Integral of the square of the Dirac delta

Question

What is the following integral?

$$\int_{-\infty}^{\infty} \delta^2(x) {\,\rm d} x$$

I think it should be one, by Parseval's Theorem.

Answer 1

There is no such thing as $\delta^2(x)$ and $\int_{-\infty}^{\infty} \delta^2(x) {\,\rm d} x$.

Nevertheless, there is a natural extension of the $L^2$ norm to the distributions, which of course assigns $\infty $ to $\delta$ and many others.

Take any $\phi \in C^\infty_c(-1,1),\phi \ge 0,\|\phi\|_{L^1}=1$, $\phi_n(x)=n\phi(nx)$ then $$\|T\|_{L^2} = \lim \sup_{n\to \infty} (\int_{-\infty}^\infty |T\ast \phi_n(x)|^2dx)^{1/2},\qquad T\ast \phi_n(x)= \langle T,\phi_n(x-.)\rangle$$

It is an interesting exercice to show (using the Fourier series of $ (T\ast \phi_n)\psi$) that $\|T\|_{L^2}$ is finite iff the distribution $T$ is (represented by) a function in $L^2$. Moreover $\|T\|_{L^2}=\|\hat{T}\|_{L^2}$.

Note that the same idea gets more complicated for $L^1$ : $\delta$ is a finite measure but it is not in $L^1$.

Answer 2

$$\int_{-\infty}^{\infty} \delta^2(x) dx=2\int_{-\infty}^\infty dx$$