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Is there a built-in function in Mathematica for this function?

\begin{equation} {\displaystyle \operatorname {Circ} (r)=\left\{{\begin{array}{rl}1,&{\text{if }}r<{1}\\{\frac {1}{2}},&{\text{if }}r={1}\\0,&{\text{if }}r>{1}.\end{array}}\right.} \end{equation} where $r=\sqrt{x^2+y^2}$

I use:

UnitBox[1/2*Sqrt[x^2 + y^2]]

and the function is equivalent, but Mathematica doesn't do the Fourier transform in my system (MMA v. 11.0).

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  • $\begingroup$ FourierTransform[UnitBox[1/2*Sqrt[x^2 + y^2]], {x, y}, {u, v}] returns a result on my machine. $\endgroup$ – Chip Hurst Jan 22 at 14:14
  • $\begingroup$ MMA version 12.1 on PC returns: 1/2 Hypergeometric0F1Regularized[2, 1/4 (-u^2 - v^2)] $\endgroup$ – Daniel Huber Jan 22 at 14:21
  • $\begingroup$ well, my computer is simple, maybe because of this he doesn´t do the transform, my wolfram version is 11.0 $\endgroup$ – user740332 Jan 22 at 14:26
  • $\begingroup$ the fourier transform of circ is a bessel function J1(2*Pi*r)/r $\endgroup$ – user740332 Jan 22 at 14:27
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$Version

(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)

The functions are equivalent for integration but they have different values for r == 1. Use PiecewiseExpand to convert the expression

Assuming[Element[{x, y}, Reals],
 UnitBox[1/2*Sqrt[x^2 + y^2]] // PiecewiseExpand]

enter image description here

Assuming[Element[{x, y}, Reals],
 FourierTransform[
  UnitBox[1/2*Sqrt[x^2 + y^2]] // PiecewiseExpand, {x, y}, {u, v}]]

enter image description here

% // FullSimplify

enter image description here

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You can obtain a result in terms of Bessel functions using FunctionExpand:

FunctionExpand@
 FourierTransform[UnitBox[1/2*Sqrt[x^2 + y^2]], {x, y}, {u, v}]

(* Out: BesselI[1, Sqrt[-u^2 - v^2]]/Sqrt[-u^2 - v^2] *)
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