5

I have been looking at the proof for $\frac{d}{dx}\ln|x|=\operatorname{p.v.}\left(\frac{1}{x}\right)$ in the context of distributions and I am having trouble understanding why in the second term after integration by parts the limits we are subbing for $x$ are $\epsilon$ and $-\epsilon$ (the bit is highlighted in blue). The reason for my confusion is that $[a,-a]\backslash[-\epsilon,\epsilon]$ means $[-a,-\epsilon]\cup[\epsilon,a]$ so I am not sure why we are not using this for limits in the integral instead.

Please note that $\operatorname{p.v.}\left(\frac{1}{x}\right)$ is Cauchy principal value of $\frac{1}{x}$ defined as: $$\left\langle \operatorname{p.v.}\left(\frac{1}{x}\right), \phi\right\rangle = \lim_{\epsilon\to0} \int_{|x|>\epsilon} \frac{1}{x}\phi(x)\,dx$$

Here is the proof I have been referring to:

For any $\phi\in\mathcal D(\mathbb R)$ (i.e. it is a test function) with $\operatorname{supp}\phi = [a,-a]$ we have:

\begin{align*} \left\langle\frac{d}{dx}\ln|x|, \phi\right\rangle &=-\langle \ln|x|, \phi'\rangle\\ &= - \lim_{\epsilon\to0} \int_{\mathbb R\backslash[-\epsilon,\epsilon]} \ln|x|\phi'(x) dx\\ &=- \lim_{\epsilon\to0}\left[\int_{[a,-a]\backslash[-\epsilon,\epsilon]} \frac{1}{x}\phi(x) dx + \color{blue}{(\ln|\epsilon|)(\phi(\epsilon)-\phi(-\epsilon))}\right]\\ &= - \lim_{\epsilon\to0} \int_{[a,-a]\backslash[-\epsilon,\epsilon]} \frac{1}{x}\phi(x) dx\\ &=\left\langle\operatorname{p.v.}\left(\frac{1}{x}\right), \phi\right\rangle \end{align*}

Stackman
  • 4,028
LiNXO
  • 345
  • Integrate by parts $\int_{\epsilon}^a \ln |x| \phi'(x)dx$ then do the same with $\int_{-a}^{-\epsilon} \ln |x| \phi'(x)dx$, both have a divergent term which cancel out in the sum (since $\phi'(x) = \phi'(0)+O(x)$) – reuns Oct 09 '18 at 22:18
  • Thank you for your comment. When integrating by parts I get to what is stated in the proof an additional term $ln|a|(\phi(a)-\phi(-a))$ but don't quite see what you mean by the divergent term there. Also I am not sure how in the Taylor expansion of $\phi'(x)$ we can just omit terms of order x or higher since for example $a$ is not necessarily $<<1$? – LiNXO Oct 10 '18 at 07:26
  • The factor $\ln |\epsilon|$ is divergent, but the term $(\ln |\epsilon|) ( \phi(\epsilon) - \phi(-\epsilon) )$ is not. That tends to zero: $$(\ln |\epsilon|) ( \phi(\epsilon) - \phi(-\epsilon) ) = 2 (\epsilon \ln |\epsilon|) \frac{\phi(\epsilon) - \phi(-\epsilon)}{2\epsilon} \to 2 \cdot 0 \cdot \phi'(0) = 0.$$ – md2perpe Oct 12 '18 at 17:25
  • @md2perpe Yes, I agree that the expression in blue tends to zero. However, what I do not understand is where does the additional term $ln|a|(\phi(a)−\phi(−a))$ disappear to from integration by parts, since the limits of the integral are $[-a,-\epsilon]\cup[\epsilon,a]$? – LiNXO Oct 12 '18 at 20:33
  • 1
    You are forgetting that $f(a)=f(-a)=0$, by definition of "support". – Giuseppe Negro Oct 25 '20 at 22:45

1 Answers1

5

We can split the integral into two parts, one on $[\epsilon, \infty)$ and one on $(-\infty, -\epsilon]$.

We rewrite both parts using integration by parts: $$ \int_\epsilon^\infty \ln |x| \, \phi'(x) \, dx = \left[ \ln |x| \, \phi(x) \right]_\epsilon^\infty - \int_\epsilon^\infty \frac{1}{x} \, \phi(x) \, dx = -\ln \epsilon \, \phi(\epsilon) - \int_\epsilon^\infty \frac{1}{x} \, \phi(x) \, dx $$ and $$ \int_{-\infty}^{-\epsilon} \ln |x| \, \phi'(x) \, dx = \left[ \ln |x| \, \phi(x) \right]_{-\infty}^{-\epsilon} - \int_{-\infty}^{-\epsilon} \frac{1}{x} \, \phi(x) \, dx = \ln \epsilon \, \phi(-\epsilon) - \int_{-\infty}^{-\epsilon} \frac{1}{x} \, \phi(x) \, dx $$

Thus, $$ \int_{|x|>\epsilon} \ln |x| \, \phi'(x) \, dx = \left( \ln \epsilon \, \phi(-\epsilon) - \ln \epsilon \, \phi(\epsilon) \right) - \int_{|x|>\epsilon} \frac{1}{x} \, \phi(x) \, dx \\ = ( \ln \epsilon ) \left( \phi(-\epsilon) - \phi(\epsilon) \right) - \int_{|x|>\epsilon} \frac{1}{x} \, \phi(x) \, dx. $$

md2perpe
  • 26,770
  • Thank you! This makes sense now, keeping $\infty$ on the limits instead of $a$ seems to make it easier to see the vanishing terms. – LiNXO Oct 14 '18 at 07:48