Finite part of $-1/x^2$

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I'm learning the basic of Distributional Theory. I ended up solving the following exercise: 'Find the distributional derivative of $P.V.1/x$'. After few computation, I arrived at the following:

$$\left\langle\left(P.V.\frac{1}{x}\right)', \phi\right\rangle = \lim_{\epsilon \rightarrow 0^{+}}\bigg(\frac{2\phi(0)}{\epsilon} + \int_{|x| \geq \epsilon} \phi(x) \bigg(-\frac{1}{x^2}\bigg)dx\bigg).$$

I've discovered that $P.V.\dfrac{1}{x}$ has a short form, without the limit:

$$\left\langle P.V.\dfrac{1}{x}, \phi\right\rangle = \int_{-1}^{1} \frac{\phi(x) -\phi(0)}{x}dx + \int_{|x| > 1}\frac{\phi(x)}{x}dx.$$

Could I find a similar form for partie finie? Thanks for your help.

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Yes, there is an analogous way to write the distributional derivative of $\text{PV}\left(\frac1x\right)$. In fact, I used a distribution for $|x|^s$, $s\in \mathbb{R}$ in THIS ANSWER to determine the Fourier Transform of $|x|^s$.

Let $\psi(x)$ be the distribution $\text{PV}\left(\frac1x\right)$. Then, for $\phi\in C_C^\infty$ we have

$$\begin{align} \langle \psi',\phi\rangle&=-\langle\psi,\phi'\rangle\tag1\\\\ &=-\text{PV}\int_{-\infty}^\infty \frac{\phi'(x)}x\,dx\tag2\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{|x|\ge\varepsilon}\frac{\phi'(x)}x\,dx\tag3\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{|x|\ge\varepsilon}\frac{\phi(x)-\phi(0)}{x^2}\,dx \tag4\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{1\ge|x|\ge\varepsilon}\frac{\phi(x)-\phi(0)}{x^2}\,dx -\int_{|x|\ge 1}\frac{\phi(x)-\phi(0)}{x^2}\,dx\tag5\\\\ &=-\lim_{\varepsilon\to 0^+}\int_{1\ge|x|\ge\varepsilon}\frac{\phi(x)-\phi(0)-\phi'(0)x}{x^2}\,dx -\int_{|x|\ge 1}\frac{\phi(x)-\phi(0)-\phi'(0)x}{x^2}\,dx\tag6\\\\ &=-\int_{|x|\le 1}\frac{\phi(x)-\phi(0)-\phi'(0)x}{x^2}\,dx-\int_{|x|\ge 1}\frac{\phi(x)-\phi(0)-\phi'(0)x}{x^2}\,dx\tag7\\\\ &=-\int_{-\infty}^\infty \frac{\phi(x)-\phi(0)-\phi'(0)x}{x^2}\,dx\tag8 \end{align}$$

And we are done.


NOTES:

Equation $(1)$ is the definition of the distributional derivative.

In going from $(1)$ to $(2)$ we applied the definition of the distribution $\text{PV}\left(\frac1x\right)$.

In going from $(2)$ to $(3)$ we used the definition of the principal value.

In going from $(3)$ to $(4)$, we integrated by parts the integral on the right-hand side of $(3)$ with $u=\frac1x$ and $v=\phi(x)-\phi(0)$. We also exploited the fact that $\phi \in C_C^\infty$.

In going from $(4)$ to $(5)$ we split the integral into the sum of integrals.

In going from $(5)$ to $(6)$ we exploited the fact$(i)$ $\frac1x$ is odd and well behaved for $x$ bounded away from $0$ and $(ii)$ the integral of an well behaved odd function around symmetric limits is zero.

In going from $(6)$ to $(7)$ we noted that the first integral on the right-hand side of $(6)$ is not improper.

In going from $(7)$ to $(8)$ we recombined the integrals.