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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.

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$\large a > 0$.

\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,

$$\color{#ff0000}{\large% \int_{-\infty}^{\infty}{\rm f}\left(x\right) \,\delta\left(\vphantom{\Large A}{\rm F}\left(\vphantom{}x\right)\right)\,{\rm d}x \color{#000000}{\ =\ } \sum_{i} {{\rm f}\left(x_{i}\right) \over \left\vert{\rm F\,}'\left(x_{i}\right)\right\vert} \quad\color{#000000}{\mbox{where}}\quad {\rm F}\left(x_{i}\right) \color{#000000}{\ =\ } 0} $$

Felix Marin
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    (+1) This is generalized to $\mathbb{R}^n$ in this answer; the standard measure for points being counting measure. – robjohn Sep 21 '13 at 01:12
  • Correct me if I'm wrong, but it looks like your proof only relies on $|a|>0$ not $a$ being strictly positive. – Alex Schiff Aug 08 '14 at 22:17
  • @AlexSchiff Yes. It's true. Note that $a = 0$ is 'special' and it will require some expected features of ${\rm f}\left(x\right)$. It would be nice to read carefully this page. In particular, the section ${\bf\mbox{As a distribution}}$ where the behavior of the integrating function is discussed. – Felix Marin Aug 09 '14 at 01:01
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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.

Andrés
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