Integral of $\exp$ over the unit ball I would like to know if the following integral has a closed form. Let $B(0,r)$ be the euclidean ball in $\mathbb{R}^d$. 
I am intered in $$\int_{B(0,r)}\exp(-i \langle x,\xi\rangle) dx_1\ldots dx_d$$
where $\langle \cdot,\cdot\rangle$ is the standard dot product. 
(I will assume that $\xi_i\ne 0$ for all $i$)
For $d=1$ it's easy but even for $d=2$ i am stuck. Using polar coordinate I get $$\int_{[0,\infty)\times (0,2\pi)} \exp(-ir(\cos\theta\xi_1 +sin\theta\xi_2))rdrd\theta.$$
Not helpful... 
 A: 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, 
$$
\int_{B_n(0,R)} \exp(-i\xi\cdot\mathbf{x})d^n\mathbf{x} = \left(\frac{2\pi R}{|\xi|}\right)^{n/2}J_{n/2}(R|\xi|),
$$
A: I guess, I have an idea.
You can use central symmetry of the problem and rotate your coordinate system.
Any rotation should have no impact on the integral.
Let me take Spherical coordinates
$$
\begin{align}
x_1 &= r \cos(\varphi_1) \\
x_2 &= r \sin(\varphi_1) \cos(\varphi_2) \\
x_3 &= r \sin(\varphi_1) \sin(\varphi_2) \cos(\varphi_3) \\
    &\vdots\\
x_{n-1} &= r \sin(\varphi_1) \cdots \sin(\varphi_{n-2}) \cos(\varphi_{n-1}) \\
x_n &= r \sin(\varphi_1) \cdots \sin(\varphi_{n-2}) \sin(\varphi_{n-1}),
\end{align}
$$
and rotate everything so that $\vec{\xi}$ coincides with the axis to which I measure $\varphi_1$.
In this case your integral can be written as
$$
\int_{\varphi_{n-1}=0}^{2\pi} \int_{\varphi_{n-2}=0}^\pi \cdots \int_{\varphi_1=0}^\pi\int_{r=0}^R
e^{-ir\cos(\varphi_1)|\xi|}
r^{n-1}\sin^{n-2}(\varphi_1)\sin^{n-3}(\varphi_2)\cdots \sin(\varphi_{n-2})\,
dr\,d\varphi_1 \, d\varphi_2\cdots d\varphi_{n-1},
$$
or integrating most angles we get (hopefully, I didn't make a mistake)
$$
\frac{(\sqrt{\pi})^{n-3}}{\Gamma\left(\frac{n-1}{2}\right)}
\int_{\varphi_{n-1}=0}^{2\pi} \int_{r=0}^R
e^{-ir\cos(\varphi_1)|\xi|}
r^{n-1}\sin^{n-2}(\varphi_1)\,
dr\,d\varphi_1.
$$
The rest is up to you.
You may check for similar technique used in Internal magnetic field distribution in plasmas for 3D case.
A: Taylor Series
Using $\Omega_{n-1}=\frac{\pi^{\frac{n-1}2}}{\Gamma\left(\frac{n+1}2\right)}$, we can compute the Taylor series of the integral for any $n$:
$$
\begin{align}
\int_{|x|\le R}e^{-ix\cdot\xi}\,\mathrm{d}x
&=\Omega_{n-1}R^n\int_{-1}^1\cos(|\xi|Rt)\,\left(1-t^2\right)^{\frac{n-1}2}\mathrm{d}t\tag1\\
&=\Omega_{n-1}R^n\int_0^1\cos\left(|\xi|R\sqrt{t}\right)\,(1-t)^{\frac{n-1}2}t^{-1/2}\,\mathrm{d}t\tag2\\
&=\Omega_{n-1}R^n\sum_{k=0}^\infty(-1)^k\frac{|\xi|^{2k}R^{2k}}{(2k)!}\frac{\Gamma\!\left(\frac{n+1}2\right)\Gamma\!\left(k+\frac12\right)}{\Gamma\!\left(\frac{n}2+k+1\right)}\tag3\\
&=\pi^{\frac{n-1}2}R^n\sum_{k=0}^\infty(-1)^k\frac{|\xi|^{2k}R^{2k}}{(2k)!}\frac{\Gamma\!\left(k+\frac12\right)}{\Gamma\!\left(\frac{n}2+k+1\right)}\tag4\\
&=\pi^{\frac{n}2}R^n\sum_{k=0}^\infty(-1)^k\frac{|\xi|^{2k}R^{2k}}{4^kk!\,\Gamma\!\left(\frac{n}2+k+1\right)}\tag5\\
\end{align}
$$
Explanation:
$(1)$: write as an integral in slices $\perp\!\xi$ of radius $R\sqrt{1-t^2}$ where $x\cdot\xi=|\xi|Rt$
$(2)$: apply symmetry to restrict to $[0,1]$ and substitute $t\mapsto\sqrt{t}$
$(3)$: use the Taylor series for cosine and the Beta Integral
$(4)$: apply the value of $\Omega_{n-1}$
$(5)$: $\frac{\Gamma\left(k+\frac12\right)}{(2k)!}=\frac{\sqrt\pi}{4^kk!}$
As a check, note that for $\xi=0$, the sum is
$$
\frac{\pi^{n/2}}{\Gamma(n/2+1)}R^n=\Omega_nR^n\tag6
$$
which is the volume of a sphere of radius $R$.

Closed Form
Applying the series expansion for the Bessel Function of the first kind
$$
J_\alpha(x)=\sum_{k=0}^\infty\frac{(-1)^k}{k!\,\Gamma(k+\alpha+1)}\left(\frac{x}2\right)^{2k+\alpha}\tag7
$$
to $(5)$ yields
$$
\int_{|x|\le R}e^{-ix\cdot\xi}\,\mathrm{d}x
=\left(\frac{2\pi R}{|\xi|}\right)^{n/2}J_{n/2}\!\left(|\xi|R\right)\tag8
$$
