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

(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.


$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

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*}


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