Using Laplace Transforms to solve $\int_{0}^{\infty}\frac{\sin(x)\sin(x/3)}{x(x/3)}\:dx$ So, I've come across the following integral (and it's expansion) many times and in my study so far, Complex Residues have been used to evaluate it. I was hoping to find an alternative approach using Laplace Transforms. I believe the method I've taken is correct, but I'm concerned there may certain theorems/tests I should have applied first
$$I = \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)}\:dx$$
The first step is to make a slight change of variable $x = 3u$, which gives us 
$$I = \int_{0}^{\infty} \frac{\sin\left(3u\right)\sin\left(u\right)}{u^2} \: du $$
Here I employ the Feynman Trick but with two variables, i.e. 
$$I(a,b) = \int_{0}^{\infty} \frac{\sin\left(3ua\right)\sin\left(ub\right)}{u^2}\: du$$
Take the Laplace Transform w.r.t '$a$'
$$\mathscr{L}_{a} \left[I(a,b)\right] = \int_{0}^{\infty} \frac{\mathscr{L}_{a}\left[\sin\left(3ua\right)\right]\sin\left(ub\right)}{u^2}\: du = \int_{0}^{\infty} \frac{3u\sin\left(ub\right)}{\left(s^2 + 9u^2\right)u^2}\: du $$
Or
$$ \overline{I}(s,b) = \int_{0}^{\infty} \frac{3\sin\left(ub\right)}{\left(s^2 + 9u^2\right)u}$$
Now apply the Laplace Transform w.r.t '$b$'. Here $\omega$ will be used as the alternate '$s$' variable. Hence we arrive at
$$ \mathscr{L}_{b}\left[\overline{I}(s,b)\right] = \int_{0}^{\infty} \frac{3\mathscr{L}_{b}\left[\sin\left(ub\right)\right]}{\left(s^2 + 9u^2\right)u}\:du =  \int_{0}^{\infty} \frac{3u}{\left(s^2 + 9u^2\right)u\left(\omega^2 + u^2\right)}\:du $$
Or 
$$\overline{\overline{I}}\left(s,\omega\right) = \int_{0}^{\infty} \frac{3}{\left(s^2 + 9u^2\right)\left(\omega^2 + u^2\right)}\:du = \frac{3\pi}{2s\omega}\left(\frac{1}{s + 3\omega} \right)$$
In no specific order we now take the Inverse Laplace Transform w.r.t. '$\omega$'
$$\overline{I}\left(s,b\right) = \mathscr{L}_{\omega}^{-1}\left[\frac{3\pi}{2s\omega}\left(\frac{1}{s + 3\omega} \right) \right] = \frac{3\pi}{2}\left[\frac{1}{s^2} - \frac{e^{\frac{sb}{3}}}{s^2}\right]$$
We now take the Inverse Laplace Transform w.r.t. '$s$'
$$I(a,b) = \frac{3\pi}{2}\mathscr{L}_{s}^{-1}\left[ \frac{1}{s^2} - \frac{e^{\frac{sb}{3}}}{s^2}\right] = \frac{3\pi}{2}\left[a - \left(a - \frac{b}{3} \right)\mathcal{H}\left(a - \frac{b}{3}\right) \right]$$
And so, 
$$I = I(1,1) = \frac{3\pi}{2}\left[1 - \left(1 - \frac{1}{3} \right)\mathcal{H}\left(1 - \frac{1}{3}\right) \right] = \frac{\pi}{2}$$
As required. 
Is this a stroke of luck? or is it just employing the Dominated Convergence Theorem and Fubini's Theorm (as I believe is valid here). 
 A: 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 $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?
A: 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 $x\mapsto x/a$
$(4)$: $\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x=\frac\pi2$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\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}
