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In my research I've recently stumbled upon integrals of the form: $$ \int_{S^2} \frac{x^T B x}{x^T A x} \,dx $$ where $A$ is a symmetric positive-definite $3\times 3$ matrix and $B$ is just a symmetric $3\times 3$ matrix.

I was wondering whether anything was known about these types of integrals?

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$\newcommand\E{\mathsf E}$You want to find $$E:=\E\frac{X^T BX}{X^T AX},$$ where $X$ is a random vector uniformly distributed on $S^2$, so that $X$ equals $$\frac{[G_1,G_2,G_3]^T}{\sqrt{[G_1,G_2,G_3]^T[G_1,G_2,G_3]}}$$ in distribution, where $G_1,G_2,G_3$ are independent standard normal random variables.

Also, by the spherical symmetry, without loss of generality the matrix $A$ is diagonal. So, for some real $b_{i,j}$'s and some positive real $a_i$'s, $$E=\E\frac{\sum_{j,k=1}^3 b_{j,k}G_jG_k}{\sum_{i=1}^3 a_i G_i^2}=\sum_{j,k=1}^3 b_{j,k}\E R_{j,k},$$ where $$R_{j,k}:=\frac{G_jG_k}{\sum_{i=1}^3 a_i G_i^2}.$$ Note that $R_{j,k}$ will turn into $-R_{j,k}$ if $j\ne k$ and $G_j$ is replaced by $-G_j$. So, if $j\ne k$, then the distribution of $R_{j,k}$ is symmetric and hence $\E R_{j,k}=0$, So, $$E=\sum_{j=1}^3 b_jE_j,$$ where $b_j:=b_{j,j}$ and $$E_j:=\E R_{j,j}=\E\frac{G_j^2}{\sum_{i=1}^3 a_i G_i^2}.$$

So, the problem is equivalent to computing $E_j$. However, it is highly unlikely that $E_j$ can be expressed in terms of elementary or even special functions, even for specific values of the $a_i$'s. For instance, Mathematica cannot do anything for $E_j$ even when $j=1$, $a_1=1$, $a_2=2$, and $a_3=3$:

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However, we have this:

enter image description here

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  • $\begingroup$ Out of curiosity, how did you reduce $\sum b_{i,j}G_i G_j$ to $\sum b_j G_j^2$? It seems to me that would be equivalent to diagonalizing $A$ and $B$ simultaneously, which is not generally possible. $\endgroup$ Commented Oct 28, 2021 at 7:19
  • $\begingroup$ @DmytroYeroshkin : I have added a detail here. Do you have any other questions about this answer? $\endgroup$ Commented Oct 28, 2021 at 11:33
  • $\begingroup$ Thanks, that makes sense. Unfortunate that the problem isn't really solvable, but I already had some suspicion of that when I asked. $\endgroup$ Commented Oct 28, 2021 at 12:14

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