Fourier Transform in polar coordinates of 1

Question

Like in the table of transforms https://en.wikipedia.org/wiki/Fourier_transform#Distributions,_one-dimensional the FT (Fourier transform) of $\delta$ is 1 and the FT of 1 is $\delta$, but in polar coordinates the expressions for transformations is quite different, but like in section 4 of this review https://www.researchgate.net/publication/241069465_Two-Dimensional_Fourier_Transforms_in_Polar_Coordinates the Dirac delta for polar coordinates is $\frac{1}{r}\delta(r-r_0)\delta(\theta-\theta_0)$ and your FT is $e^{-i\vec{w}.\vec{r}_0}$, I am interested in Dirac delta function at the origin (subsection 4.1 of the review), this delta is $\frac{\delta(r)}{2\pi r}$ and your FT is 1, but the FT of 1 is $\frac{\delta(r)}{2\pi r}$?

My question arose because the FT in polar coordinates for radially symmetric functions is $$\mathbb{F}(f(r))=2\pi\int^{\infty}_0f(r)J_0(\rho r)dr=2\pi\mathbb{H}(f(r))$$ where $\mathbb{H}$ is the Hankel transform.

But, when f(r)=1, the integral above have solution only for finite intervals.

The table https://en.wikipedia.org/wiki/Hankel_transform#Some_Hankel_transform_pairs have the Hankel transform of 1 and it is $\frac{\delta(r)}{r}$, this is the only place where I had some answer, but I am not convinced. So my questions are, the FT in polar coordinates of 1 is $\frac{\delta(r)}{2\pi r}$? How prove it?

Both formulas are correct, but the first one is not valid if the delta is located at the origin. — Ninad Munshi, Feb 08 '21 at 20:20
Are you talking about the expression with the Hankel transform? — user740332, Feb 08 '21 at 20:30
The dirac delta in polar coordinates formula is what I was talking about, but I think I misunderstood your point of confusion — Ninad Munshi, Feb 08 '21 at 21:20

score 1 · Accepted Answer · answered Feb 10 '21 at 19:51

This was mostly proved in my answer here.

One has to be a bit careful when manipulating distributions (i.e. generalized functions). The first remark is that in general, the multiplication of two distributions is not defined (one can only define the multiplication of distributions with smooth functions). However, one can still look case by case. Here for example, one can see the distribution $f(r) := \,"\frac{\delta_0}{2πr}"$ as the solution of the equation $$ 2\pi\, r\, f(r) = \delta_0. $$ However, this equation have several solutions (see again here). If we choose the only homogeneous distribution that is solution, we find (see again here for proof) $$ f(r) = -\frac{1}{2\pi}\,\delta_0'(r) $$ where $\delta_0'$ is the derivative of the Dirac delta distribution. Again in the same post, I also proved that $f(r)$ is the radial representation of the Dirac delta $\delta_0$, an so since the Fourier transform of $FT(1)=\delta_0$ (in cartesian coordinates), its (homogeneous) radial representation is $f(r)$.

Fourier Transform in polar coordinates of 1

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