$ \lim_{ \varepsilon \rightarrow 0^+ } \int_{|x| \geq \varepsilon} \frac{ \varphi(x) }{x}dx = - \int_{-\infty}^\infty \phi'(x) \ln(|x|) dx$ How do I prove that
$$ \lim_{ \varepsilon \rightarrow 0^+ } \int_{|x| \geq \varepsilon} \frac{ \varphi(x) }{x}dx  = - \int_{-\infty}^\infty \phi'(x) \ln(|x|)dx $$
for all $ \varphi \in C_0^{\infty} (\mathbb{R})?$
I was starting as follows
$$ \lim_{ \varepsilon \rightarrow 0^+ } \int_{|x| \geq \varepsilon} \frac{ \varphi(x) }{x}dx  = \lim_{ \varepsilon \rightarrow 0^+ } \left( \int_{\varepsilon}^{\infty}  +\int_{-\infty}^{- \varepsilon}  \right) \frac{ \varphi(x) }{x}dx $$
$$ = \lim_{ \varepsilon \rightarrow 0^+ } \left( \varphi(x)\ln(x)\big|_{\varepsilon}^{\infty} + \varphi(x)\ln(x)\big|_{-\infty}^{- \varepsilon} + \left( \int_{\varepsilon}^{\infty}  +\int_{-\infty}^{- \varepsilon}  \right)  \varphi'(x)  \ln(x) dx \right) $$
$$= \lim_{ \varepsilon \rightarrow 0^+ } \left(- \varphi(\varepsilon)\ln(\varepsilon) + \varphi(-\varepsilon)\ln(-\varepsilon) + \left( \int_{\varepsilon}^{\infty}  +\int_{-\infty}^{- \varepsilon}  \right)  \varphi'(x)  \ln(x) dx \right)  $$
but here I am stuck. Does anyone have any hint on how to proceed?
Thanks

When you integrate $1/x$ you should get a $\log|x|$, not $\log x$ (check it).
Your boundary term should thus be estimated like this: 
$$|(\varphi(-\epsilon)-\varphi(\epsilon))\ln(\epsilon)|\le(C\epsilon+O(\epsilon^2))|\ln(\epsilon)|\longrightarrow 0$$
as $\epsilon\rightarrow 0$, where the first inequality holds for $\epsilon$ small enough since $\varphi$ is differentiable at $0$.
Also note that $\log|x|$ is integrable at $0$, so your principal value term can be rewritten as
$$\lim_{\epsilon\rightarrow 0+}\int_{|x|\ge \epsilon}\varphi^\prime(x)\log|x|dx=\int_{-\infty}^\infty \varphi^\prime(x)\log|x|dx$$ 
