So today is perhaps the first real productive day for creative thinking; I have been engaging in mild reading the first day I walked out of the hospital. Here is some brief note on the results:

The integral mentioned in the title is the following highly symmetric and natural one:

int_{S^{n-1}} prod_{i=1}^n |x_i| sigma( dx_1ldots dx_n)

Notice the absolute value is needed because otherwise by symmetry the integral would be 0. Alternatively we could integrate only over the region of the (n-1) sphere with positive coordinates. That clearly equals 2^{-n} times the value of the integral above by symmetry again.

The following paragraph is optional reading, only intended for those who know the Kac’s model of Boltzmann gas.

It came up in my search for a Lipschitz function on the unit sphere as test function for showing slow convergence in L^1 Wasserstein distance of the Kac random walk. Computationally the easiest nontrivial such test functions are sum of squares of the coordinates, as such functions are natural invariants under Euclidean rotations. But since such functions also don’t capture the interactions between coordinates, they tend to give false fast convergence.

The way to compute the above integral is simply by disintegration. Namely we know that under the uniform probability measure on the sphere, the marginal distribution of the coordinates squared are beta distributed, hence in general if we have a function f:[-1,1] to RR, then

int_{S^{n-1}} f(x_n) sigma(dx_1 ldots dx_n) = int_{-1}^1 f(x) Gamma(n/2)/(Gamma(1/2) Gamma((n-1)/2)) (1-x^2)^{(n-2)/2} dx

If we denote our target integral by p(n,1), where 1 stands for the radius of the sphere, then we shall derive a recursive formula for p(n,r) in general, in terms of lower n’s.

The crucial observation is that when we fix x_n = c, the last coordinate on S^{n-1}, the submanifold described by the intersection of S^{n-1} and the hyperplane { x_n = c } is also a sphere, S^{n-2}(r = (1-c^2)^{1/2}).

Also we have the following scaling property (exercise):

p(n,r) = r^n p(n,1)

So p(n,1) = int_{-1}^1 |x| p(n-1,(1-x^2)^{1/2}) Gamma(n/2)/(Gamma(1/2) Gamma((n-1)/2)) (1-x^2)^{(n-2)/2} dx

= p(n-1,1) Gamma(n/2)/(Gamma(1/2) Gamma((n-1)/2)) int_{-1}^1 |x| (1-x^2)^{n/2 + (n-2)/2} dx

= p(n-1,1) Gamma(n/2)/(Gamma(1/2) Gamma((n-1)/2)) 2 int_{0}^1 x (1-x^2)^{n -1} dx

= p(n-1,1) Gamma(n/2)/(Gamma(1/2) Gamma((n-1)/2)) 1/n

Also since p(1,1) = 1/2 |1| + 1/2 |-1| = 1, we have by induction that

p(n,1) = prod_{k=2}^n Gamma(k/2)/(Gamma(1/2) Gamma((k-1)/2)) 1/k

= Gamma(1/2)^{-n} 1/(n!)

= pi^{-n/2} (n!)^{-1}

recalling that Gamma(1/2) = sqrt(pi)

Notice that the same method would allow us to compute the integral over the sphere of any polynomial function.

How did you know about my sickness? I don’t remember telling you about it. Yes I was hospitalized for kidney stone but it’s much better now. Thank you for asking.

照顾好自己，像我一样

咦,怎么之前是arena的慰问词，现在变成你的了？多谢关心拉。我会照顾好自己的。多喝水多运动少憋气。