# Derivative of $\ln |x|$ in the distributional sense

Consider the function $$\ln |x|$$, since it is locally integrable we can form the distribution

$$(\ln |x|,\phi)=\int_{-\infty}^{\infty}\ln |x|\phi(x)dx.$$

Now, I want to show that in the sense of distributions we have $$\ln |x|' = \operatorname{Pv}\frac{1}{x}$$. My obvious try was to substitute directly the definition of the derivative for distributions:

$$(\ln |x|', \phi)=-(\ln |x|,\phi')=-\int_{-\infty}^{\infty} \ln |x|\phi'(x)dx,$$

the obvious thing to do would be to split this in two integrals:

$$(\ln |x|', \phi)=-\int_{-\infty}^0 \ln(-x)\phi'(x)dx - \int_0^\infty \ln (x) \phi'(x)dx.$$

Now, I've seem the question Derivative of a distribution and it tells what to do next: we simply rewrite all of that as

$$(\ln |x|,\phi')=\lim_{\epsilon\to 0^+}\int_{-\infty}^{-\epsilon}\ln |x|\phi'(x)dx+\int_{\epsilon}^\infty \ln |x|\phi'(x)dx$$

But I can't understand where this limit comes from. I mean how does one get from where I stopped all the way to this line? Because obviously after that just inetgration by parts is enough to get what we want.

My doubt is how that $$\epsilon \to 0^+$$ really appeared.

• Did you try to use integration by parts to switch the derivative back onto $\ln x$ and $\ln(-x)$? At least away from $0$, I'd expect your distributional derivative to coincide with the derivative $\frac{1}{x}$ – Roland Mar 31 '16 at 17:06
• If we do this we get that the derivative in the sense of distributions in this case coincides with the usual derivative in the sense that $\ln' |x| = 1/x$ right? But what about the point zero? In the usual derivative it would not even make sense to ask, because $\ln |x|$ isn't defined there. But in the sense of distributions it doesn't matter? – user1620696 Mar 31 '16 at 17:07
• Possible duplicate of Derivative of a distribution – Umberto P. Mar 31 '16 at 17:21
• Thanks for pointing me to that other question. I saw your answer there but there's a minor detail I didn't understand. Something I believe to be very basic. I added an edit here about it. My whole doubt is: if $(\ln |x|,\phi)$ is defined as $\int_{\mathbb{R}}\ln |x|\phi(x)dx$, why this suddenly turned into a principal value? – user1620696 Mar 31 '16 at 17:50
• $\ln |x|$ is integrable on a neighborhood of $0$; $1/x$ is not. – Umberto P. Mar 31 '16 at 17:52

The derivative of $\ln(|x|)$ is $\text{Vp} \frac{1}{x}$:

$$\langle \ln(|x|)', \phi \rangle = -\int_{-\infty}^\infty \ln(|x|) \phi'(x) dx$$

$$=- \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \ln(|x|) \phi'(x) dx$$

$$=\lim_{\epsilon \to 0} \underbrace{-\ln(|\epsilon|) (\phi(-\epsilon)-\phi(\epsilon))}_{\to 0} + \int_{|x|>\epsilon} \frac{ \phi(x)}{x} dx$$

$$=\lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{ \phi(x)}{x} dx$$

• Thanks for answering @Tryss. I understood almost all of the reasoning. The only thing that confuses me is: why do we have $$\int_{-\infty}^\infty \ln(|x|) \phi'(x) dx=\lim_{\epsilon \to 0} \int_{|x|>\epsilon} \ln(|x|) \phi'(x) dx$$ I mean, I can't get where this limit comes from in this case. After all the integral was over the entire line. Thanks again! – user1620696 Apr 8 '16 at 0:50

To complete your computation, integrate by parts, using the fact that $\phi(x)-\phi(0)$ has the same derivative as $\phi(x)$. The result: $(\ln'|x|,\phi) =\int_{-\infty}^\infty {\phi(x)-\phi(0)\over x}\,dx$.