Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 8B (Further Complex Methods)

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 ${\pi }^{-1}\int (1-\cos y)y^{-2}dy=1$.

Splitting the exponential into the half-plane where it decays givesTheir singularities at zero are removable. The first function is analytic above the real axis and the second below it; their exponential terms are bounded in those respective half-planes, giving the required $O(1/z)$ decay. Their difference on the real axis is the prescribed jump. Moreover their sum there is $-i\sin x/x$, agreeing with the Sokhotski–Plemelj formula and the transform computed above.

For uniqueness in the class with continuous boundary values, subtract two solutions. The zero jump glues their differences into an entire function, tending to zero at infinity. Liouville's theorem makes it zero. Thus these are the unique solutions of the Riemann-Hilbert problem with the stated decay.