I've seen this identity in my electrodynamics book: $$\delta (f(x))=\sum_i{ \frac{1}{|{df\over dx}(x_i)|}\delta (x-x_i)}$$

Where $x_i$ shows the $i$th zero of $f(x)$. How can I prove it? I've tried the integral definition of delta function, but doesn't work.

  • $\begingroup$ It is quite unclear what you mean by the $x_i$'s. $\endgroup$ – zozoens Feb 3 '14 at 16:51
  • $\begingroup$ Sorry. I edited the question. $\endgroup$ – user215721 Feb 3 '14 at 16:53
  • 2
    $\begingroup$ This formula is just the change-of-variables formula for the Dirac delta function (link to relevant section of Wikipedia). $\endgroup$ – Zev Chonoles Feb 3 '14 at 16:58

Here's an informal idea:

Start with the integral

$$ \int \delta(f(x)) g(x)\,dx $$

and for every $x_i$, take disjoint neighborhoods $U_i$ where $f$ is a diffeomorphism (i.e. $f' \neq 0$). So,

$$ \int \delta(f(x)) g(x)\,dx = \sum_i \int_{U_i} \delta(f(x)) g(x)\,dx $$

use change of variables in each neighborhood: $u_i = f(x)$ so $$ \int \delta(f(x)) g(x)\,dx = \sum_i \int_{f(U_i)} \delta(u_i) \frac{g(f^{-1}(u_i))}{|f'(f^{-1}(u_i))|}\,du $$

then $u_i = 0$ exactly when $x = x_i$, so we have

$$ \int \delta(f(x)) g(x)\,dx = \sum_i \frac{g(x_i)}{|f'(x_i)|} $$


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