Proof a basic relation for the Dirac's delta with only three informations

Question

Can you prove the relation $$ \delta(x^2-a^2) = \frac{1}{|2a|}(\delta(x - a) + \delta(x + a))$$ just knowing that $$ \delta(ax) = \frac{1}{|a|}\delta(x)$$ that $$ \int_{\mathbb{R}}\delta(x) dx = 1$$ and that $$\int_{\mathbb{R}}f(x)\delta(x-x_0)dx = f(x_0)$$ ?

Answer 1

We need an additional theorem from the theory of distributions.

Let $f(x_i)=0$ for $i=1\dots,n$. If $f$ is differentiable and monotonic in a neighborhood of its zeroes, then in distribution

$$\delta(f(x))=\sum_{i=1}^n \frac{\delta(x-x_i)}{|f'(x)|} \tag1$$

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Applying this to the problem at hand, $f(x)=x^2-a^2$, $x_1=|a|$ and $x_2=-|a|$, and $|f'(|a|)|=2|a|$ and $|f'(-|a|)|=2|a|$. Using $(1)$ we find immediately that

$$\delta(x^2-a^2)=\frac{\delta(x-|a|)}{2|a|}+\frac{\delta(x-|a|)}{2|a|}=\frac{\delta(x-a)+\delta(x+a)}{2|a|}$$

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HEURISTIC DEVELOPMENT:

If one proceeds formally, as if the notation $\int_{\mathbb{R}}\phi(x)\delta(x)\,dx$ is an integral (it is NOT), then by (non-rigorously) enforcing the substitution $x=-\sqrt{t+a^2}$ for $x\le 0$ and $x=\sqrt{t+a^2}$ for $x\ge 0$, we obtain

$$\begin{align} \int_{\mathbb{R}}\phi(x)\delta(x^2-a^2)\,dx&=\int_{-\infty}^0 \phi(x)\delta(x^2-a^2)\,dx+\int_0^\infty \phi(x)\delta(x^2-a^2)\,dx\\\\ &=\int_0^\infty \phi(-\sqrt{t+a^2})\delta(t)\,\frac{1}{2\sqrt{t+a^2}}\,dt+\int_0^\infty \phi(\sqrt{t+a^2})\delta(t)\,\frac{1}{2\sqrt{t+a^2}}\,dt\\\\ &=\frac{\phi(-|a|)+\phi(|a|)}{2|a|}\\\\ &=\int_{\mathbb{R}}\phi(x)\frac{\delta(x-a)+\delta(x+a)}{2|a|} \tag 2 \end{align}$$

Since, $(2)$ is true for all test functions $\phi(x)$, we find that in distribution $\delta(x^2-a^2)=\frac{\delta(x-a)+\delta(x+a)}{2|a|}$, which recovers the result obtained by using $(1)$.