A rough idea to prove $\int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2} ?$ 
I have a rough idea of approach to prove the Borwein integral in $(1)$ via Complex-Analytic techniques, is it valid ?

$(1)$
$$ \int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2}.$$
From glancing at it seems one can consider an Indented Contour $\Gamma_{R}$, and define in $(1.2)$
$(1.2)$
$$\Gamma_{R}^{1}(t) = t  \, \, \, \, \text{if} \, \, \, -R \leq t \leq -1/R $$
$$\Gamma_{R}^{2}(t) = e^{it}/R  \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$
$$\Gamma_{R}^{3}(t) = e^{it}/R  \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$
$$ \Gamma_{R}^{4}(t) = Re^{it}  \, \, \, \, \text{if} \, \, \, 0 \leq t \leq  \pi$$
Our original integral $\Psi(x) = \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}$ can be defined on $\Gamma_{R}$ as $\Psi(z)=\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}$ so our original integral now becomes
$$I = \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz; $$
Then it seems via Cauchy's Theorem one has in $(1.3)$
$(1.3)$
$$\, \, \, \, \, \, \lim_{R \rightarrow \infty} \bigg(\oint_{\Gamma_{R}^{1}} \Psi(x)dx + \oint_{\Gamma_{R}^{2}} \Psi(z)dz + \oint_{\Gamma_{R}^{3}}\Psi(x)dx   + \oint_{\Gamma_{R}^{4}} \Psi(z)dz \bigg) = 0 $$
To finish the argument it seems one would have to invoke Jordan's Lemma on the last integral in $(1.3)$, then parametrize the integral over $\Gamma_{R}^{4}$ then in $(1.4)$ we have
$(1.4)$
$$\lim_{R \rightarrow \infty} \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz \rightarrow \int_{0}^{\pi}\frac{e^{ire^{i\theta}}}{re^{i\theta}}\frac{e^{ire^{i\theta}}}{re^{i\theta}/3}d\theta = \frac{\pi}{2}$$
 A: As an alternative to finding a suitable integration contour, the Laplace transform often deserves a thought. For any $a,b>0$ we have
$$\begin{eqnarray*} I(a,b)&=&\int_{0}^{+\infty}\frac{\sin(ax)\sin(bx)}{x^2}\,dx=\int_{0}^{+\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{2x^2}\,dx\\&=&J(a+b)-J(a-b)\end{eqnarray*} $$
where
$$ J(c)=\int_{0}^{+\infty}\frac{1-\cos(cx)}{2x^2}\,dx = \int_{0}^{+\infty}\frac{\sin^2\left(\frac{c}{2}x\right)}{x^2}\,dx\stackrel{\mathcal{L},\mathcal{L}^{-1}}{=}\int_{0}^{+\infty}\frac{c^2}{2c^2+2s^2}\,ds=\frac{\pi}{4}|c| $$
ensures:
$$ I(a,b) = \frac{\pi}{4}\left(|a+b|-|a-b|\right)=\color{red}{\frac{\pi}{2}\min(a,b).}$$
A: Perhaps you might be interested in seeing a real method used to evaluate this integral.
We begin by enforcing a substitution of $x \mapsto 3x$. This gives
\begin{align*}
\int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx &= \int_0^\infty \frac{\sin (3x) \sin x}{x^2} \, dx\\
&= 3 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 4 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx, \tag1
\end{align*}
since $\sin (3x) = 3 \sin x - 4 \sin^3 x$.
For the first of these integrals, as is well-known
$$\frac{\pi}{2} = \int_0^\infty \frac{\sin x}{x} \, dx,$$
(this is just the Dirichlet integral) enforcing a substitution of $x \mapsto 2x$ gives
\begin{align*}
\frac{\pi}{2} &= \int_0^\infty \frac{\sin (2x)}{x} \, dx = \int_0^\infty \frac{2 \sin x \cos x}{x} \, dx\ = \int_0^\infty \frac{(\sin^2 x)'}{x} \, dx\\
&= \left [\frac{\sin^2 x}{x} \right ]_0^\infty + \int_0^\infty \frac{\sin^2 x}{x^2} \, dx = \int_0^\infty \frac{\sin^2 x}{x^2} \, dx.
\end{align*}
And for the second of the integrals, as we just found above
$$\frac{\pi}{2} = \int_0^\infty \frac{\sin^2 x}{x^2}dx.$$
Enforcing a substitution of $x \mapsto 2x$ gives
$$\frac{\pi}{2} = \frac{1}{2} \int_0^\infty \frac{\sin^2 (2x)}{x^2} \, dx.$$
Since
$$\sin^2 (2x) = 4 \cos^2 x \sin^2 x = 4(1 - \sin^2 x)\sin^2 x = 4 \sin^2 x - 4 \sin^4 x,$$
we have
$$\frac{\pi}{2} = 2 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx = 2 \cdot \frac{\pi}{2} - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx,$$
or
$$\int_0^\infty \frac{\sin^4 x}{x^2}dx = \frac{\pi}{4}.$$
So returning to (1) we have
$$\int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx = 3 \cdot \frac{\pi}{2} - 4 \cdot \frac{\pi}{4} = \frac{\pi}{2},$$
as required to show.
