# Integral transform with Dirac delta

Let $f,g: \mathbb{R}^n \to \mathbb{R}$.

Let $\delta$ denote the Dirac delta function.

How can I write the integral over $\mathbb{R}^n$ (on the left hand side) as an integral over $g^{-1}(0)$

$$\int\limits_{\mathbb{R}^n} f(r) \delta(g(r))\ dr \ \ \ = \int\limits_{g^{-1}(0)} ?\ dr$$

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–  martini Oct 18 '13 at 8:56


In some particular cases ( it depends on the particular form of ${\rm g}$ ), the Dirac delta $\delta\pars{{\rm g}\pars{\vec{r}}}$ can be reduced to a product of "more simple" Dirac delta's. For example, with $n = 3$ and spherical coordinates it's like $$\delta\pars{\vec{r}} = {1 \over r^{2}}\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi}$$

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Maybe they can. But then what? –  Did Oct 19 '13 at 10:17
@Did $f$ and $g$ are known functions. It's reduced to a "technical problem". –  Felix Marin Oct 19 '13 at 10:20
@Did Thanks. You motivated the ADDENDUM. –  Felix Marin Oct 19 '13 at 10:28

See Gelfand and Shilov "Generalized Functions, vol. 1" page 222. The Delta Distribution with a function as its argument is discussed there. On page 230 there is the following definition (with $k=1$):

$\delta(P,\phi) = \int_{P=0}\phi dx$ (where $P$ is a hypersurface).

So yours is the integral of f over the hypersurface $g=0$. If by $g^{-1}(0)$ is meant $g=0$ then

$\int_{g^{-1}(0)}f(r)dx$

is the right half of your equation.

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