 Solution

Use the sign convention in the question: the Hilbert transform kernel is $(y-x{)}^{-1}$, the negative of another common convention. A contour indentation at the origin givesFor $s>0$ close the contour in the upper half-plane; for $s<0$ close it below. Thus the transform of $\sin (sx)$ is $\cos (sx)$ for $s>0$. Since $(1-\cos x)/x={\int }_0^1\sin (sx)ds$, integration of this multiplier identity, justified first with an Abel regularization, yieldsThe value at zero also follows directly from ${\pi }^{-1}\int (1-\cos y)y^{-2}dy=1$.