$\DeclareMathOperator{\vp}{v.p.}$ 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 :-/


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, $$\int_{\mathbb R} \ln |x| \phi'(x) \, dx = - \lim_{\epsilon \to 0^+} \int_{|x| \ge \epsilon} \frac 1x \phi(x) \, dx.$$

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  • $\begingroup$ well thank you very much! It seems all clear. In case I'll have any doubts I'll ask :) $\endgroup$ – Ant Jan 29 '15 at 16:49

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