Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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?
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 $$\int_{-\infty}^{\infty} f(x)\delta(x^2-a^2)\,\mathrm{d}x = \frac{1}{2|a|}\left(f(|a|)+f(-|a|)\right) = \frac{1}{2|a|}\left(f(a)+f(-a)\right).$$
