How to integrate $$\int_{\mathbb{R}^n}|x|e^{-|x|^2}dx$$ in high dimension $\mathbb{R}^n$? In one dimension, by change of variables $s=x^2$, we have \begin{equation} \int_{-\infty}^\infty |x|e^{-x^2}dx = 2\int_0^\infty xe^{-x^2}dx = \int_0^\infty e^{-s}\, ds =1, \end{equation}
But I don't know how to do change of variables in $\mathbb{R}^n$?
Change to polar coordinates to get \begin{align} \int_0^{\infty}\int_{S^{n-1}}re^{-r^2}\,r^{n-1}d\sigma\,dr=A_{n-1}\int_0^{\infty}r^{n}e^{-r^2}\,dr, \end{align} where $A_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the surface area of the $(n-1)$-dimensional unit sphere $S^{n-1}\subset\Bbb{R}^n$. Now, make the substitution $t=r^2$, and with some finessing, you’ll get an answer in terms of the Gamma function. The crucial part in evaluating this integral is that the integrand was radial.
$$\frac1{(2\pi)^{n/2}}\int_{\mathbb R^n} \lVert x \rVert e^{-\frac12\lVert x \rVert^2}dx = \sqrt 2\frac {\Gamma(\frac{n+1}{2})}{\Gamma(\frac n2)}$$
– StubbornAtom May 27 '23 at 11:18