# Dirac Delta in polar coordinates, integrating delta from 0 to $\infty$

Many sources mention how to recast Dirac's Delta function from Cartesian into polar or spherical coordinates. They say e.g. for polar: $$\int_{-\infty}^{\infty}\delta(\vec{x})\,{\rm d}^3x=\int_{-\infty}^{\infty}\int_{-\infty}^\infty\delta(x)\delta(y)\,{\rm d}x\,{\rm d}y=\int_0^\infty\int_0^{2\pi}\frac1r\delta(r)\delta(\phi)\,r\,{\rm d}\phi\,{\rm d}r$$

(while they mostly work with $\delta(x-x')\delta(y-y')$ I recast it to the special case of $x'=0=y'$).

however, if I rewrite the last integral as: $$\int_0^\infty\delta(r)\,{\rm d}r\int_0^{2\pi}\delta(\phi)\,{\rm d}\phi$$

what are these integrals? They should both evaluate to $1$ to keep $\int_{-\infty}^{\infty}\delta(\vec{x})\,{\rm d}^3x=1$ as suggested here but that is far from obvious to me (I do not see how they came to the result in the answer's last equation). Another similar topic does not really resolve the issue to my understanding. Actually, it occured to me:

$$\int_0^\infty\delta(r)\,{\rm d}r=\int_0^\infty\frac{{\rm d} H(x)}{{\rm d}x}\,{\rm d}x=H(\infty)-H(0) = 1 - H(0)$$

where $H(x)$ is the standard Heavyside function: $H(x)=1$ for $x>0$ and $H(x)=0$ for $x<0$.

They are actually not welldefined as they stand, but a common way to solve it is to identify the integrals as being taken from just below $0$, e.g. $$\lim_{\epsilon \to 0+}\int_{-\epsilon}^\infty \delta(r) \, \mathrm{d}r$$