 Solution

Integrate $e^{iz}/z$ around the upper half-disc with the upper half-disc of radius $\epsilon$ about zero removed. The real segments run from $-R$ to $-\epsilon$ and $\epsilon$ to $R$; the small semicircle is clockwise. There are no poles inside this indented contour.

The large arc contribution tends to zero: its modulus is at most ${\int }_0^{\pi }{e}^{-R\sin \theta }d\theta$, which tends to zero by dominated convergence. On the small arc $z=\epsilon e^{i\theta }$, $\theta$ decreases from $\pi$ to zero, so its integral tends to ${\int }_{\pi }^{0}id\theta =-i\pi$. The contour theorem consequently givesTaking imaginary parts is legitimate for each truncated integral. The imaginary integrand $\sin x/x$ is continuous at zero after assigning its limiting value one, so the inner indentation can be removed. Therefore