2
$\begingroup$

We can define the Dirac Delta function as a distribution satisfying

$$\int_{-\infty}^{\infty} \text{d}x\; \delta(x-a)\; f(x) = f(a) .\tag{1}$$

What if I have a product of delta functions?

$$\int_{-\infty}^{\infty} \text{d}x\; \delta(a-x)\; \delta(x-b)\; f(x) =~? \tag{2}$$

$\endgroup$
2
  • 2
    $\begingroup$ This is $0$ if $a \neq b$. $\endgroup$
    – WhatsUp
    Oct 19, 2019 at 22:55
  • 1
    $\begingroup$ What if $a = b$? $\endgroup$
    – firest
    Oct 19, 2019 at 23:09

1 Answer 1

Reset to default
2
$\begingroup$

OP's expression (2) makes sense in distribution theory: $$\color{blue}{\int_{\mathbb{R}} \!\mathrm{d}x\; \delta(a-x)\; \delta(x-b)\; f(x) ~=~\delta(a-b)\; f(b)}.\tag{A}$$ Proof: If we apply a test function $g\in C^{\infty}_c(\mathbb{R}^2)$ to both sides $$\begin{align}\iint_{\mathbb{R}^2}\!\mathrm{d}a\;\mathrm{d}b\;g(a,b)\color{blue}{\int_{\mathbb{R}} \!\mathrm{d}x\; \delta(a-x)\; \delta(x-b)\; f(x)} ~=~&\int_{\mathbb{R}} \!\mathrm{d}x\; g(x,x)\; f(x)\cr ~=~&\iint_{\mathbb{R}^2}\!\mathrm{d}a\;\mathrm{d}b\;g(a,b)\;\color{blue}{\delta(a-b)\; f(b)},\tag{B}\end{align}$$ they yield the same result. $\Box$

In other words, the LHS and RHS of eq. (A) is a notation for the same distribution $u\in{\cal D}^{\prime} (\mathbb{R}^2)$ given by $$u[g]~=~\int_{\mathbb{R}} \!\mathrm{d}x\; g(x,x)\; f(x). \tag{C}$$

$\endgroup$
2
  • $\begingroup$ Hi! Does it mean that $\int \delta^2(x-a) f(x) dx \ne f(a)$? Because somewhere on the forum I saw a post that the square of a delta function acts the same as just delta with a strict proof. So it is kinda confusing that the case $a=b$ is a somewhat special case. $\endgroup$
    – Sl0wp0k3
    Sep 23, 2022 at 15:02
  • $\begingroup$ In OP's distribution (2) the variables $a$ and $b$ are indeterminates, not fixed numbers. $\endgroup$
    – Qmechanic
    Sep 23, 2022 at 15:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .