Is this 'differentiated' form of the Sochocki-Plemelj theorem valid? For a complex function $f(z)$ continuous and on the real line, and $a, b > 0$ positive real numbers, the Sochocki-Plemelj theorem says that:
$$
\lim_{\epsilon \to 0^{+}}\int_{-a}^{b} \frac{f(x)}{x \pm i \epsilon} \ = \ \mp i \pi f(0) + \mathscr{P}\int_{-a}^{b} \frac{f(x)}{x} dx
$$
I am from physics, and often this is written as $\lim\limits_{\epsilon \to 0^{+}}\frac{1}{z \pm i \epsilon} = \mp i \pi \delta(z) + \mathscr{P}\left( \frac{1}{z} \right)$. Naiively, I have differentiated this 'physicist' expression to get $\lim\limits_{\epsilon \to 0^{+}}\frac{1}{(z \pm i \epsilon)^2} = \pm i \pi \delta'(z) + \mathscr{P}\left( \frac{1}{z^2} \right)$, and wondered if the following statement might also hold:
$$
\lim_{\epsilon \to 0^{+}}\int_{-a}^{b} \frac{f(x)}{(x \pm i \epsilon)^{2}} \ = \ \pm i \pi f'(0) + \mathscr{P}\int_{-a}^{b} \frac{f(x)}{x^2} dx
$$
Obviously this lacks rigour, but is there any chance the above identity might also hold? Maybe there would have to be some extra assumptions about the function $f$?
 A: The differentiated form is perfectly valid in the distributional sense, with the limit understood as the distributional limit as well. It says that when you apply the functionals on the lhs to $f$ and take the limit of the resulting numerical sequence, the outcome will be the same as applying the functional on the rhs to $f$.
Interchanging the limit with differentiation is valid as well.
The integral form states exactly the same thing in terms of the action of the functionals on $f$.
$\mathscr P(1/x^2)$ and $\mathscr P \int f(x)/x^2 dx$ are being used as symbolic notations for the functional and its action on $f$:
$$\left( \mathscr P \!\left( \frac 1 {x^2} \right), f \right) =
\left( \mathscr P \!\left( \frac 1 x \right), f' \right) =
\operatorname{v.\!p.} \int_{-\infty}^\infty \frac {f'(x)} x dx = \\
\int_{-\infty}^\infty \frac {f(x) - f(0) - x f'(0) [-1 < x < 1]} {x^2} dx,$$
which shows why the notation $\mathscr P \int f(x)/x^2 dx$ can be misleading: it's not $\operatorname{v.\!p.} \int f(x)/x^2 dx$ (which is to be expected, $\operatorname{v.\!p.} \int f(x)/x^2 dx$ is not defined for every $f$).
