= Solution
Let $f=\widehat\varphi$, a <Schwartz function>. Splitting the inverse <Fourier transform> at the jump in the <Hilbert-transform Fourier multiplier> gives
$$
\mathcal H\varphi(x)=-\frac i{2\pi}\left[\int_0^\infty e^{ix\xi}f(\xi)\,d\xi-\int_{-\infty}^0e^{ix\xi}f(\xi)\,d\xi\right].
$$
For $x\ne0$, <integration by parts> on each half-line shows
$$
\mathcal H\varphi(x)=\frac{f(0)}{\pi x}+\frac1{2\pi x}\left[\int_0^\infty e^{ix\xi}f'(\xi)\,d\xi-\int_{-\infty}^0e^{ix\xi}f'(\xi)\,d\xi\right].
$$
Both restricted <derivatives> are in $L^1$. The <Riemann-Lebesgue lemma> makes the bracket tend to zero as $|x|\to\infty$. Hence \b[the two-sided tail is]
$$
\boxed{\mathcal H\varphi(x)=\frac{\widehat\varphi(0)}{\pi x}+o(|x|^{-1}),\qquad |x|\to\infty.}
$$
Here $\widehat\varphi(0)=\int\varphi$. The <large-distance tail of the Hilbert transform> thus depends on the zeroth moment of the input. If that moment is nonzero, the $1/x$ tail proves that the output is not a <Schwartz function>, despite being smooth and square-integrable.
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