= Solution
The locally integrable function $\log|x|$ defines a <tempered distribution>. Its <distributional derivative> is $\operatorname{PV}(1/x)$: perform <integration by parts> outside $(-a,a)$, and observe that the boundary contribution $\log a[\varphi(a)-\varphi(-a)]$ tends to zero. By the <Fourier transform of a derivative> and the preceding result,
$$
ik\widehat{\log|x|}=-i\pi\operatorname{sgn}k.
$$
Division by $|k|$ needs a <finite-part inverse-absolute-value distribution>, since $1/|k|$ is not locally integrable at zero. One possible normalization is
$$
\left\langle\operatorname{Pf}\frac1{|k|},\varphi\right\rangle
=\int_{|k|<1}\frac{\varphi(k)-\varphi(0)}{|k|}\,dk
+\int_{|k|\ge1}\frac{\varphi(k)}{|k|}\,dk.
$$
It satisfies $k\operatorname{Pf}(1/|k|)=\operatorname{sgn}k$. The general solution of $kT=0$ is a multiple of the <Dirac delta distribution>: a <test function> vanishing at zero can be written $k\psi(k)$, so $T$ depends only on its value at zero. Consequently the <Fourier transform of the absolute logarithm> is
$$
\boxed{\widehat{\log|x|}(k)=-\pi\operatorname{Pf}\frac1{|k|}+C\delta(k)}.
$$
Changing the finite-part normalization changes $C$. The printed reciprocal expression is meaningful with this interpretation, not as an ordinary function at $k=0$.
Back to article page
