Proof a basic relation for the Dirac's delta with only three informations 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 
$$\int_{\mathbb{R}}f(x)\delta(x-x_0)dx = f(x_0)$$
?
 A: 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 
$$\delta(f(x))=\sum_{i=1}^n \frac{\delta(x-x_i)}{|f'(x)|} \tag1$$

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

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 $(1)$.
