= Sine-product integral with inverse-square weight
{title2=$\int_0^\infty\frac{\sin(ax)\sin(bx)}{x^2}dx=\frac\pi2\min(a,b),\quad a,b>0$}
By symmetry assume $a\geq b>0$ 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 $(e^{i(a+b)z}-e^{i(a-b)z})/(2iz^2)$ bounds the large arc by $\pi/R$, including when $a=b$. The <upper-half-plane indentation rule> therefore gives $\operatorname{PV}\int_{\mathbb R}f(x)dx=i\pi b$. 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.
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