Let $\mathcal C$ and $\mathcal H$ denote the Cauchy principal value and Hadamard finite part. According to the [Wiki](https://en.wikipedia.org/wiki/Hadamard_regularization):
$$
{\frac {\mathrm d}{\mathrm dx}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)={\mathcal {H}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,\mathrm dt\quad ({\hbox{for }}a<x<b).
$$
But why? By definition this is stating
$$
\frac{\mathrm d}{\mathrm dx}\lim\_{\epsilon\to 0^+}\int\_{[a,b]\setminus(x-\epsilon,x+\epsilon)}\frac{f(t)}{t-x}\,\mathrm dt=\lim\_{\epsilon\to0^+}\left(\int\_{[a,b]\setminus(x-\epsilon,x+\epsilon)}\frac{f(t)}{(t-x)^2}\,\mathrm dt-\frac{f(x+\epsilon)+f(x-\epsilon)}{\epsilon}\right).
$$
Does this also generalize to higher order derivatives? In such a case, the Cauchy principal value integral becomes a generating function for regularizing the higher order singular integrals involving $(x-t)^{-n}$. In other words:
$$
{\frac {\mathrm d^n}{\mathrm dx^n}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)=n!\,{\mathcal {P}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{n+1}}}\,\mathrm dt,
$$
with $\mathcal P$ being the principal value described by [Charles Fox](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/generalization-of-the-cauchy-principal-value/34E58E1057169989D3CCCD0BCD76B0DE).



Derivative of Cauchy PV is equivalent to Hadamard regularization?
A derivation of the relation
$${\frac {\mathrm d}{\mathrm dx}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)={\mathcal {H}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,\mathrm dt$$
is given by W.T. Ang in [Notes on Cauchy principal and Hadamard finite-part integrals](https://www3.ntu.edu.sg/home/MWTANG/hypersie.pdf) (page 3).


The generalization to higher derivatives is in [A new algorithm for Cauchy principal value and Hadamard finite-part integrals](https://doi.org/10.1016/S0377-0427(96)00142-2)
$$\frac{1}{n!}{\frac {\mathrm d^n} {\mathrm dx^n}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)={\mathcal {H}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{n+1}}}\,\mathrm dt,\;\;n=1,2,\ldots$$


