Dirac delta have the representation
$$ (2\pi)^4\delta^4(x) := \int e^{ik.x} d^4 k $$
I would like to know how such integral representation realized in cylindrical coordinates. I tried the following parametrization where $k^{\mu}$ and $x^{\mu}$ are four vectors:
$$ k:= (k_0, \, k_r \cos{\phi_k}, \, k_r \sin{\phi_k}, \, k_z),\ \qquad x^{\mu}: ( t,\, r\sin{\phi}, \, r\cos{\phi}, \, z) $$ and wrote \begin{align} \int e^{ik.x} d^4 k &= \int dk_0 \, e^{i k_0 t} \int dk_z \, e^{-i k_z z} \int k_r \, d k_r \, d\phi_k \,e^{-i k_r r \cos{(\phi-\phi_k)}}\\ & = (2\pi)^2 \delta(z)\delta(t) \int k_r \, dk_r \, d\phi_k \, e^{i k_r r \cos{(\phi-\phi_k)}}\\ &= (2\pi)^2 \delta(z)\delta(t) \int k_r \, dk_r \, d\phi_k \sum_{m} i^m e^{i m (\phi_k-\phi)} J_{m}(-k_r r) \end{align}
The integrals along $k_0$ and $k_z$ are still cartesian. But I am stuck at the remainder. I have used the Jacobi-Anger expansion in the third line. I think the remaining integrals should effectively act as
$$ \frac{1}{r}\delta(r)\delta(\phi):= \int k_r \, dk_r \, d\phi_k \sum_{m} i^m e^{i m (\phi_k-\phi)} J_{m}(-k_r r) $$
but I could not show this is to be the case. I would appreciate any insight.