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I cannot get this identity by using the condition:

$\frac{1}{x^{2}}=-\partial(\frac{1}{x})$, and the integrands are defined by continuity at $x=0$.

My reasoning goes as $$\langle -\partial\frac{1}{x},\phi\rangle=\langle \frac{1}{x},\partial \phi\rangle=\int^{\infty}_{-\infty}\frac{1}{x^{2}}\phi dx$$ and we can break it into $$\lim_{\epsilon\rightarrow 0}\int^{\infty}_{\epsilon}\frac{1}{x^{2}}\phi dx+\int^{-\epsilon}_{-\infty}\frac{1}{x^{2}}\phi dx+\int^{\epsilon}_{-\epsilon}\frac{1}{x^{2}}\phi dx$$ The first and second term adds up to $$\int^{\infty}_{\epsilon} \frac{\phi(x)+\phi(-x)}{x^{2}}dx$$ while the third term $$\int^{\epsilon}_{-\epsilon}\frac{1}{x^{2}}\phi(x) dx=\int^{\epsilon}_{0}\frac{\phi(x)+\phi(-x)}{x^{2}}$$ approximates $$2\int^{\epsilon}_{0}\frac{\phi(0)}{x^{2}}$$

But this last step is not really justified; and even if it works the result looks remarkably different from the one I wanted. So I decided to ask.

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I think there's some conditions missing in this question. Is $\phi(x)$ a continuous function or something like that? I think this identity is wrong. Consider $\phi(x)=1$ when $x\in [-1,1]$ and $\phi(x)=0$ on the rest of $\mathbb{R}$. Then the right side is 0 and the left side goes to infinity. – caozhu Jun 14 '12 at 2:04
$\phi\in C^{\infty}_{c}$ as for distributions. – Bombyx mori Jun 14 '12 at 2:05
Your $\phi(x)$ is not smooth, though. – Kerry Jun 14 '12 at 3:27
You are right, but we can do some smoothing at x=1 and x=-1, the right side will still be finite and the left side still goes to infinity. – caozhu Jun 14 '12 at 3:31
The left hand side does not go to infinity because $\langle \frac{1}{x^{2}},\phi\rangle=-\langle \frac{1}{x},\partial \phi\rangle$. The integral of $\int \frac{1}{x}\partial\phi$ should be finite by your definition. – Kerry Jun 14 '12 at 3:44
up vote 5 down vote accepted

Let $\phi$ be a smooth test function. Since $\big\langle \frac{1}{x^2}, \phi \big\rangle = \big\langle -\partial \left( \frac{1}{x} \right), \phi \big\rangle = \big\langle \frac{1}{x}, \partial \phi \big\rangle$, it suffices to show that

$$ \left< \frac{1}{x}, \partial \phi \right> = PV \int_{-\infty}^{\infty} \frac{\partial \phi (x)}{x} \; dx = \int_{0}^{\infty} \frac{\phi(x) + \phi(-x) - 2\phi(0)}{x^2} \; dx.$$

But this follows from

$$ \begin{align*} PV \int_{-\infty}^{\infty} \frac{\partial \phi (x)}{x} \; dx &= \lim_{\epsilon\downarrow 0} \left( \int_{-\infty}^{-\epsilon} \frac{\partial \phi (x)}{x} \; dx + \int_{\epsilon}^{\infty} \frac{\partial \phi (x)}{x} \; dx \right) \\ &= \lim_{\epsilon\downarrow 0} \int_{\epsilon}^{\infty} \frac{\partial \phi (x) - \partial \phi(-x)}{x} \; dx \\ &= \lim_{\epsilon\downarrow 0} \left( \left. \frac{\phi (x) + \phi(-x)}{x} \right|_{\epsilon}^{\infty} + \int_{\epsilon}^{\infty} \frac{\phi (x) + \phi(-x)}{x^2} \; dx \right) \\ &= \lim_{\epsilon\downarrow 0} \left( - \frac{\phi (\epsilon) + \phi(-\epsilon)}{\epsilon} + \int_{\epsilon}^{\infty} \frac{\phi (x) + \phi(-x)}{x^2} \; dx \right) \\ &= \lim_{\epsilon\downarrow 0} \left( \frac{2\phi(0) - \phi (\epsilon) - \phi(-\epsilon)}{\epsilon} + \int_{\epsilon}^{\infty} \frac{\phi (x) + \phi(-x) - 2\phi(0)}{x^2} \; dx \right) \\ &= \int_{0}^{\infty} \frac{\phi (x) + \phi(-x) - 2\phi(0)}{x^2} \; dx, \end{align*}$$

where we have used the fact that

$$ \lim_{\epsilon\downarrow 0} \frac{\phi (\epsilon) + \phi(-\epsilon) - 2\phi(0)}{\epsilon} = 0.$$

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This is correct. – Kerry Jun 14 '12 at 3:54
Is there good generalization of this to higher rank derivatives of $\frac{1}{x}$? – Kerry Jun 14 '12 at 4:21
@ChangweiZhou, I have not thought of that. I guess, though, that this is closely related to derivatives of the Dirac delta function, thus there seems no reason to stop at degree 2. – Sangchul Lee Jun 14 '12 at 4:27
I was not aware of that. Could you elaborate on the connection of this with the delta function? I am working details out for $x^{-3}$. Thank you. – Kerry Jun 14 '12 at 4:34
I tried calculations similar to the case $x^{-2}$, but not so successful as our example. – Sangchul Lee Jun 14 '12 at 5:14

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