The Fourier transform of the (indicator function of the) unit ball is well known to be given by the Bessel functions (see [Fourier transform of the unit sphere](https://mathoverflow.net/questions/149692/fourier-transform-of-the-unit-sphere)).


What can be said about the $\ell\_1$ ball, that is the set of points in $\mathbb{R}^n$ with $\ell\_1$ norm bounded by radius $r$? More generally one can ask about $\ell\_p$ ball for other values of $p$. The interesting fact about $\ell\_2$ is that if the function is radially symmetric, the same can be said about its Fourier transform and effectively the transform becomes a uni-variate one. It's not clear if the same be said about the $\ell\_p$ ball for $p \neq 2$. On a related note, is there an orthogonal/useful transform which is as "friendly" to the $\ell\_1$ metric as the Fourier transform is to the $\ell\_2$ metric? (probably not, by the fact that "orthogonal" is an $\ell\_2$ notion.)



Fourier transform of the unit ball in L1 metric
The Fourier transform $F(\mathbf{k})$ of the unit ball in L1 metric can be evaluated as follows$^\ast$.  

*Notation:* $\theta(x)$ is the unit step function, $\delta(x)$ is the delta function, and ${\cal P}$ denotes the principal value of the integral.


$$F(k\_1,\ldots k\_n)=\int \cdots\int e^{i\mathbf{k\cdot x}}\theta\left(1-\sum\_{p=1}^n|x\_p|\right)dx\_1\cdots dx\_n$$
$$\qquad=\frac{1}{2\pi i}{\cal P}\int\_{-\infty}^\infty \frac{ds}{s}\left(1-e^{-is}\right)\prod\_{p=1}^n\left(\pi\delta(k\_p-s)+\pi\delta(k\_p+s)+\frac{2is}{s^2-k\_p^2}\right)\qquad \mathbf{(1)}$$
$$\qquad =\frac{1}{2\pi i}{\cal P}\int\_{-\infty}^\infty \frac{ds}{s}\left(1-e^{-is}\right)\prod\_{p=1}^n\left(\frac{2is}{s^2-k\_p^2}\right)$$
$$\qquad\qquad\qquad\qquad+\,\text{Re}\,\sum\_{q=1}^n\left[\frac{1-e^{-ik\_q}}{ik\_q}\prod\_{p=1,p\neq q }^n\left(\frac{2ik\_q}{k\_q^2-k\_p^2}\right)\right].\qquad \mathbf{(2)}$$
Then finally I arrive at the result



> 
> $$F(k\_1,k\_2,\ldots k\_n)=2\,\text{Re}\,\sum\_{q=1}^n\left[\frac{1-e^{-ik\_q}}{ik\_q}\prod\_{p=1,p\neq q }^n\left(\frac{2ik\_q}{k\_q^2-k\_p^2}\right)\right].\qquad \mathbf{(3)}$$
> 
> 
> 


I checked that for $n=2$ and $n=3$ this result agrees with the direct evaluation of the integrals (expressions given in the comment).




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$^\ast$ Details of the calculation: 
I first replace $\theta(1-\sum\_p|x\_p|)$ by $\theta(\xi-\sum\_p|x\_p|)$, denote the $\xi$-dependent integral by $F\_\xi$, and take the derivative with respect to $\xi$, $F\_\xi\mapsto F'\_\xi$, so that the step function becomes the delta function $\delta(\xi-\sum\_p|x\_p|)$. Then I Fourier transform with respect to $\xi$, $F'\_\xi\mapsto G\_s$. The $n$-fold integral of $G\_s$ over $x\_1$, $x\_2, \ldots x\_n$ factorizes, so I can carry it out explicitly, using the identity $\int\_0^\infty e^{ikx}\,dx=\pi\delta(k)+i{\cal P}k^{-1}$; finally I Fourier transform back $G\_s$ to the $\xi$ variable, to obtain $F'\_\xi$. One more integration of the $\xi$ variable, using $F\_{\xi=0}=0$, and then setting $\xi=1$ gives the first integral expression (1).


For the second integral expression I may safely assume that all $k\_p$'s are distinct, so products of delta functions do not contribute. I can then integrate out the single delta functions, obtaining the second integral expression (2).


The integral can then be performed by closing the contour in the lower half of the complex plane, the poles are shifted to the lower half of the complex plane, and the residue is reduced by a factor of two to account for the principal value. It then turns out that the integral equals exactly the contribution from the delta functions, so I end up with the final result (3).
