1
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.