Integral in $n$-dimensional spherical coordinates I have to calculate the following integral:
$$\frac{c^{-2n}}{\sqrt{2n}}\int_0^\infty\cdots\int_0^\infty \left( \sqrt{x_1 ^2 +\cdots+x_n ^2}\right) (x_1\ldots x_n) \exp{ \left( -\frac{x_1^2+\cdots+x_n^2}{2c^2} \right) } \, dx_1 \cdots dx_n$$
by transforming to spherical polar coordinates. Looking here I found the transformations, but I'm not sure what good this does with this ugly factor of $x_1 \ldots x_n$ sitting there. Is this doable at all or did I screw up somewhere earlier deriving the formula?
This arises when I want to test if the estimator $\hat{c} = \frac{X_\mathrm{RMS}}{\sqrt{2}}$ for the Rayleigh distribution is unbiased, so I'm calculating $\frac{1}{\sqrt{2n}} E\left[\sqrt{X_1^2+\cdots+X_n^2}\right]$ hoping to get $c$.
 A: Note that
$$
\int_0^\infty xe^{-x^2/2}\,\mathrm{d}x=1\tag{1}
$$
Thus, multiplying $n$ copies of $(1)$ together yields
$$
\int_{\mathbb{R_+^n}}x_1x_2\dots x_n\,e^{|x|^2/2}\,\mathrm{d}x=1\tag{2}
$$
Consider the following integral over a quadrant of the surface of the unit sphere:
$$
\int_{S_+^{n-1}}x_1x_2\dots x_n\,\mathrm{d}\sigma(x)=\Lambda_{n-1}\tag{3}
$$
Integrating $(3)$ times $e^{-r^2/2}$ over quadrants of spheres of radius $r$ and thickness $\mathrm{d}r$ yields
$$
\begin{align}
\int_{\mathbb{R_+^n}}x_1x_2\dots x_n\,e^{|x|^2/2}\,\mathrm{d}x
&=\int_0^\infty\Lambda_{n-1}r^{2n-1}e^{-r^2/2}\,\mathrm{d}r\\
&=2^{n-1}\Lambda_{n-1}\Gamma(n)\tag{4}
\end{align}
$$
Equating $(2)$ and $(4)$ yields
$$
\Lambda_{n-1}=\frac1{2^{n-1}\Gamma(n)}\tag{5}
$$
The integral in question adds one more power of $r$ to the integrand in $(4)$
$$
\begin{align}
\frac{c^{2n}}{\sqrt{2n}}\int_{\mathbb{R_+^n}}|x|\,x_1x_2\dots x_n\,e^{|x|^2/(2c^2)}\,\mathrm{d}x
&=\frac{c}{\sqrt{2n}}\int_{\mathbb{R_+^n}}|x|\,x_1x_2\dots x_n\,e^{|x|^2/2}\,\mathrm{d}x\\
&=\frac{c}{\sqrt{2n}}\int_0^\infty\Lambda_{n-1}r^{2n}e^{-r^2/2}\,\mathrm{d}r\\
&=\frac{c}{\sqrt{2n}}\frac{2^{n-1/2}\Gamma(n+1/2)}{2^{n-1}\Gamma(n)}\\
&=\frac{c}{4^n}\binom{2n}{n}\sqrt{\pi n}\tag{6}
\end{align}
$$
Asymptotically, $\displaystyle\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$. Therefore, $\displaystyle\frac{c}{4^n}\binom{2n}{n}\sqrt{\pi n}\sim c$.
