# Fourier transform in polar coordinates using built-in hankel transform of the function constant 1 [closed]

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 question is, the FT in polar coordinates of 1 is $$\frac{\delta(r)}{2\pi r}$$?

When I do the Hankel Transform in mathematica it gives

In[1]:= HankelTransform[1, r, w]

Out[1]= HankelTransform[1, r, w, 0]

In[2]:= InverseHankelTransform[DiracDelta[w]/(2*Pi*w), w, r]

Out[2]= (1 - HeavisideTheta[0])/(2 \[Pi])

In[3]:= HankelTransform[DiracDelta[w]/(2*Pi*w), w, r]

Out[3]= (1 - HeavisideTheta[0])/(2 \[Pi])

In[4]:= InverseHankelTransform[1, r, w]

Out[4]= InverseHankelTransform[1, r, w, 0]


I should define "HeavisideTheta[0]=0" to get the answer, for Out[2], I should do the Hankel transform and multiply for $$2\pi$$ to get the FT, and it should be $$\delta(r)/r$$, but mathematica don't make this.

• In my opinion, this is rather math than Mathematica. Let us consider the one-dimensional case. The command FourierTransform[1, x, s] results in Sqrt[2 \[Pi]] DiracDelta[s]. However, the improper integral $$\frac 1 {\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ixs}\,dx$$ diverges for each real number $s$ and this Fourier transform is interpreted in the sense of distributions. It is impossible to explain that in a comment. Ask your question in MSE. – user64494 Feb 10 at 14:42
• It should be added in the linked article Natalie Baddour handles Fourier transforms of such kind as usual Fourier transforms and this causes contradictions. – user64494 Feb 10 at 14:56
• I asked in MSE but nobody answered math.stackexchange.com/questions/4018066/… – user740332 Feb 10 at 15:22