$ \lim_{ \varepsilon \rightarrow 0^+ } \int_{|x| \geq \varepsilon} \frac{ \varphi(x) }{x}dx = - \int_{-\infty}^\infty \phi'(x) \ln(|x|) dx$

Question

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

Answer 1

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 $$\lim_{\epsilon\rightarrow 0+}\int_{|x|\ge \epsilon}\varphi^\prime(x)\log|x|dx=\int_{-\infty}^\infty \varphi^\prime(x)\log|x|dx$$