= Hilbert transform of the sinc function
{c}
{title2=$\mathcal H(\sin t/t)=(1-\cos t)/t$}
With convention $\mathcal Hf(t)=\pi^{-1}\operatorname{PV}\int f(\tau)/(t-\tau)\,d\tau$, 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 $\operatorname{PV}\int F(\tau)/(\tau-t)\,d\tau=i\pi F(t)$. Taking imaginary parts after reversing the denominator proves the displayed transform, with value zero at $t=0$ by continuous extension.
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