= Solution
Although $x\mapsto|x|^{-1}$ is not integrable on all of $\mathbb R^3$, it defines a <tempered distribution>. For $x\ne0$, the <Gamma function> integral gives
$$
\frac1{|x|}
=\frac1{\sqrt\pi}\int_0^\infty
t^{-1/2}e^{-t|x|^2}\,dt.
$$
The three-dimensional form of the <Fourier transform of a Gaussian> is
$$
\mathcal F(e^{-t|x|^2})(\xi)
=\left(\frac\pi t\right)^{3/2}
e^{-|\xi|^2/(4t)}.
$$
Therefore, up to a nonzero constant depending on the Fourier-transform convention,
$$
\widehat{|x|^{-1}}(\xi)
=C\int_0^\infty t^{-2}e^{-|\xi|^2/(4t)}\,dt.
$$
For $\xi\ne0$, the substitution $u=|\xi|^2/(4t)$ turns the last integral into
$$
\frac4{|\xi|^2}\int_0^\infty e^{-u}\,du.
$$
Thus, as asserted by the <Fourier transform of inverse distance in three dimensions>,
$$
\boxed{\widehat{|x|^{-1}}(\xi)=C'|\xi|^{-2}}
$$
as a tempered distribution, where $C'\ne0$.
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