= Solution
The removable singularities at zero are understood in
$$
\boxed{F^+(z)=\frac{e^{iz}-1}{2iz},\qquad F^-(z)=\frac{e^{-iz}-1}{2iz}}.
$$
These 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>, $|e^{iz}|\leq1$, so $|F^+(z)|\leq1/|z|$ 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 written
$$
F^\pm(x)=\pm\frac{\sin x}{2x}+\frac{\cos x-1}{2ix}.
$$
Indeed 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 $\int_{\mathbb R}\sin t/t\,dt=\pi$ and the oddness of $\cos t/t$. The limiting formula at zero follows continuously.
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