Derivative of a distribution $\DeclareMathOperator{\vp}{v.p.}$
We define $\vp \frac 1x \in \mathcal D'(\mathbb R)$ (the principal value of $\frac 1x$) as 
$$\left\langle \vp \frac 1x, \varphi \right \rangle = \lim_{\varepsilon \to 0_+} \int_{|x| > \varepsilon} \frac{\varphi(x)}{x} \ dx = \vp \int_\mathbb R \frac{\varphi(x)}x \ \ \forall \varphi \in \mathcal D(\mathbb R)$$
Now I am required to show that  $\displaystyle \frac{d}{dx}\ln |x| = \vp \frac 1x$
This is the solution, which I do not understand;

As $\ln |x| \in L^1_{loc} (\mathbb R) \subset \mathcal D'(\mathbb R)$,
  for all $\varphi \in \mathcal D(\mathbb R)$ we have 
$$\langle d / dx \ln |x|, \varphi \rangle = - \langle \ln |x|, \varphi
 '\rangle  = - \int_\mathbb R \ln|x|\varphi'(x)\ dx =$$$$= -
 \lim_{\varepsilon \to 0_+, R \to \infty} \int_{-R}^{-\varepsilon}
 \ln|x| \varphi'(x) \ dx + \int_{\varepsilon}^R \ln|x| \varphi'(x) \ dx
  = \vp \int_\mathbb R \frac{\varphi(x)}x$$$$= \langle \vp \frac 1x,
 \varphi \rangle$$

And the only comment to the above calculations is :
We have made an integration by parts, remembering that $\text{spt }\varphi$ is compact and $\lim_{\varepsilon \to 0} (\varphi(\varepsilon) - \varphi(-\varepsilon))\ln \varepsilon = 0$
Can somebody explain to me what is going on here? I do not understand this solution nor the comment :-/
 A: First of all note that $\ln |x|$ is integrable near $0$ so that for any $\phi \in \mathcal{D}'$ the integral $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx$$ is defined, and that $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx.$$
Now,
$$\int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx = \int_{-\infty}^{-\epsilon} \ln(-x) \phi'(x) \, dx + \int_{\epsilon}^\infty \ln x \phi'(x) \, dx$$
so that (taking into account that $\phi$ is compactly supported means the limits at infinity do not contribute to the boundary terms) an integration by parts on each integral separately gives
\begin{align*}\int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx &= \ln(\epsilon) \phi(-\epsilon) - \int_{-\infty}^{-\epsilon} \frac 1x \phi(x) \, dx - \ln(\epsilon) \phi(\epsilon) - \int_\epsilon^\infty \frac 1x \phi(x) \, dx\\ &= -\ln(\epsilon) [\phi(\epsilon) - \phi(-\epsilon)] - \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.\end{align*}
As for the first term, note that
$$ \lim_{\epsilon \to 0^+} \ln(\epsilon) [\phi(\epsilon) - \phi(-\epsilon)] = \lim_{\epsilon \to 0^+} 2\epsilon \ln(\epsilon) \cdot \frac{\phi(\epsilon) - \phi(-\epsilon)}{2\epsilon} = 0 \cdot \phi'(0) = 0.$$
Thus
$$ \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \ln|x| \phi'(x) \, dx = - \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.$$
That is,
$$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = - \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.$$
