= 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 $|e^{iz}|\leq1$. 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> gives
$$
\operatorname{PV}\int_{-\infty}^{\infty}\frac{G(\tau)}{\tau-t}\,d\tau=i\pi G(t).
$$
With the stated <Hilbert transform> convention this is $\mathcal HG(t)=-iG(t)$. But
$$
G(t)=\frac{\sin t}{t}+i\frac{1-\cos t}{t}.
$$
Taking real parts yields the <Hilbert transform of the sinc function>:
$$
\boxed{\mathcal H\!\left(\frac{\sin t}{t}\right)=\frac{1-\cos t}{t}.}
$$
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.
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