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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.

Math Attack
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Nick
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    Partie Finie is (French?) not English, would ‘principal part’ be a good translation? Use \langle,\rangle to get angled brackets. – FShrike May 10 '23 at 20:20
  • I don't think so. I suppose 'principal part' denotes a different concept. I've never found an English translation of 'Partie finie' before – Nick May 10 '23 at 20:36
  • I think you mean this: https://en.wikipedia.org/wiki/Cauchy_principal_value – Moishe Kohan May 10 '23 at 21:19
  • Yes, I do. This is the definition of Cauchy-principal-value. For 'partie finie' I mean exactly the limit that I wrote in the question. In Distribution theory sources, they call it $P.F. -\frac{1}{x^2}$ – Nick May 10 '23 at 21:26
  • So, $\langle P.V. 1/x, \phi \rangle = \lim_{\epsilon \rightarrow 0^{+}} \int_{|x| \geq \epsilon} \frac{\phi(x)}{x}dx $ – Nick May 10 '23 at 22:00
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    The translation of "partie finie (de Hadamard)" is usually "(Hadamard) finite part" or Hadamard regularization https://en.m.wikipedia.org/wiki/Hadamard_regularization – LL 3.14 May 10 '23 at 22:05
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    And yes, there are similar formulas without limits, such as the one I used here https://math.stackexchange.com/questions/3723136/the-fourier-transform-of-1-p3/3724502#3724502 – LL 3.14 May 10 '23 at 22:09
  • https://www.cambridge.org/core/services/aop-cambridge-core/content/view/34E58E1057169989D3CCCD0BCD76B0DE/S0008414X00044448a.pdf/generalization_of_the_cauchy_principal_value.pdf – Aaron Hendrickson May 10 '23 at 23:50
  • https://math.stackexchange.com/questions/4654690/principal-value-integral-on-half-interval/4659798#4659798 – Aaron Hendrickson May 10 '23 at 23:52
  • I don't think any of the previous answers is the exact one to my problem – Nick May 11 '23 at 05:36
  • The Fourier Transform of $|x|^s$ for all real $s$ was answer HERE. – Mark Viola May 15 '23 at 02:33

1 Answers1

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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.

Mark Viola
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