Differential of Dirac Delta Function

Question

By considering $\int_{-\infty}^{\infty}f(x)\frac{d(\delta(x))}{dx}dx$ and $\int_{-\infty}^{\infty}f(x)\frac{\delta(x)}{x}dx$ show that $\frac{d(\delta(x))}{dx}=-\frac{1}{x}\delta(x)$

The hint that I've been given is to take the macluarin expansion of $f(x)$ and note that $\int_{-\infty}^{\infty}\frac{\delta(x)}{x}dx=0$ since the integral is an odd function of $x$.

I know I'm meant to provide some working that I've done, but I really don't even know where to start on this, I've looked elsewhere online and that didn't help. Any help on this would be greatly appreciated.

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 $f$

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

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

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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 as user1952009 points out, 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 )

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

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PUTTING IT ALTOGETHER

Since for all test functions $f$,

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

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

Answer 2

To add some details on what Dr.Mv wrote, with the principal value removed :

Take $\chi \in C^\infty_c([1,2])$ such that $\int_1^2 \chi(x)dx = 1$ and let $\phi(x) = \frac{\chi(x)+\chi(-x)}{2}, \phi_n(x) = n \phi(nx)$. The typical exercice is to show that $\phi_n \to \delta$ in the sense of distribution, that is for every $\varphi \in C^\infty_c$ : $$\lim_{n \to \infty}\langle \phi_n,\varphi \rangle \overset{def}= \lim_{n \to \infty} \int_{-\infty}^\infty \phi_n(x) \varphi(x)dx = \varphi(0) = \langle \delta,\varphi \rangle$$

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Now look at the sequence of distributions $\Phi_n(x) = \frac{\phi_n(x)}{x}$, well-defined since $\frac{\phi_n(x)}{x} \in L^1$. Take an even test function $\psi \in C^\infty_c$ such that $\psi(0) = 1$. Since $\Phi_n$ is odd, we have : $$\langle \Phi_n, \psi \rangle = \int_{-\infty}^\infty \Phi_n(x) \psi(x)dx = 0$$ and for any $\varphi \in C^\infty_c$ : $$\lim_{n \to \infty}\langle \Phi_n,\varphi \rangle = \lim_{n \to \infty}\langle \Phi_n,\varphi-\varphi(0) \psi \rangle= \lim_{n \to \infty} \langle \phi_n,\frac{\varphi-\varphi(0)\psi}{x} \rangle = \langle \delta,\frac{\varphi-\varphi(0)\psi}{x} \rangle$$ $$ = \lim_{x \to 0} \frac{\varphi(x)-\varphi(0)\psi(x)}{x} = \varphi'(0) = -\langle \delta', \varphi \rangle$$ i.e. $\Phi_n \to -\delta'$ in the sense of distributions

(and tell your teacher that $\frac{1}{x} \delta$ is really an abuse of notation for referring to $\lim_{n \to \infty} \Phi_n$ in the sense of distributions !)