It is known (Mathematica) that
$$\int_0^\infty\int_{-1}^1 r^{n-1} e^{-\alpha^2(r^2 - r x)} (1-x^2)^{\frac{n-3}{2}} \, \mathrm{d}x\, \mathrm{d}r = \frac{\sqrt{\pi}}{2\alpha^n} \Gamma\left(\frac{n-1}{2}\right) \exp\left(\frac{\alpha^2}{4}\right)$$
I was trying to rederive this result elementary but either integration in any of the two variables would result in some intermediate integral with modified Bessel functions I was not able to compute by hand. Is there an elementary way how to derive the result using for example some clever substitution?
Additional question: Can one generalise the result changing $(n-3)/2$ to general $m$?
I am sorry to everyone! After just posting the question that bothered me a quite a long time, I just have found the answer and it is very elegant. I hope you don't mind posting this answer myself...
First in inner integral the substitution $x = \cos\theta$ is performed, then the integral is equal to
$$I =\int_0^\infty \int_0^\pi r^{n-1} \sin^{n-2}\theta e^{-\alpha^2 (r^2 - r \cos\theta)} d\theta dr $$
This integral looks like an integral in polar coordinates with $x=r\cos\theta$ and $y=r\sin\theta$ with the usual Jacobian $r$. Therefore
$$I =\int_{-\infty}^\infty \int_0^\infty y^{n-2} e^{-\alpha^2 (x^2 + y^2 - x)}dy dx = \frac{\Gamma\left(\frac{n-1}{2}\right)}{\alpha^{n-1}}\int_{-\infty}^\infty e^{-\alpha^2 (x^2 - x)}dx$$
The last step is trivial.