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Firstly, apologies for abusing the notation by placing the Dirac measure inside an integral for which I was told that this should not be done from a previous question asked by me. But given the circumstances, I have no choice.

This is almost a word by word copy of the derivation given on page 2 of this paper on the Dirac delta:

Starting from $\displaystyle\int_{-\infty}^{\infty} f(x) \delta(x) \, \mathrm{d}x = f(0)$ and by changing variables and noting that $\delta(x)$ is symmetric we can derive a more general formula:

$$\space \space \space \space \space \space \space \space \begin{align}\color{red}{\tag{1}}\quad f(0)=\int_{-\infty}^{\infty} f(\xi) \delta(\xi) \, \mathrm{d}\xi \end{align}$$

$$\begin{align}\color{blue}{\tag{2}}\quad f(x)=\int_{-\infty}^{\infty} f(\xi+x) \delta(\xi) \, \mathrm{d}\xi\end{align}$$

$$\begin{align}\color{#F80}{\tag{3}}\quad f(x)=\int_{-\infty}^{\infty} f(\xi) \delta(\xi - x) \, \mathrm{d}\xi\end{align}$$

$$\begin{align}\color{#080}{\tag{4}}\quad f(x)=\int_{-\infty}^{\infty} f(\xi) \delta(x-\xi) \, \mathrm{d}\xi\end{align}$$

$\color{red}{(1)}$ was the only part I understood. So does this mean the change of variables was such that $x \rightarrow \xi$?

Going from part $\color{#F80}{(3)}$ to $\color{#080}{(4)}$ can only mean that $\delta(\xi - x)=\delta(x-\xi)$. Why is this true? I know that $\delta(x)$ is symmetric but still don't get why this makes $\delta(\xi - x)=\delta(x-\xi)$?

Now, lastly and most importantly, I have no idea how they went from $\color{blue}{(2)}$ to $\color{#F80}{(3)}$ and would be very grateful if anyone could explain/show all the intermediate steps.

Thanks in advance.

Community
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BLAZE
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2 Answers2

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Details of the steps:

$\color{red}{(1)}$ to $\color{blue}{(2)}$, function substitution $f(t)\mapsto f(t+x)$.

$\color{blue}{(2)}$ to $\color{#F80}{(3)}$, substitute $\xi\mapsto\xi-x$.

$\color{#F80}{(3)}$ to $\color{#080}{(4)}$, $\delta$ is even; i.e. $\delta(-x)=\delta(x)$.

