Sine-product integral with inverse-square weight (${\int }_0^{\infty }\frac{\sin (ax)\sin (bx)}{x^2}dx=\frac{\pi }{2}\min (a,b),a,b>0$)

By symmetry assume $a\ge b>0$ and integrate $f(z)=e^{iaz}\sin (bz)/z^2$ around an upper semicircle indented clockwise above zero. The residue at zero is $b$. The expansion $(e^{i(a+b)z}-e^{i(a-b)z})/(2i{z}^2)$ bounds the large arc by $\pi /R$, including when $a=b$. The upper-half-plane indentation rule therefore gives $\mathrm{PV}{\int }_{\mathbb{R}}f(x)dx=i\pi b$. Its imaginary part is even and ordinarily integrable, yielding the displayed value. At zero the sine-product integrand tends to $ab$; at infinity it is bounded by $1/x^2$. The principal value is needed for the auxiliary complex integral, not for the real sine-product integral.