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
$(1.4)$
$$\lim_{R \rightarrow \infty} \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz \rightarrow \int_{0}^{\pi}\frac{e^{ire^{i\theta}}}{re^{i\theta}}\frac{e^{ire^{i\theta}}}{re^{i\theta}/3}d\theta = \frac{\pi}{2}$$