How to interpret $|x|^{-1}$ as a distribution Currently I'm reading a little about regularization of distributions which usually have annoying singularities like $x^{-1}$ or $|x|^{-1}$. To the best of my understanding,
$$
\frac{1}{x}
$$
is interpreted as a distribution by considering it's Cauchy principal value instead of doing the bruteforce integration. This way, it's possible to recover a meaningful result by carefully cancelling contributions to the singularity. So if $g$ is, say, a function from Schwartz space, then
$$
\left\langle \text{p.v.}\, \frac{1}{x}, \, g(x) \right\rangle = \int _{0}^{\infty} \frac{g(x) - g(-x)}{x} \, dx.
$$
Intuitively, this makes total sense to me. However, when reading about $|x|^{-1}$ I get slightly confused. The definition states that $|x|^{-1}$ (or, rather, it's regularized cousin $\mathcal{P}|x|^{-1}$) acts on test functions as per
$$
\left\langle \mathcal{P}\frac{1}{|x|}, \, g(x) \right\rangle = \int _{|x|<1} \frac{g(x)-g(0)}{|x|} \, dx + \int_{|x|>1} \frac{g(x)}{|x|} \, dx
$$
The second integral is pretty straightforward; no singularities to sneak around. But the first one...
Does it solve the singularity problem? Yes, I can see that, since
$$
\frac{g(x)-g(0)}{|x|}
$$
remains bounded as $x \rightarrow0$. But how do we get to it? What's the reasoning behind it? I've been struggling to see it for a little while. Any and all help will be appreciated :).

Since $u(x):=\operatorname{sign}(x)\ln|x| \in L^{1}_{\text{loc}}(\mathbb{R})$ we can take
$$
\frac{1}{|x|}:=u'(x),
$$
i.e. for $\varphi\in C_c^\infty(\mathbb{R})$ we set
$$\begin{align}
\left< \frac{1}{|x|}, \varphi(x) \right>
&:= - \int_{-\infty}^{\infty} \operatorname{sign}(x)\ln|x| \, \varphi'(x) \, dx
\\&= \lim_{\epsilon\to 0} \left( \int_{-\infty}^{-\epsilon} \ln|x|\,\varphi'(x)\,dx - \int_{\epsilon}^{\infty} \ln|x|\,\varphi'(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( \left[\ln|x|\,\varphi(x)\right]_{-\infty}^{-\epsilon} - \int_{-\infty}^{-\epsilon} \frac{1}{x} \varphi(x)\,dx - \left[\ln|x|\,\varphi(x)\right]_{\epsilon}^{\infty} + \int_{\epsilon}^{\infty} \frac{1}{x}\,\varphi(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( \ln\epsilon\,\varphi(-\epsilon) - \int_{-\infty}^{-\epsilon} \frac{1}{x} \varphi(x)\,dx + \ln\epsilon\,\varphi(\epsilon) + \int_{\epsilon}^{\infty} \frac{1}{x}\,\varphi(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( \ln\epsilon\,(\varphi(-\epsilon)+\varphi(\epsilon)) + \int_{|x|>\epsilon} \frac{1}{|x|}\,\varphi(x)\,dx \right)
\\&= \lim_{\epsilon\to 0} \left( 2\ln\epsilon\,\varphi(0) + \int_{|x|>\epsilon} \frac{1}{|x|}\,\varphi(x)\,dx \right)
.\end{align}$$
