Prove that for $\vec{x},\vec{\xi} \in \mathbb{R}^D$, $\mathbb{R} \ni c, \alpha = \mathrm{const.}$ the $D$-dimensional Fourier transform of $ f(\vec{x}) = \exp{(-c|\vec{x}|^\alpha})$ \begin{equation} \hat{f}(\vec{\xi}) = \int \exp{(-c|\vec{x}|^\alpha}) e^{-i\vec{\xi} \cdot \vec{x}} \, d^D \vec{x} \end{equation} can be transformed into the integral over just one variable $x = |\vec{x}|$ \begin{equation} \hat{f}(\xi) = (2\pi)^{D/2} \xi^{1-D/2} \int_0^\infty J_{D/2 - 1} (\xi x) x^{D/2} \exp{(-cx^\alpha)} \, dx \end{equation} where $\xi = |\vec{\xi}|$ and $J_\nu$ is a Bessel function. Are hyperspherical coordinates of use here? Maybe the integral definition of the Bessel function \begin{equation} J_\nu(x) = \frac{1}{2\pi} \int_{-\pi}^\pi \exp{[i(\nu t - x \sin t)]} \, dt \end{equation} is useful for the proof. I am thankful for any ideas.
Edit: I now noticed that this integral definition cannot be used here since it is only defined for integer values of $\nu$, whereas $D/2 -1$ is a multiple of 1/2 so it can be a rational value.
Since $f$ is rotation symmetric, so is $\hat f$. Therefore we can use a fixed $\vec\xi$; lets take $\vec\xi=(0,\ldots,0,\xi).$
Then, to parameterize the space, we start with writing $\vec x = r\vec n,$ where $r=|\vec x|$ and $\vec n\in S^{D-1},$ the unit sphere in $D$-dimensional space. Then we parameterize $S^{D-1}$ by $n_D=\cos\theta$ and $(n_1,\ldots,n_{D-1})=\sin\theta\ \vec\nu,$ where now $\vec\nu\in S^{D-2},$ the unit sphere in $(D-1)$-dimensional space, and $\theta\in[0,\pi].$ Our parameterization of the space thus is $$ (x_1,\ldots,x_{D-1},x_D)=(r\sin\theta\ \nu_1,\ldots,r\sin\theta\ \nu_{D-1}, r\cos\theta). $$
With this parameterization the volume element becomes $d^D\vec x = r^{D-1}\sin^{D-2}\theta \, dr \, d\theta \, d\omega,$ where $d\omega$ is the surface measure on $D^{D-2}.$
Thus, $$ \hat{f}(\vec{\xi}) = \int e^{-c|\vec{x}|^\alpha} e^{-i\vec{\xi} \cdot \vec{x}} \, d^D \vec{x} = \int_{r=0}^\infty \int_{\theta=0}^\pi \int_{\vec\nu\in S^{D-2}} e^{-cr^\alpha} e^{-i\xi r\cos\theta} \, r^{D-1}\sin^{D-2}\theta \, dr \, d\theta \, d\omega(\vec\nu). $$
Since the integrand has no dependency on $\vec n$ the integral over that variable can be extracted and we get $$ \begin{align} \hat{f}(\vec{\xi}) &= \omega(S^{D-2}) \int_{r=0}^\infty \int_{\theta=0}^\pi e^{-cr^\alpha} e^{-i\xi r\cos\theta} \, r^{D-1}\sin^{D-2}\theta \, dr \, d\theta \\ &= \omega(S^{D-2}) \int_{0}^\infty e^{-cr^\alpha} r^{D-1} \left(\int_{0}^\pi e^{-i\xi r\cos\theta} \, \sin^{D-2}\theta \, d\theta \right) \, dr, \end{align} $$ where $\omega(S^{D-2})$ is the total surface size of $S^{D-2}.$
Can you finish from here?