 Solution

Consider $G(z)=(e^{iz}-1)/(iz)$, extending it at zero by $G(0)=1$. It is holomorphic in the upper half-plane and on its boundary, and on large upper semicircles $G(z)=O(1/|z|)$ because $|e^{iz}|\le 1$. For real $t$, indent the upper-half-plane contour above the pole of $G(z)/(z-t)$. The clockwise indentation contributes $-i\pi G(t)$, while the large semicircle contributes zero in the limit. The Cauchy integral theorem givesWith the stated Hilbert transform convention this is $\mathcal{H}G(t)=-iG(t)$. ButTaking real parts yields the Hilbert transform of the sinc function:At $t=0$ the right side has limit zero; the same value follows by oddness of the principal-value integrand. The apparent singularity of $G$ at zero is removable, so the contour calculation is valid there too.