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

Question

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.

Answer 1

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

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

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

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

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

Answer 2

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