Integration involving the Delta function.

Question

We are familiar with the general form of the Delta function, which takes the form: $$\delta(f(x))=\sum_{i}\frac{\delta(x-x_i)}{f'(x_i)}$$

It is also well established that the integral involving Delta function is straighaway takes the following form: $$\int_{-\infty}^{\infty} \delta(x-a)f(x)dx=f(a)$$

Now, If we havve the integral which is of the form $$\int_{-\infty}^{\infty} \delta((x-a)^2)f(x)dx$$ then how will it be done? Can it be done by residue method, or is there some other way out? I was trying but got stucked in the midway.

Thanks in advance.

Answer 1

So as you can always move your coordinates, we can assume that $a=0$, and so your question is: what meaning can we give to $\delta_0(x^2)$. As Saleh comments, it is not defined in the standard theory of distributions.

One of the property one usually wants with distributions is the compatibility with the change of variable in integrals. For general functions $u$ and continuous tests functions $\varphi$ it holds $$ \int_{\Bbb R} u(x^2)\,\varphi(x)\,\mathrm d x = \int_0^\infty u(x)\,\frac{\varphi(\sqrt x)+\varphi(-\sqrt x)}{2\,\sqrt x}\,\mathrm d x $$ so one might want to define $$ "\int_{\Bbb R}\varphi(x)\, \delta_0(x^2)" = \lim_{x\to 0} \frac{\varphi(\sqrt x)+\varphi(-\sqrt x)}{2\,\sqrt x} = \lim_{x\to 0_+} \frac{\varphi(x)+\varphi(-x)}{2\,x} $$ This is however infinite if $\varphi(0)\neq 0$ (so it does not define a measure or a distribution). If $\varphi(0)=0$ and $\varphi$ is differentiable at $x=0$, then from a first order Taylor expansion, this limit is $0$. The only thing that remains is the case of functions that are continuous but not diferentiable at $x=0$. For example, $\int_{\Bbb R} |x|\, \delta_0(x^2) = 1$, and more generally, if $\varphi(x) = \psi(|x|)$ for a function $\psi \in C^1(\Bbb R_+)$ such that $\psi(0)=0$, then $$\int_{\Bbb R} \psi(|x|)\, \delta_0(x^2) = \lim_{x\to 0} \frac{\psi(x)}{x} =\psi'(0). $$