# Integrating the dirac delta function multiplied with another function

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!

• the Dirac delta is a measure so the expression $\delta (g(x))$ doesn't have sense Aug 30, 2022 at 5:23
• this answer might be useful.
– robjohn
Aug 30, 2022 at 7:57
• @robjohn I approve of the fact that you wait for other opinions. But I would also approve of closing this as a duplicate of the question you suggested. I don't frequent the diract-delta tag, so you may still want to wait for more users to chime in :-) Aug 30, 2022 at 16:19
• @JyrkiLahtonen: I do think it is a possible duplicate of this question, but it seems better to have a consensus rather than a unilateral decision.
– robjohn
Aug 30, 2022 at 16:36

$$\space\space\space$$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;

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

• We only need $g'(x)\ne0$ when $g(x)=0$.
– robjohn
Aug 30, 2022 at 8:06