# Principal value of 1/x

I'm trying to show that p.v.$\frac{1}{x}(\varphi):= \lim_{\epsilon \to 0^{+}}\int_{|x|> \epsilon} \frac{\varphi(x)}{x}dx, \varphi \in \mathcal{S}$ exists and defines a tempered distribution, where $\mathcal{S}$ denotes Schwartz space.

First step is I split the integral up:

p.v.$\frac{1}{x}(\varphi) = \int_{|x|<1}\frac{\varphi(x)-\varphi(0)}{x}dx + \int_{|x|>1} \frac{\varphi(x)}{x}dx$

Now I notice that the first integral on the right hand side is essentially the mean value theorem so is equal to some derivative of $\varphi$.

Question:

• How do i deal with the second integral?
• I know that $\int \frac{1}{x}dx$ is not integrable near the origin due to it have a singularity, so are my steps justified so far and if so why?

First question. \begin{align*} \int_{|x|>1}\dfrac{|\varphi(x)|}{x}dx&=\int_{|x|>1}\dfrac{|x\varphi(x)|}{x^{2}}dx\\ &\leq\sup_{x\in{\bf{R}}}|x\varphi(x)|\int_{|x|>1}\dfrac{1}{x^{2}}dx\\ &=2\sup_{x\in{\bf{R}}}|x\varphi(x)|. \end{align*} Second question. You should write $\lim_{\epsilon\rightarrow 0^{+}}\displaystyle\int_{\epsilon<|x|<1}\dfrac{\varphi(x)-\varphi(0)}{x}dx$. Now we see that \begin{align*} \chi_{\epsilon<|x|<1}\left|\dfrac{\varphi(x)-\varphi(0)}{x}\right|\leq\chi_{|x|<1}\|\varphi'\|_{L^{\infty}({\bf{R}})}, \end{align*} and $\chi_{|x|<1}\|\varphi'\|_{L^{\infty}({\bf{R}})}\in L^{1}({\bf{R}})$, so Lebesgue Dominated Convergence Theorem applies.
• why is $\chi_{|x|<1}||\varphi'||_{L^{\infty}(\mathbb{R})} \in L^{1}(\mathbb{R})$ – VBACODER Mar 26 '18 at 9:23
• That $\|\varphi'\|_{L^{1}({\bf{R}})}$ is a constant, now $\displaystyle\int_{{\bf{R}}}\chi_{|x|<1}dx=v_{n}<\infty$, the volume of the unit ball. – user284331 Mar 26 '18 at 17:06