Identity with Dirac delta function: $\delta (x^2-a^2) = \frac{1}{2|a|}[\delta(x-a)+\delta(x+a)]$ How can I show that 
$\delta (x^2-a^2) = \frac{1}{2|a|}[\delta(x-a)+\delta(x+a)]$? I'm suppose to integrate it by a differentiable function and integrate, but I can't figure this one out.
 A: $\large a > 0$.
\begin{align}
\int_{-\infty}^{\infty}
{\rm f}\left(x\right)\,
\color{#ff0000}{\large\delta\left(x^{2} - a^{2}\right)}\,{\rm d}x
&=
\int_{-\infty}^{0}
{\rm f}\left(x\right)\,\delta\left(x^{2} - a^{2}\right)\,{\rm d}x
+
\int^{\infty}_{0}
{\rm f}\left(x\right)\,\delta\left(x^{2} - a^{2}\right)\,{\rm d}x
\\[3mm]&=
\int_{\infty}^{0}
{\rm f}\left(-\sqrt{y + a^{2}}\right)\,\delta\left(y\right)\,
\left(-\,{1/2 \over \sqrt{y + a^{2}}}\right){\rm d}y
\\[3mm]&+
\\[3mm]&
\int^{\infty}_{0}
{\rm f}\left(\sqrt{y + a^{2}}\right)\,\delta\left(y\right)\,
{1/2 \over \sqrt{y + a^{2}}}\,{\rm d}y
\\[3mm]&=
\int^{\infty}_{0}\left\lbrack%
{\rm f}\left(-\sqrt{\vphantom{\large A}y + a^{2}\,}\,\right)
+
{\rm f}\left(\sqrt{\vphantom{\large A}y + a^{2}\,}\,\right)
\over
2\sqrt{\vphantom{\large A}y + a^{2}\,}
\right]\,
\delta\left(y\right)\,{\rm d}y
\\[3mm]&=
{{\rm f}\left(-\left\vert a\right\vert\right) \over 2\left\vert a\right\vert}
+
{{\rm f}\left(\left\vert a\right\vert\right) \over 2\left\vert a\right\vert}
\\[3mm]&=
\int_{-\infty}^{\infty}{\rm f}\left(x\right)\,\left\{%
\color{#ff0000}{\large{1 \over 2\left\vert a\right\vert}\,
\left[\vphantom{\LARGE A}
\delta\left(x + \left\vert a\right\vert\right)
+
\delta\left(x - \left\vert a\right\vert\right)\,
\right]}\right\}\,{\rm d}x
\end{align}
In general,
$$\color{#ff0000}{\large%
\int_{-\infty}^{\infty}{\rm f}\left(x\right)
\,\delta\left(\vphantom{\Large A}{\rm F}\left(\vphantom{}x\right)\right)\,{\rm d}x
\color{#000000}{\ =\ }
\sum_{i}
{{\rm f}\left(x_{i}\right)
 \over
 \left\vert{\rm F\,}'\left(x_{i}\right)\right\vert}
\quad\color{#000000}{\mbox{where}}\quad {\rm F}\left(x_{i}\right)
\color{#000000}{\ =\ } 0}
$$
A: I think that when you change the x variable to y variable you are not keeping the same integral limits. You are actually changing them also. Because the first one goes up to zero, but once the change is made it goes up to -a, regarding to what it is inside the function.
Moreover, I don't know if it makes sense to integrate just until when what it is inside the delta function becomes zero. If you integrate up to that point and you count it, then you can integrate straight up to infinite because it wouldn't add anything to the result.
