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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.

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    This results sounds me **very very strange**. On which book are you working ? – 2017-01-28
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    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. – 2017-01-28
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    This question has been set as the 'challenge' question for my homework from my university prof – 2017-01-28
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    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 $f(x)\delta=f(0)\delta$... – 2017-01-28
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    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 – 2017-01-28
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    @Tom The distributions $f(x)\delta(x)$ and $f(0)\delta(x)$ are equivalent for suitable test functions $f$. But $\delta(x)$ is not a function, it is a Generalized function. – 2017-01-28
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    @Dr.MV That is what I was thinking, so how could this solution be true for all possible $\delta(x)$. – 2017-01-28
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    By definition, for a suitable test function $f$, we have $\langle f,\delta'\rangle =-f'(0)$. – 2017-01-28
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    Yes, for a suitable function, however the question implies that it's true for all functions $f$ doesn't it? – 2017-01-28
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    @Tom The assumption is that $f$ can be represented by its Maclaurin series (and implicitly, $f$ is of compact support). See my solution. – 2017-01-28
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    I don't know what is $\frac{1}{x} \delta$ in the context of distributions, that's what you should answer to your teacher – 2017-01-28

2 Answers 2

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

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    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) – 2017-01-28
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    @CalvinKhor Yes, that is a regularized representation. – 2017-01-28
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    $\frac{1}{x} \delta$ 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 – 2017-01-28
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    @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 – 2017-01-28
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    @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 $\phi(0) =0$) – 2017-01-28
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    Thank you so much for this. I think this question must have been set with the intention that none of us would be able to do it. We haven't covered the Cauchy Principal Value or the Peano form of the remainder. My lack of knowledge is going far enough that I'm not understanding the angled bracket notation, however if I just accept it I can understand the RHS mostly. I am not understanding the jump between $\langle d,f \rangle=f'(0)$ and the line previous. – 2017-01-28
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    @Tom You're welcome. My pleasure. -Mark – 2017-01-28
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    @Dr.MV Sorry I am having trouble understanding the jump between $\langle d,f \rangle=f'(0)$ and the line previous – 2017-01-29
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    I added an answer explaining what I was thinking to @Tom – 2017-01-29
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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 !)