What is $\int_{-\infty }^{\infty }|x| \delta \left(x^2\right) \, dx$?

Question

In my impression it should be $1$ (or maybe, 2) but Wolfram Mathematica returns $0$. Who is wrong?

In general, in my impression $\int_a^b \delta(f(x))|f'(x)|dx$ should return the number of roots of the function $f(x)$ on the interval.

Answer 1

Let $f(x)=|x|$, $g(x)=x^2$, and $\delta_n(x)$ be a nascent Dirac Delta. We wish to determine the existence of the functional $\langle \delta\circ g,f\rangle$.

We now proceed to show that such a functional fails to exist. Note that we can write

$$\begin{align} \lim_{n\to \infty}\langle \delta_n\circ g,f\rangle&=\lim_{n\to \infty}\int_{-\infty}^\infty |x|\delta_n(x^2)\,dx\\\\ &=\lim_{n\to \infty}2\int_{0}^\infty x\delta_n(x^2)\,dx\\\\ &=\lim_{n\to \infty}\int_{0}^\infty \delta_n(x)\,dx\tag1 \end{align}$$

If $\delta_n(x)=n(H(x)-H(x-1/n))$, then $\lim_{n\to \infty}\langle \delta_n\circ g,f\rangle=1$.

However, if $\delta_n(x)=n(H(x+1/2n)-H(x-1/2n))$, then $\lim_{n\to \infty}\langle \delta_n\circ g,f\rangle=1/2$

Inasmuch as the value of the limit depends on the nascent Dirac Delta, the functional of interest fails to exist.