From $\displaystyle\int_{-\infty}^{\infty}f(x)\delta(x)\mathrm{d}x=f(0)$ derive $\displaystyle\int_{-\infty}^{\infty}f(x)\delta(x-a)\mathrm{d}x=f(a)$

Question

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.

Answer 1

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. $\delta(-x)=\delta(x)$.

Answer 2

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 $$\int_{\mathbb R} f(x)\>\delta(x-a))\>dx=f(a)\ .$$