Integral of $\prod_{i=1}^n (\mathbf{x}\cdot \mathbf{u_i})\ \mathrm{d}\mathbf{x}$ over the ($n-1$)-sphere.

Question

In a physics problem I encountered, it was necessary to calculate

$$\int_{S} (\mathbf{x}\cdot\mathbf{a})(\mathbf{x}\cdot\mathbf{b})d\mathbf{x},$$

where the integral ranges over the surface of the sphere. This can be bashed out with spherical coordinates, and it equals

$$\frac{4\pi}{3} (\mathbf{a}\cdot\mathbf{b}).$$

Is there a "nice" way to do this integral in the general case where we have $m$ vectors in $\mathbb{R^n}$ and integrate over the $(n-1)$-sphere? My hunch is that, if $m=2$, we should have that the integral should be the volume of the unit $n$-ball times $\mathbf{a}\cdot\mathbf{b}$, but I'm not sure what it should be with more than $2$ vectors.

Answer 1

Suppose $Z_1, \ldots, Z_n$ are iid normal random variables with mean $0$ and standard deviation $1$, and let ${\bf Z} = [Z_1, \ldots, Z_m]$. Then $$ \eqalign{\mathbb E[({\bf Z} \cdot {\bf u}_1) \ldots ({\bf Z} \cdot {\bf u}_m)] &= const \int_{0}^\infty dr\; n r^{m+n-1} e^{-r^2/2} \int_S d{\bf s}\; ({\bf s} \cdot {\bf u}_1) \ldots ({\bf s} \cdot {\bf u}_m)\cr &= const\; \int_S d{\bf s}\; ({\bf s} \cdot {\bf u}_1) \ldots ({\bf s} \cdot {\bf u}_m)}$$

But $\mathbb E[({\bf Z} \cdot {\bf u}_1) \ldots ({\bf Z} \cdot {\bf u}_m)]$ can also be computed using Isserlis's theorem. Note that $X_1, \ldots, X_m = {\bf Z} \cdot {\bf u}_1, \ldots, {\bf Z} \cdot {\bf u}_m$ are multivariate normal with mean $0$ and covariances $\text{Cov}(X_i, X_j) = {\bf u}_i \cdot {\bf u}_j$. Thus for $m$ even what we should get (up to the appropriate constant) is the sum over all partitions of $1\ldots m$ into pairs $\{i_1,j_1\},\ldots,\{i_{m/2},j_{m/2}\}$ of the products of the dot products $\prod_{k=1}^{m/2}{\bf u}_{i_k} \cdot {\bf u}_{j_k}$, while for $m$ odd we get $0$.

Answer 2

First note that $$\mathrm{Area}(S)=\int_{S} 1\,d\mathbf x=\int_S (x_1^2+\cdots+x_n^2)\,d\mathbf x=n\int_S x_1^2\,d\mathbf x$$

because of symmetry, so $\int_S x_i^2\,d\mathbf x$ is independent of $i$.

Also, recall that $\frac{1}{n}\mathrm{Area}(S)$ is the volume of a unit sphere in $\mathbb R^n$.

Now $$(\mathbf a\cdot \mathbf x)(\mathbf b\cdot \mathbf x)=\sum_{i,j} a_ib_jx_ix_j$$

But by parity arguments, if $i\neq j$ we see that:

$$\int_{S} x_ix_j\,d\mathbf x=0$$

So you get that:

$$\int_{S} (\mathbf a\cdot \mathbf x)(\mathbf b\cdot \mathbf x)\,d\mathbf x=(\mathbf a\cdot\mathbf b)\int_S x_1^2\,d\mathbf x$$

This gives your result for $m=2$.

When $m$ is odd, you always get zero, because if $f(\mathbf x)=\prod_{i=1}^{m}(\mathbf x\cdot \mathbf a_i)$ then $f(-\mathbf x)=-f(\mathbf x)$.

The general case of $m$ even is worse.

If $m=4$, then the terms that do not cancel are like:

$$a_ib_ic_jd_j x_i^2d_j^2$$

So you get the value is something like:

$$\sum_{i\neq j} \left(a_ib_ic_jd_j+a_ib_jc_id_j+a_ib_jc_jd_j\right)\int_S x_1^2x_2^2\,d\mathbf x + \sum_{i} a_ib_ic_id_i \int_S x_1^4\,d\mathbf x$$

The case $m=6$ is gonna be something of the form:

$$F \int_S x_1^2x_2^2x_3^2\,d\mathbf x + G\int_S x_1^4x_2^2\,d\mathbf x + H\int_S x_1^6\,d\mathbf x$$ where $F,G,H$ will be horrible forms linear in each of the $\mathbf u_i$. For example, $$H=\sum_{i=1}^n u_{1,i}u_{2,i}\cdots u_{6,i}$$

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An interesting view is to think of the case where $\mathbf a,\mathbf b$ are unit vectors. Then the above means that given two points $A,B$ on the unit sphere, the average, over points $X$ on the unit sphere of $\cos\angle XOA\cdot \cos \angle XOB$ is $\frac{1}{n}\cos\angle AOB$.

I'm not sure how to prove that, geometrically, but we might use: $$\cos u\cos v = \frac{1}{2}\left(\cos(u+v)+\cos(u-v)\right)$$