Let $O(N)$ be the orthogonal group, and $a,b,c\in\mathbb N$. The question is:
$$\int\_{O(N)}U\_{11}^aU\_{22}^bU\_{33}^cdU=?$$


This is quite a tricky question:


(1) The first thought would go to probability, because with $N\to\infty$ the variables $U\_{11},U\_{22},U\_{33}$ become Gaussian and independent; however, there doesn't seem to be any good analytic method for computing the correlations at $N$ fixed.


(2) The second thought would go to combinatorics, and to the Weingarten function. But that doesn't work either: we spent some time with Collins and Schlenker on this question, and just got a kind of very long (and especially unusable!) formula.


So, a new point of view on all this would be probably needed.



Integration over the orthogonal group
This is a response to your question (1), for an analytic method to compute the integral over $O(N)$ as a power series in $1/N$. This method was developed by Prosen, Seligman and Weidenmüller in J. Math. Phys. **43**, 5135-5144 (2002) [[arXiv:math-ph/0203042](http://arxiv.org/abs/math-ph/0203042)]. Their key result can be written, in the context of your integral, as


$$\int\_{O(N)}dU\;U\_{11}^{a}U\_{22}^{b}U\_{33}^{c}=\int d\mu\; w\_{\kappa}(M)M\_{11}^{a}M\_{22}^{b}M\_{33}^{c}+{\cal O}(1/N^{z+1})$$


The order of the approximation is $z=\mbox{Int}[(\kappa+a+b+c)/2]$. The $N\times N$ real matrix $M$ has Gaussian measure $d\mu\propto\exp(-N\;\mbox{Tr}\;MM^{T})\prod\_{ij}dM\_{ij}$, so the integral on the right-hand-side is simply a Gaussian integral over the matrix elements $M\_{ij}$ of $M$. The orthonormality constraints of the integral on the left-hand-side are accounted for by a weight function $w\_{\kappa}$ of linearly independent invariants of $M$ up to order $2\kappa$. Explicit expressions for $w\_{\kappa}$ for $\kappa=1,2,3,4$ are given in the cited reference.


To lowest order, one has $\kappa=1$, when $w\_{1}\equiv 1$. This amounts to treating $U\_{11}$, $U\_{22}$, and $U\_{33}$ as independent Gaussians, which is the lowest-order approximation you mentioned in your question. Increasing $\kappa$ gives you the higher order corrections in a systematic way.


