integral over ball in $\mathbb R^d$

Question

How can I determine the integral
$$\int_{\|x\|<1}e^{-2\pi i\langle x,y\rangle}\mathrm dx$$ for $x\in \mathbb R^d$? I thought about using polar coordinates in $\mathbb R^d$ but I think this is very hard. Thanks for any ideas.

Answer 1

As pointed out in the comments we are free to assume that $y=(|y|,0,0,\ldots)$, so the integral equals $$ \int_{x_1^2+\ldots+x_d^2\leq 1} e^{-2\pi i |y| x_1}\,d\mu = \int_{-1}^{1}e^{-2\pi i |y| z}\mu(x_2^2+\ldots+x_d^2\leq 1-z^2)\,dz$$ or $$ \frac{\pi^{(d-1)/2}}{\Gamma\left((d+1)/2\right)}\int_{-1}^{1}e^{-2\pi i |y| z}(1-z^2)^{\frac{d-1}{2}}\,dz = |y|^{-d/2} J_{d/2}(2\pi|y|)$$ where the Bessel function appearing in the RHS simplifies into elementary functions if $d$ is odd.
For $d=3$ and moderately large values of $|y|$ the RHS is almost indistinguishable from $-\frac{\cos(2\pi |y|)}{\pi|y|^2}$.