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I am working through some distribution theory notes, and was specifically working on this example Derivative of principal value distribution $1/x^2$ is equal to finite part distribution $-1/x^2$?.

You can reformulate the parte finie as $\lim\limits_{\epsilon \to 0} \int\limits_{|x| \geq \epsilon} \frac{\phi(x) - \phi(0)}{x^2} dx$.

I'm fine with everything except the very first statement, as I do not understand why this integral is finite for $\phi \in \mathscr{D}(\mathbb{R})$ (the class of infinitely differentiable functions with compact support). I attempted to prove it, but of course because of the fact that there is no control of $\frac{1}{x^2}$ in a neighborhood of $0$ we can't control the integral at that particular singularity in the usual way (bounding the quantity $|\phi(x) - \phi(0)|$ by something small). I attempted to consider a similar proof to that of proving the Riemann integral of $\frac{\sin(x)}{x}$ converges, but that doesn't work either, which got me thinking of a counterexample. We can most certainly find a $\phi \in \mathscr{D}(\mathbb{R})$ such that $\phi(x)|_{[0,1]} = \sin(x)$. Then $\int_0^1 \frac{\phi(x)-\phi(0)}{x^2} dx = \int_0^1 \frac{\sin(x)}{x^2}$, but this is infinite. Throw some absolute values on, bound above by the integral over all of $\mathbb{R}$, it seems to indicate that the parte finie should be infinite. I feel like I'm either missing something to do with the Cauchy integral theorem (although with a generic function like $\phi \in \mathscr{D}(\mathbb{R})$ I'm not entirely sure how to use it) or something else entirely. Any help would be appreciated. Thanks!

mathishard.butweloveit
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1 Answers1

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Actually for $\varphi \in \mathscr{D}(\mathbb{R})$ we have that \begin{align} \left\langle {{\text{Pf}}\frac{1}{{{x^2}}},\varphi } \right\rangle &= \lim_{\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)\\ &= \lim_{\varepsilon \to 0^+ } \int_\varepsilon ^{ + \infty } {\frac{{\varphi \left( x \right) + \varphi(-x) - 2\varphi(0)}}{{{x^2}}}dx}\\ &= \int_0^{+\infty}{\frac{{\varphi \left( x \right) + \varphi(-x) - 2\varphi(0)}}{{{x^2}}}dx} \end{align}

where the last integral converges since $$x \mapsto {\frac{{\varphi \left( x \right) + \varphi(-x) - 2\varphi(0)}}{{{x^2}}}} $$ is a continuous function on $[0, \infty\rangle$ with compact support, and hence integrable. Indeed, we have $$\lim_{x\to 0^+} \frac{{\varphi \left( x \right) + \varphi(-x) - 2\varphi(0)}}{{{x^2}}} \mathop = \limits^{{\text{L'H}}} \lim_{x\to 0^+} \frac{\varphi' \left( x \right) - \varphi'(-x)}{2x} \mathop = \limits^{{\text{L'H}}} \lim_{x\to 0^+} \frac{\varphi'' \left( x \right) + \varphi''(-x)}{2} = \varphi''(0)$$

mechanodroid
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