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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 $$\int \frac{\delta(x)}{x} f(x) = f'(0)?$$

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    The integral evaluates as $\frac{f(0)}{0}$ which, if we take this to mean $\lim_{x\to 0}\frac{f(x)}{x}$, is only $f'(0)$ when $f(0)=0$. It is undefined otherwise. – 2017-02-05
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    I suppose if you take the sign shift into account, you can consider the $\int \frac{\delta(x)}{x}\,dx$ to be a certain "principal value" of the integral, and thus get $0$ for the constant part. – 2017-02-05
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    Well, is $-x\delta' = \delta$? Check: $$-x\delta'[\varphi] = -\delta'[x\varphi(x)] = \delta[(x\varphi(x))'] = \delta[\varphi(x) + x\varphi'(x)] = \varphi(0) + 0\varphi'(0) = \varphi(0) = \delta[\varphi]$$ – 2017-02-05
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    @ThomasAndrews Yes, I was also thinking of some sort of principal value of the integral. It's usual to do this in physics when the original integral does not converge. – 2017-02-05
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    @DanielFischer It may be a silly question, but does the equality $-x \delta' = \delta$ necessarily imply $- \delta' = \delta/x$ for distributions? – 2017-02-05
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    How do you define $\dfrac{T}{f}$ where $T$ is a distribution and $f$ a nice enough function? The natural thing is to say that _if_ there is a distribution $S$ such that $T = f\cdot S$, then $S$ would be a candidate for $\dfrac{T}{f}$. However, if $f$ has zeros, then if there is one such $S$, there are infinitely many. In our case, since $x\delta = 0$, we have $-\delta' + c\delta$ as candidates for $c\in \mathbb{R}$. So while I'd say that $\dfrac{\delta}{x}$ isn't well-defined, interpreting it as $-\delta'$ makes some sense. By simplicity, more sense than $-\delta' + e\pi^2\delta$. – 2017-02-05

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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 $f$

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

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

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 $\lim_{x\to 0}h(x)=0$.

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.

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

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

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

PUTTING IT ALTOGETHER

Since for all test functions $f$,

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

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

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    Nice answer, thanks! Do you think it easily generalized to the case $\delta/x^n$? Could we connect it to $\delta^{(n)}$? – 2017-02-05
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    You're welcome. My pleasure. This won't generalize even for $n=2$. Follow the development and see where and why it breaks. – 2017-02-05
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    Yes, I'm aware that the principal value integration breaks down with $1/x^2$ and other even-power singularities. However, I have a feeling that such a nice identity should generalize in some way... – 2017-02-05
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    I don't see how it generlizes since test functions belong to the space of smooth function of compact support. – 2017-02-05