Is $\delta(x)/x$ identical to $-\delta'(x)$ as a distribution?

Question

I am calculating a certain integral of the form $$\int \frac{\delta(x)}{x} f(x),$$ where $f(x)$ is well-behaved test-function. The expression, taken at face value, has no meaning, however, it shows up in a physical context, and, therefore, I have to assign it some value. It would make physical sense to take $\delta(x)/x \equiv - \delta'(x)$, but I am not sure how to mathematically justify this step. So, my question is, is there any sense in saying $$\int \frac{\delta(x)}{x} f(x) = f'(0)?$$

Answer 1

In THIS ANSWER and THIS ONE, I discuss some regularizations of the Dirac Delta.

Let $\delta_n$ be a regularization of the Dirac Delta such that for a suitable test function $f$

$$\langle \delta,f\rangle =\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)f(x)\,dx=f(0)$$

where $\delta_n(x)$ is an even function of $x$.

Note that the restriction that $\delta_n$ is even renders the following development heuristic. Any conclusion must hold true for any regularization, not only those that are even functions. So, with this disclaimer, we proceed.

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TAYLOR'S THEOREM

Since $f$ is smooth, Taylor's Theorem with the Peano form of the remainder guarantees that $f$ can be written $f(x)=f(0)+f'(0)x+h(x)x$ where $\lim_{x\to 0}h(x)=0$.

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THE DISTRIBUTION $\displaystyle d(x)=\frac{\delta(x)}{x}$

Denoting the distribution $d(x)=\frac{\delta(x)}{x}$, which is an abuse of notation, we have

$$\begin{align} \langle d,f\rangle &=\lim_{n\to \infty}\text{PV}\left(\int_{-\infty}^\infty \frac{\delta_n(x)}{x}f(x)\,dx\right)\\\\ &=\lim_{n\to \infty}\text{PV}\left(\int_{-\infty}^\infty \delta_n(x)\left(\frac{f(0)}{x}+f'(0)+h(x)x\right)\,dx\right)\\\\ &=f'(0) \end{align}$$

where $\text{PV}\int_{-\infty}^\infty f(x)\,dx=\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon}f(x)\,dx+\int_{\epsilon}^\infty f(x)\,dx\right)$ is the Cauchy Principal Value.

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THE DISTRIBUTION $\displaystyle \delta'(x)$

In addition, we have by definition (SEE THIS ANSWER )

$$\langle \delta',f\rangle =-f'(0)$$

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PUTTING IT ALTOGETHER

Since for all test functions $f$,

$$\langle d,f\rangle=-\langle \delta',f\rangle$$

then $\delta'(x)=-\frac{\delta(x)}{x}$.

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Note that since this development fails for any nascent Dirac Delta that is not an even function, we cannot say that $\delta'(x)=-\frac{\delta(x)}{x}$. We can say, however, that $x\delta'(x)=\delta(x)$. To see this note that for any test function $f$ we have

$$\begin{align} \langle x\delta',f\rangle&=\langle \delta',xf\rangle\\\\ &=-\langle \delta, (xf)'\rangle\\\\ &=-\langle \delta, xf'+f\rangle\\\\ &=\langle -\delta, f\rangle \end{align}$$

from which we see that the distributions $x\delta'$ and $-\delta$ are equivalent.

The reason that we cannot simply divide $x\delta'(x)=-\delta(x)$ by $x$ is that multiplication by distributions is not defined. Here $\frac1x$ is a distribution.