In physics and engineering you may see \[\int_\mathbb{R} \delta dx = 1,\] which is an abuse of notation because \(\delta\) is not a function. But why is this done anyway and how can we reason about this rigorously?

First we must come to rigorous definition of \(\delta\) so we can do some mathematics. If \(C_c^\infty(\mathbb{R})\) is the set of infinitely differentiable compactly supported functions on \(\mathbb{R}\) then we define \(\delta\) in the dual space, \(\delta \in (C_c^\infty)^*\) and \(\delta(f) = f(0)\). If we consider \(C_c^\infty\) as a Banach sub-space \(C_C^\infty \subset L^\infty\) then we can define a norm \(||\gamma|| = \sup\{|\gamma(f)|:||f||_\infty=1\}\) for \(\gamma \in (C_c^\infty)^*\). This means \(||\delta|| = 1\).

Also to note is that \(\exists V \cong L^1\) such that \(V \subset (L^\infty)^*\) and if \(f\in L^1\) then \(||f|| = \int_\mathbb{R} f dx\), which is what we want \(||\gamma||\) to look like.

Now how do we create an integral form? First \(\gamma \in C_c^\infty\) is positive if \(\forall f \geq 0\) we have \(\gamma(f) \geq 0\). Now applying the Riesz–Markov–Kakutani representation theorem to positive functionals we get \(\exists \mu_\gamma\) such that \[\gamma(f) = \int_\mathbb{R} f d\mu_\gamma.\] Next applying the Lebesgue decomposition theorem with respect to the Lebesgue measure we obtain \(\mu_\gamma = \nu_1 + \nu_2\) where \(\nu_1 << lm\) and \(\nu_2 \perp lm\). As \(\nu_1 << lm\) there exists a \(g\) such that \(\nu_1(E) = \int_E g lm\), by the Radon–Nikodym theorem. So we can write \(\gamma(f) = \int_\mathbb{R} f g dlm + \int_\mathbb{R}f \nu_2\). If \(\nu_2 << cm\) we can we can again apply Radon-Nikodym and get \(\gamma(f) = \int_\mathbb{R} f g dlm + \sum_{n\in \mathbb{R}} h(n)\) where \(h\) has countable support. Furthermore we can “graph” \(\gamma\) by first plotting the function \(g\) then adding impulses at each non-zero \(h\).

If \(1\) a constant function then \(||1||_\infty = 1\) then we can write \(||\gamma||\) as an integral \[||\gamma|| = |\gamma(1)| = \int_\mathbb{R} g dlm + \sum_{n\in \mathbb{R}} h(n).\] Finally we write \(||\delta||\) as an integral \[||\delta|| = \sum_{n\in \mathbb{R}} \delta_{0, n}.\]

Edit: 2026-11-05

I had a brief talk with a physics student about why they would define \(\int_\mathbb{R} \delta dx = 1\), he first said that states are defined via special unitary matrices on \(\mathbb{C}^2\) and that all the integrals they do must equal \(1\) as they have a probability space. Also that the \(\delta\) function comes from the Fourier transform of \(e^{i\pi x}\).

So I am left more confused than before, I think I will leave this until I’ve done more functional analysis and harmonic analysis and just focus on the courses I’m doing currently.