 Cosine-difference inverse-square integral

For $a,b>0$, integrate $(e^{iaz}-e^{ibz})/z^2$ over an upper semicircle with a small clockwise indentation excluding zero. The large arc vanishes; the small arc tends to $-i\pi$ times the residue $i(a-b)$, namely $\pi (a-b)$. The contour encloses no poles, so the remaining real-axis principal value is $\pi (b-a)$. Taking real parts gives the displayed ordinary integral, whose cancellation at zero removes its apparent singularity. Evenness of the cosine extends the formula to arbitrary real $a,b$, with zero frequencies obtained by a limit. The individual integrals of $\cos (ax)/x^2$ cannot be subtracted as separately convergent improper integrals.