Consider the following integral over a 2D plane,
$$\iint \mathrm{d}^2\mathbf{k}\ e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi^2\delta^2(\mathbf{r})$$
This is a Fourier transform of a distribution which is rotationally symmetric around the origin, and thus the result should also be rotationally symmetric around the origin. So it should be expressible in terms of the magnitude of $\mathbf{r}$ only, i.e. in terms of $r \equiv \lVert\mathbf{r}\rVert$.
$$\iint \mathrm{d}^2\mathbf{k}\ e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi^2 f(r)$$
What is the proper mathematical expression for $f(r)$, if it exists?
Clearly $f(r) = \delta(r)$ is unsuitable because it doesn't have the right units. Dimensional analysis suggests that it might be something like $f(r) = \frac{1}{r}\delta(r)$, but I'd like to have some sort of mathematical justification rather than just a guess.
I've looked at Delta function in curvilinear coordinates but that question is somewhat more abstract, plus it doesn't seem to eliminate the dependence on $\theta$ as I would like to do.
Best Answer
Expressions like $\delta(r)/r$ do not generally define a distribution. But if we define a distribution $T$ by setting
$$T(\phi)=\int_0^{2\pi}\int_0^{\infty} \phi(r\cos \theta,r\sin\theta) \delta(r) \mathrm dr \mathrm d\theta=\int_0^{2\pi}\int_0^{\infty} \phi(r\cos \theta,r\sin\theta) (\delta(r)/r) r \mathrm dr \mathrm d\theta$$
for test functions $\phi$, then $T=2\pi \delta(\mathbf x)$. So, informally, $\delta(\mathbf x)=\delta(r)/2\pi r$.