A seemingly simple integral How would one show the following equality,
$$\int_{\mathbb{R}^d}|x|e^{-\frac\beta 2 |x|^2}dx = c_d\beta^{-\frac{d+1}{2}}.$$
Where $c_d$ is a constant only depending on $d$. This is problem is relatively easy to do when $d=1$ by either substitution or integration by parts, when we know the value of the Gaussian integral. My problem is that I'm not aware of any substitution techniques for higher dimensions and integration by parts (i.e. Greens identities) leads me to having to compute a higher dimensional surface integral, which, in arbitrary dimensions I do not know how to compute.
Finally tried brute forcing this and splitting it into $d$ integrals, each integral having the form,
$$ \int_\mathbb R \sqrt{(x^2+a)}e^{-x^2}dx.$$
Which annoyingly wolfram alpha doesn't give me a useful answer for...
 A: As @A rural reader and @projectilemotion mentioned, the appropriate way to evaluate this integral is to switch to spherical coordinates
$$I=\int_{\mathbb{R}^d}|x|e^{-\frac\beta 2 |x|^2}dx = \int..\int_{-\infty}^\infty\sqrt{x_1^2+...+x_d^2}\,e^{-\frac\beta 2 (x_1^2+...+x_d^2)}dx_1...dx_d=$$ $$= \int{d}\Omega\int_{0}^\infty{r}\,e^{-\frac\beta 2 r^2}r^{d-1}dr$$
where $\int{d}\Omega$ is integration over all angles in spherical coordinates, and $\int_0^\infty...{d}r$ is integration over radius.
We can find $\int{d}\Omega$ from
$\int..\int_{-\infty}^{\infty}e^{-x_1^2-x_2^2-...-x_d^2}dx_1...dx_d=$$\int{d}\Omega\int_0^{\infty}e^{-r^2}r^{d-1}dr \Rightarrow \pi^{\frac{d}{2}}=$$\frac{1}{2}$$\int{d}\Omega$$\int_0^{\infty}$$e^{-t}t^{\frac{d}{2}-1}dt$
Integral over all angles $\int{d}\Omega=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}$
$$I=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}\int_{0}^\infty{r}\,e^{-\frac\beta 2 r^2}r^{d-1}dr=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}\int_{0}^\infty{e}^{-t^2}t^ddt=$$ $$=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}\frac{1}{2}\int_{0}^\infty{e}^{-x}x^{\frac{d-1}{2}}dx$$
$$I=\pi^{\frac{d}{2}}\frac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}(\frac{\beta}{2})^{-\frac{d+1}{2}}$$
