I thought that it might be instructive to present an approach to deriving the Fourier transform of $\log(|x|)$. The result includes a distributional interpretation of $\frac1{|x|}$. Finally, we show that the distributional interpretation of $\frac1{|x|}$ is non-unique and that it differs from other interpretations by a multiple of the Dirac Delta distribution. With that introduction, we now proceed.

**PRELIMARIES**

Let $\psi(x)=\log(|x|)$ and let $\Psi$ denote its Fourier Transform . Then, we write

$$\Psi(x)=\mathscr{F}\{\psi\}(x)\tag 1$$

where $(1)$ is interpreted as a Tempered Distribution. That is, for any $\phi \in \mathbb{S}$, we can write

$$\langle \mathscr{F}\{\psi\}, \phi\rangle =\langle \psi, \mathscr{F}\{\phi\}\rangle$$

Now, let $\psi_\epsilon(k) =e^{-\varepsilon|k|}\log(|k|)$. Therefore, $\psi(k)=\lim_{\varepsilon\to 0^+}\psi_\varepsilon(k)$ and we see that

$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\}, \phi\rangle&=\lim_{\varepsilon\to 0^+}\langle \psi_\varepsilon, \mathscr{F}\{\phi\}\rangle \\\\
&=\langle \psi,\mathscr{F}\{\phi\}\rangle\\\\
&=\langle \mathscr{F}\{\psi\}, \phi\rangle
\end{align}$$

Next, we evaluate the Fourier transform of $\psi_\varepsilon$.

**EVALUATING THE FOURIER TRANSFORM OF $\displaystyle \psi_\varepsilon$**

Denote by $\Psi_\epsilon$, the Fourier transform of $\psi_\varepsilon$. Then, we have

$$\begin{align}
\Psi_\varepsilon(x)&=\mathscr{F}\{\psi_\epsilon\}(x)\\\\
&=\int_{-\infty}^\infty e^{-\varepsilon|k|}\log(|k|) e^{ikx}\,dk\\\\
&=2\text{Re}\left(\int_0^\infty e^{-(\varepsilon -ix)k}\log(k) \,dk\right)\\\\
&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma -\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\\\\
&=\psi^{(1)}_\varepsilon(x)+\psi^{(2)}_\varepsilon(x)+\psi^{(3)}_\varepsilon(x)\tag2
\end{align}$$

where

$$\begin{align}
\psi^{(1)}_\varepsilon(x)&=-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma\\\\
\psi^{(2)}_\varepsilon(x)&=-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2)\\\\
\psi^{(3)}_\varepsilon(x)&=-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)
\end{align}$$

Next, we will find the distributional limits of $\psi^{(1)}_\varepsilon$, $\psi^{(2)}_\varepsilon$, and $\psi^{(3)}_\varepsilon$.

**DISTRIBUTIONAL LIMITS OF $\displaystyle \psi^{(1)}_\varepsilon$, $\displaystyle
\psi^{(2)}_\varepsilon$, and $\displaystyle \psi^{(3)}_\varepsilon$**

Again, let $\phi\in \mathbb{S}$. Then,

$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \psi^{(1)}_\varepsilon,\phi \rangle &=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \psi^{(1)}_\varepsilon(x)\phi(x)\,dx\\\\
&=\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \left(-\frac{2\varepsilon}{\varepsilon^2+x^2}\gamma \right)\phi(x)\,dx\\\\
&=-2\gamma\lim_{\varepsilon\to 0^+}\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx\\\\
&=-2\pi \gamma \phi(0)\tag3
\end{align}$$

$$\begin{align}
\langle \psi^{(2)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{\varepsilon}{\varepsilon^2+x^2}\log(\varepsilon^2+x^2) \right)\phi(x)\,dx\\\\
&=-2\log(\varepsilon)\int_{-\infty}^\infty \frac{\phi(\varepsilon x)}{x^2+1}\,dx-\int_{-\infty}^\infty \frac{\log(1+x^2)}{1+x^2}\phi(\varepsilon x)\,dx\\\\
&= -2\pi \log(\varepsilon)\phi(0)-2\pi \log(2) \phi(0)+o(\varepsilon)
\end{align}\tag4$$

$$\begin{align}
\langle \psi^{(3)}_\varepsilon,\phi \rangle &=\int_{-\infty}^\infty \left(-\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon)\right)\phi(x)\,dx\\\\
&-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\
&=-\phi(0)\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \,dx\\\\
&-\int_{|x|\le 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) (\phi(x)-\phi(0))\,dx-\int_{|x|\ge 1}\frac{2x}{\varepsilon^2+x^2}\arctan(x/\varepsilon) \phi(x)\,dx\\\\
&= \left(2\pi \log(\varepsilon) +2\pi \log(2)\right)\phi(0)+o(\varepsilon)\\\\
&-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\tag5
\end{align}$$

**FINAL RESULTS**

Substituting $(3)$, $(4)$, and $(5)$ into $(2)$, we find that

$$\begin{align}
\lim_{\varepsilon\to 0^+}\langle \mathscr{F}\{\psi_\varepsilon\},\phi\rangle =-2\pi \gamma \phi(0)-\pi \int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx-\pi\int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx\\\\
\end{align}$$

from which we assert that in distribution

$$\mathscr{F}\{\psi\}(x)=-2\pi \gamma \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$

where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$,

$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le 1}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge 1}\frac{\phi(x)}{|x|}\,dx$$

**NOTE:**

It was arbitrary to split the integration in $(5)$ into intervals $|x|\le 1$ and $|x|\ge 1$. Had we chosen instead the intervals $|x|\le \nu$ and $|x|\ge \nu$ for any $\nu>0$, we would have obtained

$$\mathscr{F}\{\psi\}(x)=-2\pi (\gamma+\log(\nu)) \delta(x)-\pi \text{PV}\left(\frac1{|x|}\right)$$

where we interpret $\text{PV}\left(\frac1{|x|}\right)$ to mean that for any $\phi\in \mathbb{S}$,

$$\int_{-\infty}^\infty \phi(x) \text{PV}\left(\frac1{|x|}\right)\,dx=\int_{|x|\le \nu}\frac{\phi(x)-\phi(0)}{|x|}\,dx+ \int_{|x|\ge \nu}\frac{\phi(x)}{|x|}\,dx$$

2more comments