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Finite part (Partie finie) of the mapping $x \mapsto \frac{1}{{{x^2}}}$ is a regular distribution defined by $$\left\langle {{\text{Pf}}\frac{1}{{{x^2}}},\varphi } \right\rangle = \mathop {\lim }\limits_{\varepsilon \to 0 + } \left( {\int_{ - \infty }^{ - \varepsilon } {\frac{{\varphi \left( x \right)}}{{{x^2}}}dx} + \int_\varepsilon ^{ + \infty } {\frac{{\varphi \left( x \right)}}{{{x^2}}}dx} - \frac{{2\varphi \left( 0 \right)}}{\varepsilon }} \right)$$

Principal value of the mapping $x \mapsto \frac{1}{x}$ is a regular distribution defined by

$$\left\langle {{\text{vp}}\frac{1}{x},\varphi } \right\rangle = \mathop {\lim }\limits_{\varepsilon \to 0 + } \left( {\int_{ - \infty }^{ - \varepsilon } {\frac{{\varphi \left( x \right)}}{x}dx} + \int_\varepsilon ^{ + \infty } {\frac{{\varphi \left( x \right)}}{x}dx} } \right)$$

I should prove that ${\left( {{\text{vp}}\frac{1}{x}} \right)^\prime } = {\text{Pf}}\frac{1}{{{x^2}}}$.

However, $$\int_{ - \infty }^{ - \varepsilon } {\frac{{\varphi \left( x \right)}}{{{x^2}}}dx} + \int_\varepsilon ^{ + \infty } {\frac{{\varphi \left( x \right)}}{{{x^2}}}dx} = \frac{{\varphi \left( { - \varepsilon } \right)}}{\varepsilon } + \int_{ - \infty }^{ - \varepsilon } {\frac{{\varphi '\left( x \right)}}{x}dx} + \frac{{\varphi \left( \varepsilon \right)}}{\varepsilon } + \int_\varepsilon ^{ + \infty } {\frac{{\varphi '\left( x \right)}}{x}dx} $$ so $$\left\langle {{\text{Pf}}\frac{1}{{{x^2}}},\varphi } \right\rangle = \mathop {\lim }\limits_{\varepsilon \to 0 + } \left( {\frac{{\varphi \left( { - \varepsilon } \right)}}{\varepsilon } + \int_{ - \infty }^{ - \varepsilon } {\frac{{\varphi '\left( x \right)}}{x}dx} + \frac{{\varphi \left( \varepsilon \right)}}{\varepsilon } + \int_\varepsilon ^{ + \infty } {\frac{{\varphi '\left( x \right)}}{x}dx} - \frac{{2\varphi \left( 0 \right)}}{\varepsilon }} \right) = $$ $$\mathop {\lim }\limits_{\varepsilon \to 0 + } \left( {\int_{ - \infty }^{ - \varepsilon } {\frac{{\varphi '\left( x \right)}}{x}dx} + \int_\varepsilon ^{ + \infty } {\frac{{\varphi '\left( x \right)}}{x}dx} } \right) + \mathop {\lim }\limits_{\varepsilon \to 0 + } \left( {\frac{{\varphi \left( { - \varepsilon } \right) + \varphi \left( \varepsilon \right) - 2\varphi \left( 0 \right)}}{\varepsilon }} \right)\mathop = \limits^{{\text{L'H}}} $$ $$\left\langle {{\text{vp}}\frac{1}{x},\varphi '} \right\rangle + \mathop {\lim }\limits_{\varepsilon \to 0 + } \left( { - \varphi '\left( { - \varepsilon } \right) + \varphi '\left( \varepsilon \right)} \right) = - \left\langle {{{\left( {{\text{vp}}\frac{1}{x}} \right)}^\prime },\varphi } \right\rangle = \left\langle { - {{\left( {{\text{vp}}\frac{1}{x}} \right)}^\prime },\varphi } \right\rangle $$ which implies ${\left( {{\text{pv}}\frac{1}{x}} \right)^\prime } = - {\text{Pf}}\frac{1}{{{x^2}}}$.

Did I make a mistake somewhere, or is the problem wrongly stated?

EDIT: Test functions are real valued on $\mathbb{R}$ ($\varphi \in \mathcal{D}\left( \mathbb{R} \right)$), so there's no conjugation.

mechanodroid
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Alen
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    There is an error: you are diving the $\lim_{\varepsilon\to 0^+}$ in the second line after "so". This is wrong since that's legit only when the two limit quantities exist finite, which is not the case for the pv$\frac{1}{x^2}$ – Tommaso Seneci Nov 17 '17 at 15:24

1 Answers1

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Your computation is correct. Someone dropped the minus sign when stating the exercise. Remembering the calculus formula $(1/x)'=-1/x^2$ would help prevent that. The calculus formula doesn't prove the result for distributions, but it must be consistent with it in the sense that differentiation away from singularities is the same as classical derivative.

Philippe Malot
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