# Differential of Dirac Delta Function

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.

• This results sounds me very very strange. On which book are you working ? – Jean Marie Jan 28 '17 at 20:01
• The integral of an odd function is zero only if the function is integrable, and that is not the case of $x\mapsto\frac1{x}$; the integral of $\delta(x)/x$ isn't defined since $1/x$ is undefined at $0$. You might be looking for the distributional derivative of the delta. – yellowquark Jan 28 '17 at 20:05
• This question has been set as the 'challenge' question for my homework from my university prof – Tom Jan 28 '17 at 20:19
• There are more or less rigorous approaches to $\delta$ distributions. Sometimes in lectures delivered to future engineers, the objective is to have an immediate operational use, justifying "accomodations". But, in most cases, when you begin to do theory with $\delta$ functions, it can be very slippery if you don't start with rigor. What surprizes me is that you are authorized to speak about $f(x)\delta$ if $f(0)$ is defined, and in this case $f(x)\delta=f(0)\delta$... – Jean Marie Jan 28 '17 at 21:15
• Thanks @JeanMarie. Just to confirm the last statement is a contradiction correct? I have only had one lecture on the delta function, I'm really quite lost with this – Tom Jan 28 '17 at 21:36

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 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.

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

• Are you taking as a definition $d(x) := \lim_{n→ ∞} \text{PV} δ_n(x) /x$? Otherwise I don't know how to make sense of $d$ (which could certainly be my fault) – Calvin Khor Jan 28 '17 at 23:35
• @CalvinKhor Yes, that is a regularized representation. – Mark Viola Jan 28 '17 at 23:39
• $\frac{1}{x} \delta$ is really an abuse of notation. Take a sequence of even $C^\infty$ functions $\phi_n$ such that $\phi_n(0) = 0$ and $\phi_n \to \delta$ in the sense of distributions, then you can look safely at the sequence of distributions $\frac{1}{x} \phi_n$ and show (as you did) that it converges to $-\delta'$ in the sense of distributions – reuns Jan 28 '17 at 23:41
• @user1952009 I agree completely. It is a complete abuse of notation. Just curious ... have you seen an alternative way of expressing besides the regularization? -Mark – Mark Viola Jan 28 '17 at 23:46
• @Dr.MV Take $\varphi \in C^\infty_c([1,2])$ and let $\phi(x) = \frac{\varphi(x)+\varphi(-x)}{2C}$ where $C = \int_1^2 \varphi(x)dx$. Then $n\phi(nx) \to \delta$ and $\frac{n \phi(nx)}{x} \to -\delta'$ in the sense of distributions (without any principal value, since $\phi(0) =0$) – reuns Jan 28 '17 at 23:50

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 $\varphi \in C^\infty_c$ : $$\lim_{n \to \infty}\langle \phi_n,\varphi \rangle \overset{def}= \lim_{n \to \infty} \int_{-\infty}^\infty \phi_n(x) \varphi(x)dx = \varphi(0) = \langle \delta,\varphi \rangle$$

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 !)