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I wish to calculate the integral $$I=\frac{1}{\omega_n}\int_{\mathbb{S}^n}\! x_{n+1}^2\,\mathrm{d}x$$ where $\mathbb{S}^{n}$ denotes the unit sphere in $\mathbb{R}^{n+1}$ with surface area $\omega_n$, and $\mathrm{d}x$ denotes integration w.r.t. the usual measure on $\mathbb{S}^{n}$, and in a slight abuse of notation, $x_{n+1}$ refers to the coordinate in $\mathbb{R}^{n+1}$. This is my approach as a solution.

We may rewrite this integral in stereographic coordinates (see my previous question for the relevant formulae) as $$ I= \frac{1}{\omega_n}\int_{\mathbb{R}^n}\,\left(\frac{2y_i}{1+|y|^2}\right)^2\frac{2^n}{(1+|y|^2)^n}\,\mathrm{d}y = \frac{1}{\omega_n}\int_{\mathbb{R}^n}\,\frac{2^{n+2} y_i^2}{(1+|y|^2)^{n+2}}\,\mathrm{d}y $$ but then \begin{align} nI &= \frac{1}{\omega_n}\int_{\mathbb{R}^n}\,\frac{2^{n+2}|y|^2}{(1+|y|^2)^{n+2}}\,\mathrm{d}y = 2^{n+2}\frac{\omega_{n-1}}{\omega_n}\int_{\mathbb{R}^n}\,\frac{r^{n+1}}{(1+r^2)^{n+2}}\,\mathrm{d}y \\ &= 2^{n+1}\frac{\omega_{n-1}}{\omega_n} \int_0^\infty{\frac{s^{\frac n2}}{(1+s)^{n+2}}}\,\mathrm{d}s \tag*{ $r^2=s$ } \\ &= 2^{n+1}\frac{\omega_{n-1}}{\omega_n}\int_0^1 \left(\frac{1-t}{t}\right)^{\frac n2}t^n\,\mathrm{d}t \tag*{ $ t=1/(1+s) $ } \\ &= 2^{n+1}\frac{\omega_{n-1}}{\omega_n} B\left(\frac n2+1,\frac n2+1\right) \\ &= 2^{n+1}\frac{\omega_{n-1}}{\omega_n}\frac{\Gamma(\tfrac n2 + 1)^2}{\Gamma(n+2)} \end{align} where $B$ is the beta function, so that $$ \frac{1}{\omega_n}\int_{\mathbb{S}^n}\! x_{n+1}^2\,\mathrm{d}x = \frac{2^{n+1}\omega_{n-1}\Gamma(\tfrac n2 + 1)^2}{n \omega_n\Gamma(n+2)}. $$

This seems awfully complicated. Is there any way to get a simpler answer to this? Expanding out $\omega_{n-1}$ and $\omega_n$ doesn't seem to help.

Ted Shifrin
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Dominic Wynter
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1 Answers1

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Yes! The fact that's it's $x_{n+1}$ is not important: by applying rotations to the coordinates, we find that $$ \int_{S^n} x_1^2 \, dx = \int_{S^n} x_2^2 \, dx = \dotsb = \int_{S^n} x_{n+1}^2 \, dx. $$ Adding all $n+1$ of these together gives $$ (n+1)\int_{S^n} x_{n+1}^2 \, dx = \int_{S^n} (x_1^2+x_2^2+\dotsb+x_{n+1}^2) \, dx = \int_{S^n} \, dx = \omega_n, $$ since the sum of the squares of the coordinates is $1$ on the unit sphere. Hence $$ \frac{1}{\omega_n}\int_{S^n} x_{n+1}^2 \, dx = \frac{1}{n+1}. $$

Chappers
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  • Never before have I wasted so much effort on a result so simple... thanks for the answer though. – Dominic Wynter Oct 17 '17 at 22:28
  • Happens to all of us :) – Chappers Oct 17 '17 at 22:31
  • Conversely, is there anything we can say about $$ \frac{1}{\omega_n}\int_{S^n} f(x_1^2,\ldots,x_{n+1}^2), dx $$ for a polynomial $f$? – Dominic Wynter Oct 18 '17 at 00:14
  • Probably very little, unfortunately. One can replace the $x_{n+1}^2$ with $1-\sum_{i=1}^n x_{i}^2$ (or one of the others if more suitable), and then the terms collect into various types depending on the various powers involved. It's also $1/n!$ of $\int_{S^n} \sum_{\sigma \in S_n} f(x_{\sigma(1)}^2,\dotsc,x_{\sigma(n+1)}^2) , dx $, which is symmetric and so can be written in terms of the elementary symmetric polynomials, but $\sum_{i=1}^{n+1} x_i^2=1 $ is the only relation between the coordinates, so that may not be that much help. Probably worth asking a separate question. – Chappers Oct 18 '17 at 01:03