 Hilbert transform of the sinc function

With convention $\mathcal{H}f(t)={\pi }^{-1}\mathrm{PV}\int f(\tau )/(t-\tau )d\tau$, the analytic function $F(z)=(e^{iz}-1)/(iz)$ has removable singularity at zero and decays on large upper semicircles. Its boundary real part is $\sin t/t$ and imaginary part is $(1-\cos t)/t$. The upper-half-plane contour indented above $t$ gives $\mathrm{PV}\int F(\tau )/(\tau -t)d\tau =i\pi F(t)$. Taking imaginary parts after reversing the denominator proves the displayed transform, with value zero at $t=0$ by continuous extension.