It seems to me that
$$
\int_{-\infty}^\infty \int_{-\infty}^{\infty} \delta(x^2 + y^2 – R^2) dx dy
$$
should evaluate to $2\pi R$, the perimeter of a circle of radius $R$, but I'm having trouble getting this answer. I tried using the identity
$$
\delta(x^2 + y^2 – R^2) = \frac{1}{2\sqrt{R^2 – y^2}}\left[\delta\left(x – \sqrt{R^2 – y^2}\right) + \delta\left(x + \sqrt{R^2 – y^2} \right)\right]
$$
But it wasn't clear to me how to evaluate the $x$ integral after making this substitution.
How do you evaluate the above integral? Does the integral represent the perimeter of a circle of radius $R$? If not, how should such an integral be written?
Best Answer
Keep in mind the dirac delta is not a function, but rather a distribution. In fact, its precise definition (in this case) is that for all test functions $f(x,y)$, $$ \int_{\mathbb R^2}f(x,y)\delta(x^2+y^2-R^2)\ dx\ dy :=\int_{\theta=0}^{2\pi}f(R\cos\theta,R\sin\theta)\ d(R\theta). $$ The left side has no independent meaning - it is defined to equal the right side.
In your case, take $f(x,y)$ to be the constant $1$ function and you get your answer.
EDIT: After the discussion in the comments, I see this question really boils down to asking
More generally, one may ask for the definition of $\delta(g)$ where $\delta$ is the $n$-dimensional delta function and $g\colon \mathbb R^n\to \mathbb R^n$ is a smooth function. And the answer turns out to be that there is no such definition that has all the desired properties in general, even for $n=1$. An explicit counterexample is furnished in Remark 2 of these notes by Terry Tao.