TITLE: Using Laplace Transforms to solve $\int_{0}^{\infty}\frac{\sin(x)\sin(x/3)}{x(x/3)}\:dx$
QUESTION [6 upvotes]: 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).

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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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}