Square equivalent of $circ(r)$

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I would like to know if there is a similar function to

$$circ(\sqrt{x^2+y^2})=1 , 0\leq \sqrt{x^2+y^2}\leq 1$$

but with a square domain $0\leq x\leq 1$ and $0\leq y\leq 1$.

If yes, which is its fourier transform?

I know that $\mathscr{F}circ(\sqrt{x^2+y^2})=\frac{J_1(2\pi\sqrt{\xi_x^2+\xi_y^2})}{\sqrt{\xi_x^2+\xi_y^2}}$

Thanks!