Computing 2d radially symmetric Fourier transforms (with Wolfram Alpha)

Question

Let's assume that I "know"

• $\mathcal{F}\{\operatorname{circ}(r)\}(\rho)=\frac{J_1(2\pi\rho)}{\rho}$
• $\mathcal{F}\{(1-r^2)\operatorname{circ}(r)\}(\rho)=\frac{J_2(2\pi\rho)}{\pi\rho^2}$

because I have looked up the first formula in a book, and derived the second formula by laborious manual computations. I have verified both formulas by actually using them in some computer code for "real" work. I "suspect" that $\mathcal{F}\{(1-r^2)^n\operatorname{circ}(r)\}(\rho)=\frac{n!J_{n+1}(2\pi\rho)}{\pi^n\rho^{n+1}}$. Because of issues related to my computer code, it would be nice if I also "knew"

• $\mathcal{F}\{(1-r^2)^2\operatorname{circ}(r)\}(\rho)=\frac{2J_3(2\pi\rho)}{\pi^2\rho^3}$
• $\mathcal{F}\{(1-r^2)^3\operatorname{circ}(r)\}(\rho)=\frac{6J_4(2\pi\rho)}{\pi^3\rho^4}$

I tried to tell Wolfram Alpha that I want to know the 2d radially symmetric Fourier transform of $\operatorname{circ}(r)$, but it neither understands "$\operatorname{circ}$" nor "2d Fourier transform". Finally I wrote

• integral of 2 pi x besselj(0,2 pi r x) from 0 to 1
• integral of 2 pi x (1-x^2) besselj(0,2 pi r x) from 0 to 1

The first query gave the correct answer $\frac{J_1(2\pi r)}{r}$, but the second gave a cryptic $\pi{}_0\tilde{F}_1(;3;-\pi^2r^2)$. But

• integral of x (1-x^2)^2 besselj(0,r x) from 0 to 1
• Integrate[x (1 - x^2)^3 BesselJ[0, r x], {x, 0, 1}]

where I omitted the $2\pi$ factors, gave the helpful answers $\frac{8J_3(r)}{r^3}$ and $\frac{48J_4(r)}{r^4}$. In a way this is all that I need for my computer code, but I'm still left puzzled how to communicate with Wolfram Alpha (or other computer algebra systems) about 2d Fourier transforms.

So what is my question? For example, I can now let Wolfram Alpha verify my "suspected" formula for many different small $n$, but that is not the same as verifying the formula for all $n\in\mathbb{N}$. Also, if I hadn't "suspected" the simple formula, I have the impression that the results of my failed attempts would have easily led me to the conclusion that Wolfram Alpha can't help me with my (specific) problem. Would using a standalone computer algebra system and digging into its documentation lead to more predictable results?

Answer 1

Since Wolfram Alpha is closely related to Mathematica software, it is reasonable to look up Mathematica documentation on Fourier transform and use its syntax in Wolfram Alpha. For example,

FourierTransform[(x*y)^6*Exp[-x^2-y^2],{x,y},{u,v}]

produces 2D Fourier transform in Wolfram Alpha.

However, you have a piecewise constant function, which would be appropriately expressed as HeavisideTheta[1-x^2-y^2] in Mathematica syntax. Unfortunately, computation with this piecewise defined function, e.g.,

FourierTransform[(1-x^2-y^2)*HeavisideTheta[1-x^2-y^2],{x,y},{u,v}] 

times out in Wolfram Alpha, whose capability is intentionally reduced compared to commercial product (Mathematica).

Would using a standalone computer algebra system and digging into its documentation lead to more predictable results?

Yes, using more powerful software (as compared to online demo) and reading its documentation is likely to improve the outcome of computation.