Using Laplace Transforms to solve $\int_{0}^{\infty}\frac{\sin(x)\sin(x/3)}{x(x/3)}\:dx$

Not sure what you are asking with your "Is this a stroke of luck?" question...

Just in case, here's a different approach to the integral:

$$I=\int_0^\infty \frac{\sin (3x) \sin(x)}{x^2} dx=\frac{1}{2}\int_0^\infty \frac{\cos (2x)-\cos(4x)}{x^2} dx$$

$$I=\int_0^\infty \frac{\cos (x)-\cos(2x)}{x^2} dx$$

Now we also introduce a parameter, though we only need one:

$$I(a)=\int_0^\infty \frac{\cos (ax)-\cos(2ax)}{x^2} dx$$

$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx$$

The integral can now be safely separated into two terms, and each has a well known value:

$$\int_0^\infty \frac{\sin(x)}{x} dx=\frac{\pi}{2}$$

So:

$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx=\frac{\pi}{2}$$

Integrating (the constant of integration is determined by $I(0)$), we have:

$$I(a)=\frac{\pi}{2}a$$

$$I(1)=\frac{\pi}{2}$$

The proofs of the $\text{sinc}$ integral can be found elsewhere, including this site. Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

This starts out similarly to Yuriy S's answer, but the execution seems a bit simpler.

Substituting $x\mapsto3x$, the integral is equal to $$ \begin{align} \int_0^\infty\frac{\sin(3x)\sin(x)}{x^2}\,\mathrm{d}x &=\int_0^\infty\frac{\cos(2x)-\cos(4x)}{2x^2}\,\mathrm{d}x\tag1\\ &=\int_0^\infty\int_2^4\frac{\sin(ax)}{2x}\,\mathrm{d}a\,\mathrm{d}x\tag2\\ &=\frac12\int_2^4\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x\,\mathrm{d}a\tag3\\[6pt] &=\frac\pi2\tag4 \end{align} $$ Explanation:
$(1)$: $\cos(3x-x)-\cos(3x+x)=2\sin(3x)\sin(x)$
$(2)$: $\int_2^4\sin(ax)\,\mathrm{d}a=\frac{\cos(2x)-\cos(4x)}x$
$(3)$: swap order of integration then substitute $x\mapsto x/a$
$(4)$: $\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x=\frac\pi2$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\sin\pars{x}\sin\pars{x/3} \over x\pars{x/3}}\,\dd x} = {1 \over 2}\int_{-\infty}^{\infty}{\sin\pars{x} \over x} {\sin\pars{x/3} \over x/3}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{-\infty}^{\infty} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\dd k} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic qx/3}\dd q}\,\dd x \\[5mm] = &\ {\pi \over 4}\int_{-1}^{1}\int_{-1}^{1}\int_{-\infty}^{\infty} \expo{\ic\pars{k - q/3}x}{\dd x \over 2\pi}\,\dd k\,\dd q = {\pi \over 4}\int_{-1}^{1}\int_{-1}^{1} \delta\pars{k - {q \over 3}}\,\dd k\,\dd q \\[5mm] = &\ {\pi \over 4}\int_{-1}^{1}\bracks{-1 < {q \over 3} < 1}\,\dd q = {\pi \over 4}\int_{-1}^{1}\,\dd q = \bbx{\pi \over 2} \end{align}