I am trying to solve the following integral in the 4D space \begin{equation} \int\exp ({-i \vec{w}^T \vec{x}}) \exp(-\frac{||\vec{x}||_2}{2}) d \vec{x} \end{equation}
I tried to follow the similar strategy shown in 3D Fourier transforms of $e^{-\beta r} $ and $re^{-\beta r} $ and solved the problem in the 4D spherical coordinate system. But the result didn't seem right.
I don't have a good idea about the polar direction in the 4D coordinate system. One thing I think I did wrong is I wrote $\vec{w}^T\vec{x}=||\vec{w}||_2 ||\vec{x}||_2 cos \theta, 0\leq \theta \leq \pi$ assuming $\vec{w}$ is parallel to the polar direction.
Since $e^{-\frac{||x||_2}{2}}$ is a radial function you are free to rotate your coordinate system such that $w$ is pointing to the "North Pole" in 4D. I will use the 4D spherical coordinates I derived from this post. So our integral becomes
$$4\pi \int_0^\pi \int_0^\infty e^{-(i||w||_2 \cos\theta + \frac{1}{2})r}r^3\sin^2\theta dr d\theta = 24\pi \int_0^\pi \frac{\sin^2\theta}{(i||w||_2\cos\theta + \frac{1}{2})^4}d\theta$$
where the complex coefficient exponential can be shown to have the same value as its counterpart on the real line by looking at a pie slice shaped contour in the complex plane, with the corner at the origin.
Recognize the fact that the integrand is even and can be turned into an integral over $[-\pi,\pi]$ with a factor of $\frac{1}{2}$, which lets us convert it into an integral over the unit circle in the complex plane. A long residue later:
$$\mathcal{F}(e^{-||x||_2}) = \frac{6\pi^2}{(\frac{1}{4}+||w||_2^2)^{\frac{5}{2}}}$$