robjohn
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    Was there something else? – robjohn Oct 07 '15 at 04:43
  • Yes, please elaborate on $\color{red}{(1)}$ to $\color{blue}{(2)}$. Can you show all the steps for the function substitution $f(x)\mapsto f(x + \xi)$? As I don't understand why you can just substitute for $f$ and not $\delta$. In other words why does the substitution not change $\int_{-\infty}^{\infty} f(x) \delta(x) , \mathrm{d}x$ to $\int_{-\infty}^{\infty} f(x+\xi) \delta(x+\xi) , \mathrm{d}\xi$? Thanks again. – BLAZE Oct 07 '15 at 12:34
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    @BLAZE: $(1)$ is true for all functions in place of $f$. Suppose we used $g(t+x)=f(t)$, then we get $$g(x)= f(0)= \int_{-\infty}^\infty f(\xi),\delta(\xi),\mathrm{d}\xi= \int_{-\infty}^\infty g(\xi+x),\delta(\xi),\mathrm{d}\xi$$ Now just rename $g$ to $f$ – robjohn Oct 07 '15 at 12:53
  • Why is $\color{red}{(1)}$ true for all functions in place of $f$? – BLAZE Oct 07 '15 at 13:26
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    That is the definition of $\delta$. If we only knew that $f(0)=\int_{-\infty}^\infty f(x),\delta(x),\mathrm{d}x$ for one particular function, that wouldn't be very useful. There would be lots of functions $\delta$ satisfy that. – robjohn Oct 07 '15 at 13:34
  • The reason I'm confused is because for $\color{blue}{(2)}$ to $\color{#F80}{(3)}$, making the substitution $\xi\mapsto\xi-x$ results in $\int_{-\infty}^{\infty} f(\xi+x) \delta(\xi) , \mathrm{d}\xi=\int_{-\infty}^{\infty} f(\xi) \delta(\xi - x) , \mathrm{d}\xi$ but for $\color{red}{(1)}$ to $\color{blue}{(2)}$, you only substitute $f(x)\mapsto f(x+\xi)$ for $f(x)$ but $\delta(x)$ is left alone, why is that? I thought an integral substitution must be carried out on the entire integrand $f(x)$ and $\delta(x)$? – BLAZE Oct 07 '15 at 13:44
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    @BLAZE: that is because in $\color{blue}{(2)}$ to $\color{#F80}{(3)}$ we are substituting the variable $\xi$ and leaving $f$ alone, whereas in $\color{red}{(1)}$ to $\color{blue}{(2)}$ we are substituting the function $f$ and leaving $\xi$ alone. – robjohn Oct 07 '15 at 14:06
  • For $\color{blue}{(2)}$ to $\color{#F80}{(3)}$, we can't be leaving $f$ alone as $\int_{-\infty}^{\infty} f(\xi+x) \delta(\xi) , \mathrm{d}\xi=\int_{-\infty}^{\infty} f(\xi) \delta(\xi - x) , \mathrm{d}\xi$: The function $f$ argument changed from $f(\xi+x)$ to $f(\xi)$. Also, in $\color{red}{(1)}$ to $\color{blue}{(2)}$ do you mean we are leaving $\delta$ alone instead of $\xi$? – BLAZE Oct 07 '15 at 14:39
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    No. In $\color{blue}{(2)}$ to $\color{#F80}{(3)}$, we are leaving $f$ and $\delta$ alone and only changing $\xi$ everywhere we see it to $\xi-x$. We never do anything to $\delta$ in any step. $\delta$ is the Dirac delta function at all points of the proof. – robjohn Oct 07 '15 at 14:59
  • Okay, you've been a massive help on this question. Just one final question: What do you mean by "we see it to $\xi-x$"? – BLAZE Oct 07 '15 at 15:11
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    That phrase is being taken out of context. What the whole phrase means is that "everywhere we see $\xi$, we change it to $\xi-x$". – robjohn Oct 07 '15 at 15:16
  • Thank you for your time, it is appreciated. – BLAZE Oct 07 '15 at 15:18
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You cannot expect a correct proof of a formula which makes no sense within the formalism at your disposal: There is no function $x\mapsto\delta(x)$ such that for "all" $f$ one has $$\int_{\mathbb R} f(x)\>\delta(x)\>dx=f(0)\ .\tag{1}$$

Instead we should view $(1)$ as shorthand for the following mental process: We consider a sequence of functions $\delta_n:\> x\mapsto \delta_n(x)\geq0$ with $\int_{\mathbb R}\delta_n(x)\>dx=1$ and $\delta_n(x)\equiv0$ when $|x|\geq{1\over n}$. For such a sequence $(\delta_n)_{n\geq1}$ one has $$\lim_{n\to\infty}\int_{\mathbb R} f(x)\>\delta_n(x)\>dx=f(0)\tag{2}$$ for all suitable $f$. Now replace the $\delta_n$ by the translated functions $\delta_{a.n}: \>x\mapsto \delta_n(x-a)$. Ordinary substitution gives $$\int_{\mathbb R} f(x)\>\delta_n(x-a))\>dx=\int_{\mathbb R} f(x'+a)\>\delta_n(x')\>dx'\ .$$ According to $(2)$, applied to the function $g(x'):=f(x'+a)$, we get that $$\lim_{n\to\infty} \int_{\mathbb R} f(x)\>\delta_n(x-a))\>dx=\lim_{n\to\infty}\int_{\mathbb R} g(x')\>\delta_n(x')\>dx'=g(0)=f(a)\ .$$ The "shorthand version" of the last formula is $$\int_{\mathbb R} f(x)\>\delta(x-a))\>dx=f(a)\ .$$

Christian Blatter
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    I believe that the confusion was not because $\delta$ is a distribution, but because of the substitutions. The same substitutions that were bothering the OP would be there for any even $\delta_n$ in place of $\delta$, and the questions asked in the comments to my answer would most likely be the same. However, I could be mistaken. – robjohn Oct 07 '15 at 17:31