# Integral over product of Dirac delta functions

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?

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

• This is $0$ if $a \neq b$. Oct 19, 2019 at 22:55
• What if $a = b$? Oct 19, 2019 at 23:09

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 $$u[g]~=~\int_{\mathbb{R}} \!\mathrm{d}x\; g(x,x)\; f(x). \tag{C}$$
• 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. Sep 23, 2022 at 15:02
• In OP's distribution (2) the variables $a$ and $b$ are indeterminates, not fixed numbers. Sep 23, 2022 at 15:38