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?


  • $\begingroup$ the Dirac delta is a measure so the expression $\delta (g(x))$ doesn't have sense $\endgroup$
    – Masacroso
    Aug 30, 2022 at 5:23
  • 2
    $\begingroup$ this answer might be useful. $\endgroup$
    – robjohn
    Aug 30, 2022 at 7:57
  • $\begingroup$ @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 :-) $\endgroup$ Aug 30, 2022 at 16:19
  • $\begingroup$ @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. $\endgroup$
    – robjohn
    Aug 30, 2022 at 16:36

1 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;


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;


  • 3
    $\begingroup$ We only need $g'(x)\ne0$ when $g(x)=0$. $\endgroup$
    – robjohn
    Aug 30, 2022 at 8:06

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