Integral of $\|x-y\|$ over the $n$ sphere

Question

Let $\sigma$ be the normalized surface measure on $S=\{x\in\mathbb{R}^n: \|x\|=1\}\subseteq \mathbb{R}^n$. Fix an element $x\in S$. Find $$\int_S \|x-y\|\,d\sigma(y).$$ The original problem is like this: Let $f\colon S\to \mathbb{R}$ be $1$-Lipschitz. Suppose that $\int_S f=0$, then $|f(x)|\leqslant 1$ for all $x$. I then realized that calculating the above integral (if it equals $1$ then it) solves this problem.

Answer 1

Up to rotations you may assume that $x=(1,0,0,\ldots)$, so the wanted integral is

$$ \int_{x_1^2+\ldots+x_n^2=1}\sqrt{(x_1-1)^2+x_2^2+\ldots+x_n^2}\,d\mu=\int_{x_1^2+\ldots+x_n^2=1}\sqrt{2-2x_1}\,d\mu $$ which in spherical coordinates becomes

$$\int_{0}^{2\pi}\underbrace{\int_{0}^{\pi}\ldots\int_{0}^{\pi}}_{n-2\text{ integral signs}}\sqrt{2-2\cos\varphi_1}\sin^{n-2}(\varphi_1)\sin^{n-3}(\varphi_2)\cdot\ldots\cdot\sin(\varphi_{n-2})d\varphi_1d\varphi_2\ldots d\varphi_{n-2}d\varphi_{n-1}$$ and can be factored through $$ \int_{0}^{\pi}(\sin\theta)^k\,d\theta = B\left(\tfrac{1}{2},\tfrac{k+1}{2}\right)=\frac{\Gamma\left(\frac{k+1}{2}\right)}{\Gamma\left(k+\frac{1}{2}\right)}\sqrt{\pi},$$ $$\begin{eqnarray*} \int_{0}^{\pi}\sqrt{2-2\cos\phi}\sin^{n-2}(\phi)\,d\phi &=& 2\int_{0}^{\pi}\sin\left(\tfrac{\phi}{2}\right)\sin^{n-2}(\phi)\,d\phi\\&=&4\int_{0}^{\pi/2}\sin(\psi)\left(2\sin(\psi)\cos(\psi)\right)^{n-2}d\psi\\&=&2^n\int_{0}^{\pi/2}\sin^{n-1}(\psi)\cos^{n-2}(\psi)\,d\psi\\&=&2\sqrt{\pi}\cdot\frac{\Gamma(n-1)}{\Gamma\left(\frac{n-1}{2}\right)}.\end{eqnarray*}$$