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Let $\sigma_{n, m}$ denote the uniform measure over matrices $U \in \mathbb{R}^{n \times m}$ satisfying $U^T U = I_m$. Let $1_k \in \mathbb{R}^k$ denote the vector with all entries equal to $1$. I am trying to calculate, for $m < n$, the following quantity $$ \mathcal{I}(n, m) = \int \frac{1}{1 - \tfrac{1}{n}\langle U^T 1_n, U^T 1_n \rangle} \, d \sigma_{n,m}(U) = \int\frac{1}{1 - \langle U^T v, U^T v \rangle } d\sigma_{n,m}(U), $$ where the final equality holds for any $v \in \mathbb{R}^n$ with unit Euclidean norm, simply using that $\sigma_{n, m}$ is unitarily invariant.

Is there a convenient way to calculate this Haar integral?

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  • $\begingroup$ isn't this average divergent? $\langle U^\top v,U^\top v\rangle$ will equal to 1 with a nonzero probability $\endgroup$
    – Carlo Beenakker
    Commented Mar 12 at 21:02
  • $\begingroup$ Sorry why is that the case? $\endgroup$
    – Drew Brady
    Commented Mar 12 at 22:26
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    $\begingroup$ Have you tried the case $n=m=2$? $\endgroup$
    – Nemo
    Commented Mar 13 at 5:41
  • $\begingroup$ I have entered an explicit calculation for $n=2$, $m=1$, to demonstrate the divergence $\endgroup$
    – Carlo Beenakker
    Commented Mar 13 at 9:30

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The average diverges. To check this, try $n=2$, $m=1$; the matrix $U={u_1\choose u_2}$ has probability density $$P(u_1,u_2)=\frac{2}{\pi}\delta(1-u_1^2-u_2^2).$$ Then, taking $v={1\choose 0}$, $$\int \frac{1}{1-\langle U^\top v,U^\top v\rangle}\,d\sigma(U)=\int_{-1}^{1}du_1\int_{-1}^1 du_2 \, \frac{P(u_1,u_2)}{1-u_1^2}$$ $$\qquad=\frac{1}{\pi}\int_{-1}^1 \frac{1}{(1-u_1^2)^{3/2}}du_1=\infty.$$

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