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The question ‘what proportion $p_2$ of a square is closer to the centre than the outside?’ has been asked here before and the answer shown to be $\frac{4\sqrt{2}-5}{3}$. In another question asking the same for an equilateral triangle an answer is given generalizing this to a regular $n$-gon.

I once extended the solution method of this answer to computing the proportion $p_3$ of a cube closer to the centre than the outside, by considering a $2\times2\times2$ cube centred at the origin and considering by symmetry only the square pyramid whose base is the side $y=1$ and apex is the origin $(0,0,0)$ and then further restricting this by requiring by symmetry $z>0$ and $x>z$ (the volume under consideration will then be $\frac{1}{48}$th of the cube). The integrand of our problem is then $\frac{1-x^2-z^2}{2}-x$ but the limits of the integration become more complex than the two-dimensional case, and the solution to the problem is now given (unless I’ve made an error) by:

$$p_3=6\int_{0}^{\frac{\sqrt{3}-1}{2}}\int_{z}^{\sqrt{2-z^2}-1}\frac{1-x^2-z^2}{2}-x\;dx\;dz$$

where the upper limits of the integrations are found by solving $\frac{1-x^2-z^2}{2}=x$ and then $\sqrt{2-z^2}-1=z$ and the lower limits follow by the definition of the region we are integrating on; the factor of $6$ comes from dividing $48$ by $2^3$ (the volume of the whole cube). I once managed to integrate this by hand to:

$$p_3=\frac{5}{4}-\frac{9\sqrt{3}}{8}+\frac{\pi}{4}$$

and Wolfram Alpha agrees giving $p_3\approx0.086841$. However, if we step this up to a 4-cube then the integral becomes correspondingly more complicated and I have no idea how to explicitly compute the resulting integral. I do not know how this method could possibly be extended easily either.

My question is this: the problem of what proportion $p_n$ of an $n$-cube is closer to the centre than the periphery seems to be one that might have a simpler or deeper way of solving it (perhaps just because it can be stated so easily); is there an alternative solution method to just doing a brute force integration over the bounding region? If not, is there any way the integration can be transformed to be more tractable so that for instance it can be calculated explicitly for a 4-cube?

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Anon
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    The displayed form for $p_3$ is much greater than the numerical approximation shown just below it, and cannot be correct. – John Bentin Nov 11 '16 at 08:49
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    I'm not sure whether this is useful or not. An alternate expression for $p_n$ is $$p_n = \int_{[0,1]^{n-1}}\left(\frac{1}{\sqrt{1+u_1^2+ \cdots + u_{n-1}^2} + 1}\right)^n du_1\ldots du_{n-1}$$ This can be derived by splitting the $n$-cube $[-1,1]^n$ into $2n$ "pyramids" with apex at origin and bases at the $2n$ facets of $(n-1)$-cube, One then parameterize the pyramids with change of coordinates of the form: $(x_1, \ldots, x_{n-1}, x_n ) = (\lambda u_1, \lambda u_2, \ldots, \lambda u_{n-1}, \lambda )$ and then integrate over $\lambda$. – achille hui Nov 11 '16 at 09:11
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    Using above integral, I get $p_3 = \frac{10 - 9\sqrt{3} + 2\pi}{8} \approx 0.08684100488246149$. Look like there is a typo in your expression. – achille hui Nov 11 '16 at 09:13
  • Thanks for noting the error guys! – Anon Nov 19 '16 at 02:39
  • @achillehui I'm still a bit confused from your comment about how exactly you get that integral formula. Where does the $(n-1)$-cube you've mentioned go exactly? – Anon Nov 19 '16 at 05:41
  • Let's take $n = 3$, a $n$-cube is the ordinary cube in 3D. It has six facets which are squares. The squares are the $(n-1)$-cubes. – achille hui Nov 19 '16 at 05:53
  • Oh, I see what you mean. – Anon Nov 19 '16 at 06:02
  • @achillehui I wonder whether using spherical coordinates would help? – Anon Dec 01 '16 at 03:26
  • I doubt that. for one layer of sine/cosine in a square root, we may get rid of it by trigonometic identities. having two layer is already problematic. for $p_4$, there are 3 layers of sine/cosine ! IMHO, a better direction is figure out the moment generating function of $r = \sqrt{x_1^2 + x_2^2 + \cdots + x_m^2}$ for the $m$-cube $[0,1]^m$. – achille hui Dec 01 '16 at 07:25
  • @achillehui What I was thinking was that it seems like in spherical coordinates the surface of equal distance between centre and face seems to be able to be expressed as $r=\frac{1}{1+\cos\theta}$ where $\theta$ is one of the angular coordinates, so that the integral would just have simple bounds on the angles and the integrand would not be much more complex than the Jacobian, and simplify down to powers of sin and cosine. I don't have the integration oomph to attack that though. – Anon Dec 09 '16 at 22:43

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