Integral of $\exp$ over the unit ball

Use $n$-dimensional spherical coordinates and choose the polar axis to be parallel to $\xi$. Then the integral is $$ \int_{\Omega} \int_0^R\int_0^\pi\exp(-i|\xi| r\cos\theta)r^{n-1}\sin^{n-2}\theta\,dr\,d\theta\,d^{n-2}\Omega, $$ where $d^{n-2}\Omega$ represents the integral over all other spherical angles. This has the same value as the surface area of the unit sphere in $n-1$ dimensions, given by $2\pi^{(n-1)/2}/\Gamma[(n-1)/2]$. So the integral is given by $$ \frac{2\pi^{(n-1)/2}}{\Gamma[(n-1)/2]}\int_0^R\int_0^\pi\exp(-i|\xi| r\cos\theta)r^{n-1}\sin^{n-2}\theta d\theta\,dr. $$ Applying some substitutions gives \begin{multline} \frac{2\pi^{(n-1)/2}}{\Gamma[(n-1)/2]}\int_0^R\int_0^\pi\exp(-i|\xi| r\cos\theta)r^{n-1}\sin^{n-2}\theta d\theta\,dr \\= \frac{2\pi^{(n-1)/2}}{|\xi|^n\Gamma[(n-1)/2]}\int_0^{R|\xi|} u^{n-1}\int_{-1}^1 \exp(-iu\zeta)(1-\zeta^2)^{(n-3)/2} d\zeta\,du. \\ = \frac{(2\pi)^{n/2}}{|\xi|^n}\int_0^{R|\xi|}u^{n/2}J_{n/2-1}(u)du = \left(\frac{2\pi R}{|\xi|}\right)^{n/2}J_{n/2}(R|\xi|), \end{multline} where $J_\nu$ is the Bessel J function and I used these two Bessel function identities to simplify the integral.

In conclusion, $$ \int_{B_n(0,R)} \exp(-i\xi\cdot\mathbf{x})d^n\mathbf{x} = \left(\frac{2\pi R}{|\xi|}\right)^{n/2}J_{n/2}(R|\xi|), $$