An "asymmetric" Cauchy principal value distribution I'm familiar with the usual Cauchy principal value distribution:
$$
PV[\phi(x)] := \lim_{\epsilon\to 0^+}\int_{\mathbb{R}\setminus[-\epsilon,\epsilon]}{\dfrac{\phi(x)}{x}}\,dx = \int_{0}^{\infty}{\dfrac{\phi(x)-\phi(-x)}{x}}\,dx.
$$
But I was wondering how one might analyze an "asymmetric" version:
$$
PV_{a}^{b}[\phi(x)] := \lim_{\epsilon\to 0^+}\int_{\mathbb{R}\setminus[-a\epsilon,b\epsilon]}{\dfrac{\phi(x)}{x}}\,dx, \quad a,b>0.
$$
Even playing around with a simple example of $a=1$ and $b=2$ has left me with a distribution that I'm unsure of how to simplify:
\begin{align}
PV_{1}^{2}[\phi(x)] &:= \lim_{\epsilon\to 0}\int_{\mathbb{R}\setminus[-\epsilon,2\epsilon]}{\dfrac{\phi(x)}{x}}\,dx \\
&= PV[\phi(x)] - \lim_{\epsilon\to 0^+}\int_{\epsilon}^{2\epsilon} {\dfrac{\phi(x)}{x}}\,dx.
\end{align}
Obviously $\int_{\epsilon}^{2\epsilon}f(x)\,dx \to 0$ if $f(x)$ is integrable at $x=0$, but when $f(x) = \dfrac{\phi(x)}{x}$ it's not clear to me what this integral should converge to. Perhaps there is some way to formulate this action on $\phi(x)$ in terms of the delta distribution? Or maybe in terms of the Fourier transform of some other familiar distribution? I feel like this shouldn't be that hard to analyze, but I'm currently drawing a blank.
 A: Note that we can write
$$\begin{align}
\int_{-\infty}^{-|a|\varepsilon}\frac{\phi(x)}{x}\,dx+\int_{|b|\varepsilon}^\infty \frac{\phi(x)}{x}\,dx&=\int_{-\infty}^{-|a|\varepsilon}\frac{\phi(x)}{x}\,dx+\int_{|a|\varepsilon}^\infty \frac{\phi(x)}{x}\,dx-\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{\phi(x)}{x}\,dx
\end{align}$$
We assume that the Cauchy Principal Value exists and wish to know whether the limit $\displaystyle \lim_{\varepsilon\to 0^+}\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{\phi(x)}{x}\,dx$ exists.
While it is not a necessary condition, it is straightforward to show that if $\lim_{x\to 0^+}\phi(x)=C$ then
$$\lim_{\varepsilon\to 0^+}\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{\phi(x)}{x}\,dx=C\log(|b/a|)\tag1$$
To see that this condition is not necessary, suppose $\phi(x)=\sin(1/x)$.  Then, we have
$$\begin{align}
\lim_{\varepsilon\to 0^+}\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{\sin(1/x)}{x}\,dx&=\lim_{\varepsilon\to 0^+}\int_{1/|b|\varepsilon}^{1/|a|\varepsilon}\frac{\sin(x)}{x}\,dx\\\\
&=0
\end{align}$$


A question was asked about the limit in $(1)$.  So, let's proceed with its proof.
If $\lim_{x\to 0^+}\phi(x)=C$, then for any $\nu>0$, there exists a $\delta>0$ such that whenever $0<x<\delta$, $|\phi(x)-C|<\nu$.
So, select any $\nu>0$ and find any appropriate $\delta>0$ that works.  Then, for $\varepsilon>0$ such that $\max(|b|,|a|)\varepsilon<\delta$ we have
$$\begin{align}
\left|\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{\phi(x)}{x}\,dx-C\log(|b/a|)\right|&=\left|\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{\phi(x)-C}{x}\,dx\right|\\\\
&\le \text{sgn}(|b|-|a|)\int_{|a|\varepsilon}^{|b|\varepsilon}\frac{|\phi(x)-C|}{x}\,dx\\\\
&<\nu |\log(|b/a|) |
\end{align}$$
And we are done.  Note that if $\phi$ is continuous from the right-hand side at $0$, then $\phi(0)=C$.
A second question asked how to show that if $\phi$ is continuous, then $\lim_{\varepsilon\to 0}\left(\phi(|b|\varepsilon)\log(|b|\varepsilon)-\phi(|a|\varepsilon)\log(|a|\varepsilon)\right)=\phi(0)\log(|b/a|)$.
To proceed, we simply note that if $\phi$ is differentiable, then $\phi(|a|\varepsilon)=\phi(0)+o(|a|\varepsilon)$.  And the rest is straightforward.
