I don't know if the formula is what you want, but it seems that they are indeed working with distributions (generalized functions). The use of the ##P.V.## notation is what made it very confusing since there is no principle value here. To me it appears the finite part of the integral is what is desired, as is done when defining pseudofunctions in distribution theory. If the notation had been ##Pf## (for pseudofunction) or ##Fp## (for finite part) it would have been clear, although to be fair some authors (eg Gelfand) don't use any special notation to indicate such things. Assuming this interpretation, I first rewrite this as the simpler
$$
\begin{eqnarray*}
\frac{1}{2} Fp \int_{-\infty}^{\infty} \left(\frac{1}{x-x^\prime} - \frac{1}{\left|x-x^\prime\right|}\right) f(x^\prime) \, dx^\prime & = & - Fp \int_x^\infty \frac{f(x^\prime)}{x^\prime-x} dx^\prime
\end{eqnarray*}
$$
where I am using the symbol ##Fp## to denote the fact that integral as written diverges but that we will be regularizing it to make it convergent. Now we use the standard approach and define the singular distribution above as some finite-order derivative of a regular distribution. In this case, since for normal functions ##\frac{d}{dx^\prime}\ln(x^\prime-x)=1/(x^\prime-x)##, we just need to compute the distributional derivative of the distribution that is ##\ln(x^\prime-x)## when ##x^{\prime}>x## and zero otherwise. So we have
$$
\begin{eqnarray*}
Fp \int_x^\infty \frac{f(x^\prime)}{x^\prime-x} dx^\prime & = &- \int_x^\infty \ln(x^\prime-x) \frac{df(x^\prime)}{dx^\prime} \, dx^\prime \\
& = & -\lim_{\epsilon\rightarrow 0^+} \int_{x+\epsilon}^\infty \ln(x^\prime-x) \frac{df(x^\prime)}{dx^\prime} \, dx^\prime \\
& = & \lim_{\epsilon\rightarrow 0^+} \left[f(x) \, \ln\epsilon + \int_{x+\epsilon}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime\right] \\
& = & \lim_{\epsilon\rightarrow 0^+} \left[-f(x)\int_{x+\epsilon}^{x+1} \frac{1}{x^\prime-x} dx^\prime + \int_{x+\epsilon}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime \right] \\
& = & \lim_{\epsilon\rightarrow 0^+} \int_{x+\epsilon}^{x+1} \frac{f(x^\prime)-f(x)}{x^\prime-x} dx^\prime + \int_{x+1}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime \\
& = & \int_{x}^{x+1} \frac{f(x^\prime)-f(x)}{x^\prime-x} dx^\prime + \int_{x+1}^\infty \frac{f(x^\prime)}{x^\prime - x} dx^\prime
\end{eqnarray*}
$$
Here the first line simply uses the definition of the distributional derivative (it should be familiar to anyone who has seen how the derivative of the delta function behaves), the third line results from integration by parts, and the fourth line is simply rewriting the logarithm in a useful way. Besides the first integral with the ##Fp## in front of it, every expression above is convergent and all of the steps are justified as long as ##f## is "nice" enough. For sure if ##f## is a Schwartz function (
https://en.wikipedia.org/wiki/Schwartz_space) then this holds, but a weaker set of requirements should also be sufficient.
jason
