# 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 $c$.

• For $n=1$ I get $c\sqrt{\pi}/2$ ... – user26872 Jun 14 '13 at 20:45
• @oen: and now my answer agrees :-) – robjohn Jun 14 '13 at 21:45
• So this does not equal $c$, but it tends to $c$ as $n\to\infty$. – robjohn Jun 15 '13 at 0:04

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, $\displaystyle\frac{c}{4^n}\binom{2n}{n}\sqrt{\pi n}\sim c$.
• The modulus in (2) and after is shorthand for $\sqrt{x_1 ^2 + \cdots x_n ^2}$ right? And what precisely is meant by $d \sigma (x)$ in (3) ? This is all kind of new territory for me :) – Spine Feast Jun 14 '13 at 22:00
• @DepeHb: yes, $|x|=\sqrt{x_1^2+\dots+x_n^2}$ . $\mathrm{d}\sigma(x)$ is surface area on the sphere. – robjohn Jun 14 '13 at 22:05
• What is the exact definiton of the set $S^{n-1}_{+}$ ? – Spine Feast Jun 15 '13 at 16:13
• $\mathbb{R_+^n}$ is the quadrant of $\mathbb{R^n}$ with non-negative coordinates. $S_+^{n-1}$ is the quadrant of $S^{n-1}$ (the unit sphere) with non-negative coordinates: $\left\{(x_1,x_2,\dots,x_n):x_j\ge0\text{ and }\sum\limits_{j=1}^nx_j^2=1\right\}$. – robjohn Jun 15 '13 at 17:03