My undergraduate system textbook has this property in the appendix $$\delta(x^2-a^2)=\frac{1}{2|a|}[\delta(x-a)+\delta(x+a)]$$ and I can't seem to derive the result

I tried the following:

$\int_{-\infty}^{\infty}f(x)\delta(x^2-a^2)\, dx$

let $u^2=x^2-a^2$ and $x=\sqrt{u^2-a^2}$ however I couldn't resolve the new limit of integral as $u$ must be strictly positive under this substitution. What is another way I can approach this?

  • 4
    $\begingroup$ Informally: $\delta(x^2−a^2)=\delta((x+a)(x-a))$ which only "weights" at $x=-a$ and $x=a$. Around $x \to a$, this behaves as $\delta(2a (x-a) )$ which by the scaling property is $ \frac{1}{|2 a|}\delta(x-a)$. Summing up the other term, at $x= -a$, we get your formula. $\endgroup$
    – leonbloy
    Sep 23 '13 at 23:36

Let $x=\sqrt{u}$, so $\mathrm{d}x = \frac{\mathrm{d}u}{2\sqrt{u}}$, then $$\int_0^\infty f(x)\delta(x^2-a^2)\,\mathrm{d}x = \int_0^\infty \frac{f(\sqrt{u})}{2\sqrt{u}}\delta(u-a^2)\,\mathrm{d}u=\frac{f(|a|)}{2|a|}.$$ Let $x=-\sqrt{u}$, so $\mathrm{d}x = -\frac{\mathrm{d}u}{2\sqrt{u}}$, then $$\int_{-\infty}^0 f(x)\delta(x^2-a^2)\,\mathrm{d}x = \int^{\infty}_0 \frac{f(-\sqrt{u})}{2\sqrt{u}}\delta(u-a^2)\,\mathrm{d}u=\frac{f(-|a|)}{2|a|}.$$ Hence $$\int_{-\infty}^{\infty} f(x)\delta(x^2-a^2)\,\mathrm{d}x = \frac{1}{2|a|}\left(f(|a|)+f(-|a|)\right) = \frac{1}{2|a|}\left(f(a)+f(-a)\right).$$

  • $\begingroup$ How do you change the integration boundaries from $-\infty, 0$ to $ 0,+\infty $ on the second line $\endgroup$
    – Ruvik
    Mar 18 '20 at 5:53
  • 1
    $\begingroup$ By changing the limits he inverts the sign of the function. $\endgroup$
    – mnuizhre
    Feb 12 at 21:21

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