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What is the Fourier transform of the indicator of the unit ball in $\mathbb R^n$? I think it is known as one of special functions, so I would be happy to know which one.

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    $\begingroup$ The unit ball for the Euclidian norm I guess. $\endgroup$ Commented Sep 10, 2013 at 12:39

1 Answer 1

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Let $\alpha_d = \dfrac{\pi^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$ the volume of the $d$-dimensional unit ball. Since the characteristic function of the unit ball is rotationally symmetric, so is its Fourier transform, hence let's compute it at the point $\xi = (0,\,\dotsc,\,0,\,\rho)$ with $\rho > 0$:

$$\begin{align} \hat{\chi}_B(\xi) &= \frac{1}{(2\pi)^{n/2}} \int_{\lVert x\rVert < 1} e^{-i x_n\rho}\, dx_1\, \dotsc\, dx_n\\ &= \frac{1}{(2\pi)^{n/2}} \int_{-1}^1 \left(1-x_n^2\right)^{(n-1)/2}\alpha_{n-1} e^{-i x_n\rho}\,dx_n\\ &= \frac{1}{2^{n/2}\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)} \int_0^\pi \sin^n \varphi e^{-i\rho\cos\varphi}\,d\varphi. \end{align}$$

Recalling that we have the Bessel functions

$$J_p(x) = \frac{(x/2)^p}{\sqrt{\pi}\Gamma\left(p + \frac12\right)}\int_0^\pi \sin^{2p} \varphi e^{-ix\cos\varphi}\,d\varphi,$$

we see that

$$\hat{\chi}_B(\xi) = \lVert \xi\rVert^{-n/2}J_{n/2}(\lVert\xi\rVert).$$

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