How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$ Recently I found a claim saying that 
$$ 
\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2 
$$
from what I can see this seems to be true. I already know that
$\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above. 
Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg
$$
\int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x
$$
Where $a_k, \cdots, a_N$ are arbitrary positive constants.
It seems related to the Borwein Integral, but there are some subtle differences.

First of all, we use compound angle formula and IBP.
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\sin (a x) \sin (b x)}{x^{2}} d x \\
=& \frac{1}{2} \int_{0}^{\infty}[\cos (a-b) x-(\sin (a+b) x] d\left(-\frac{1}{x}\right) \quad \text{(Via compound angle formula)}\\
\stackrel{IBP}{=}  &-\left[\frac{1}{2 x}(\cos (a-b) x-\cos (a+b) x)\right]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{-(a-b) \sin (a-b) x}{x} d x \\\quad &+\frac{1}{2} \int_{0}^{\infty} \frac{(a+b) \sin (a+b) x}{x} d x
\end{aligned}
$$
Then using the famous formula $$
\int_{0}^{\infty} \frac{\sin k x}{x} d x=(\operatorname{sgn} k) \frac{\pi}{2},
$$
we can now conclude that $$
\begin{aligned}
I &=\left\{\begin{array}{ll}
\frac{1}{2}\left[\frac{-(a-b) \pi}{2}+\frac{(a+b) \pi}{2}\right] & \text { if } a \geqslant b \\
\frac{1}{2}\left[\frac{(a-b) \pi}{2}+\frac{(a+b) \pi}{2}\right] & \text { if } b>a
\end{array}\right.\\
&=\left\{\begin{array}{ll}
\frac{b \pi}{2} & \text { if } a \geqslant b \\
\frac{a \pi}{2} & \text { if } b>a
\end{array}\right.\\
&=\frac{\min (a, b) \pi}{2}
\end{aligned}
$$
