Why the Hilbert transform make sense with limit form

Question

I am learning Hilbert transform. A tempered distribution called the principal value of $\frac{1}{x}$, abbrevaited $p.v.\frac{1}{x}$, is defined by $$p.v.\frac{1}{x}(\phi)=\lim_{\epsilon\rightarrow0}\int_{|x|>\epsilon}\frac{\phi(x)}{x}dx,$$ $\phi$ belongs to Schwartz class. I couldn't figure out why this limit exists.

Answer 1

Split the integral over $|x|>\epsilon$ into the integrals over $|x|> 1$ and $\epsilon<|x|\le 1$. The former clearly converges and doesn't depend on $\epsilon$. For the latter, use the identity $$ \int_{\epsilon<|x|\le 1}\frac{\phi(x)}{x}dx = \int_{\epsilon<|x|\le 1}\frac{\phi(x)-\phi(0)}{x}dx $$ The integrand on the right has a finite limit at $0$, namely $\phi'(0)$. So it can be extended to a continuous function on $[-1,1]$, and integrating such a function presents no difficulties.