Integrating the dirac delta function multiplied with another function

Question

I'm trying to evaluate

$$ \int_{-\infty}^{\infty}dxf(x)\delta(g(x)) $$ Where $\delta$ is the dirac delta function, and $g(x)$ has zeros at $\{x=x_n, n = 1,...,N\}$. Here're some of my steps: $$ \int_{-\infty}^{\infty}dxf(x)\delta(g(x)) = \sum_{n = 1}^N\int_{x_n-\epsilon}^{x_n+\epsilon}dxf(x)\delta(0+g'(x_n)(x-x_n)+O(\epsilon^2)) \\= \sum_{n = 1}^N\int_{-\infty}^{\infty}dxf(x)\delta(g'(x_n)(x-x_n)) $$ I'm not exactly sure if these steps look correct. For the first equality, I Taylor expanded $g(x)$ around each of the zeros, but for the remaining terms, is it on the order of $\epsilon^2$? Is dropping the higher order terms necessary to change the integrand back to the improper integral （the next line）?

Also generally speaking, do we need to assume $f(x)$ and $g(x)$ to be real functions?

Thanks!

Answer 1

$\space\space\space$The Dirac delta can, in fact, be composed with a function $g(x)$ provided that $g(x)$ is continuously differentiable and that $g'(x)\not=0$ for all $x\in\mathbb{R}$ where $g(x)=0$. By the change of variables formula we have;

$$\int_{-\infty}^{\infty}f(x)\delta(x)dx=\int_{-\infty}^{\infty}f(h(y))\delta(h(y))\cdot|h'(y)|dy$$

Therefore, if we define $\delta(h(y))$ as;

$$\delta(h(y))={\delta(y-y_0)\over |h'(y_0)|}$$

where $y_0$ is the single real zero of the function $h(y)$, then the equality of the two integrals above will, in fact, hold.

From this, the following is arrived at by definition as well;

$$\delta(g(x))=\sum_{n=0}^{N}{\delta(x-x_n)\over |g'(x_n)|}$$

where $\{x_n : 1\leq n\leq N\}$ are the $N$ real zeroes of the function $g(x)$...

So the integral in question will be;

$$\int_{-\infty}^{\infty}f(x)\delta(g(x))dx=\sum_{n=0}^{N}{f(x_n)\over|g'(x_n)|}$$