Integral w.r.t. the uniform measure on the unit sphere I need to compute the following integral:
$$I=\int_{(\mathbb{S}^{n-1})^2}\arccos|x^Ty|d\gamma(x)\otimes\gamma(y),$$
where $\gamma$ is the uniform distribution on $\mathbb{S}^{n-1}$.
My attempt:
since ${\gamma}$ is uniform over $\mathbb{S}^{d-1}$, the integral $\int\arccos|x^Ty|d{\gamma}(x)$ is the same for all $y$. Then,
$$I=\int\arccos|x^Ty|d{\gamma}(x).$$
Set $y=(1,0\dots0)\in\mathbb{R}^n$, the integral reduces to:
$$I=\int\arccos|x_1|d{\gamma}(x)=\frac{1}{S}\int\arccos|x_1|dS,$$
where $x_i$ is the $i$-th entry of $x$, and  $S$ is the surface area of $\mathbb{S}^{n-1}$.
I don't know how to proceed from there, any ideas?
 A: (Note: to make things easier on myself I instead choose $y=(0,\dotsc, 0,1)$ in this problem).
You must introduce coordinates for the $(n-1)$-sphere. Note that for $x=(x_1,\dotsc, x_n)\in S^{n-1}$ the coordinates are given in terms of angular parameters $\phi_1\in [0,2\pi)$ and $\phi_2,\dotsc, \phi_{n-1}\in[0,\pi)$ by the formulas
\begin{align}
x_n &= \cos(\phi_{n-1}) \\
x_{n-1} &= \sin(\phi_{n-1})\cos(\phi_{n-2}) \\
x_{n-2} &= \sin(\phi_{n-1})\sin(\phi_{n-2})\cos(\phi_{n-3}) \\
&\,\,\vdots\\
x_2 &= \sin(\phi_{n-1})\sin(\phi_{n-2})\dotsm\sin(\phi_2)\cos(\phi_1)\\
x_1 &= \sin(\phi_{n-1})\sin(\phi_{n-2})\dotsm\sin(\phi_2)\sin(\phi_1).
\end{align}
It is not hard to see that this agrees with the usual circular/spherical coordinate system in the $n=2,3$ cases (up to swapping the first two coordinates).
Now using this information we can define the "(hyper)surface area" element as follows: let
$$g_{ij} = \sum_{k=1}^n\frac{\partial x_k}{\partial \phi_i}\frac{\partial x_k}{\partial \phi_j}$$
where $i,j=1,\dotsc, n-1$ be the metric tensor. Then the form $dS = \sqrt{|\det(g)|}d\phi_1\dotsm d\phi_{n-1}$ is the volume form associated to the $(n-1)$-dimensional manifold $S^{n-1}$, which coincides with the notion of "(hyper)surface area" form that we desire.
The calculations are a little arduous, but eventually you will find
$$dS=\sin ^{n-2}(\phi _{n-1})\sin ^{n-3}(\phi _{n-2})\cdots \sin(\phi _{2})\,d\phi _{1}\,d\phi _{2}\cdots d\phi _{n-1}.$$
Applying Fubini's theorem we obtain
$$I = \frac{1}{S}\left(\int_0^{\pi}{\rm arccos}|\cos(\phi_{n-1})|\sin^{n-2}(\phi_{n-1})\,d\phi_{n-1} \right)\bigg(\text{integral of the other terms}\bigg).$$
In fact, since we have
$$S = \left(\int_0^{\pi}\sin^{n-2}(\phi_{n-1})\,d\phi_{n-1} \right)\bigg(\text{integral of the other terms}\bigg),$$
the desired integral has value
$$I = \left(\int_0^{\pi}{\rm arccos}|\cos(\phi_{n-1})|\sin^{n-2}(\phi_{n-1})\,d\phi_{n-1} \right) \large/\small\left(\int_0^{\pi}\sin^{n-2}(\phi_{n-1})\,d\phi_{n-1} \right).$$
I'll leave this last calculation for you to do.
Note: in the circle case $(n=2)$ the above calculations seem to imply $I=\pi/4$ (rather than dealing with any sine terms we simply find the average value of $\arccos(|\cos(\theta)|)$ for $\theta\in[0,2\pi]$), which is different from what @Mason claimed. Hopefully this is not evidence of an error. Please comment if you spot anything incorrect.
