Formally, if I want to measure the length of a closed curve $f(x,y) = 0$, I presumed I could write:
$$ L = \int^\infty_{-\infty}\int^\infty_{-\infty} \delta( f(x,y) )\, dx\, dy, $$
but trying this out I don't think this works. What is wrong with this formula?
Edit: Am I missing a measure like a Jacobian or something? How can you prove this?
For the integral $L := \iint_{\mathbb R^2} \delta(f(x,y)) \, dx \, dy$ to measure the length of the curve $f(x,y) = 0$ we must have $\|\nabla f(x,y)\| = 1$ where $f(x,y) = 0.$ This is related to the scaling property.
So in the case of the circle with center in $(0,0)$ and radius $R,$ we could take $f(x,y) = \frac{1}{2R} (x^2+y^2-R^2).$ This would then give $$ L = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \delta\left( \frac{1}{2R}(x^2 + y^2 - R^2) \right) \, dx \, dy \\ = \{ \text{ polar coordinates } \} \\ = \int_0^{2\pi} \left( \int_0^\infty \delta\left( \frac{1}{2R}(r^2 - R^2) \right) \, r \, dr \right) \, d\theta \\ = \left\{ u = \frac{1}{2R}r^2 \right\} \\ = \int_0^{2\pi} \left( \int_0^\infty \delta\left( u-R/2 \right) \, R \, du \right) \, d\theta \\ = \int_0^{2\pi} R \, d\theta = 2\pi \, R. $$