t)|\,|g(u)|\,dt\, du
 +\int\_0^{2\pi} d\thh \int\_0^1 \frac{r^{2p}}r\,r\,dr \ll1. 
\end{multline\*}
So, by the dominated convergence theorem,
\begin{equation\*}
 J^+\_b\underset{b\to\infty}\longrightarrow\frac\pi2\,
 \int\_0^{\infty} \int\_0^{\infty} \frac{f(t) g(u)}{t+u} dt\, du. 
\end{equation\*}
Thus, in view of (3), the limits in (1) and (2) do exist and are equal to each other, so that 
\begin{equation\*}
 I^+=J^+=\frac\pi2\,
 \int\_0^{\infty} \int\_0^{\infty} \frac{f(t) g(u)}{t+u} dt\, du.
\end{equation\*}


