 Solution

The removable singularities at zero are understood inThese are entire functions, with values $F^{+}(0)=1/2$ and $F^{-}(0)=-1/2$. On the real axis their difference is $\sin x/x$, including its continuous value at zero. In the complex upper half-plane, $|e^{iz}|\le 1$, so $|F^{+}(z)|\le 1/|z|$ away from zero; the analogous bound for $F^{-}$ holds in the lower half-plane. This verifies analyticity, the required jump and the specified decay.

In the Sokhotski–Plemelj formula, the boundary values may also be writtenIndeed the Cauchy principal value of $\sin t/[t(t-x)]$ is $\pi (\cos x-1)/x$: split $1/[t(t-x)]=x^{-1}((t-x{)}^{-1}-t^{-1})$, translate the first integral, and use ${\int }_{\mathbb{R}}\sin t/tdt=\pi$ and the oddness of $\cos t/t$. The limiting formula at zero follows continuously.