I know how to estimate the integral* (see the update) \begin{gather} \int f(Ub)d\mu(U), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2] \end{gather} where $f:S^{n-1}(\mathbb{R})\to \mathbb{R}$ and $f(x)=\sum\limits_{i=1}^{n}\frac{1}{x_i^2}$ and $d\mu$ is the Haar measure over orthogonal group $O(n)$. My trouble is the following:

Suppose we multiply $b$ with a matrix of complex values (in my case DFT matrix) $M$, i.e. $$\int g(UMb)d\mu(U),$$ where $g:S^{n-1}(\mathbb{C})\to \mathbb{R}$ and $g(x)=\sum\limits_{i=1}^{n}\frac{1}{|x_i|^2}$ where $|.|$ is the absolute value. It is intuitively obvious for me that this integral should be equal to [2], specially considering that if $M$ was orthogonal this would have been true (due to properties of Haar measure), yet here $M$ is unitary.

Since my calculus knowledge is not that amazing --as you see-- I would appreciate some guidance to prove this. I imagine the title is not the best but that is what I could came up with.

UPDATE: This integral (as mentioned by @CarloBeenakker) is infinity (and I was aware of that). The integral I was referring (and mistakenly didn't mention) to, and have estimation of it is $\int h(Ub)\mu(U)$ where $h(x)=\sum_{i=1}^{m} \frac{1}{x_i^2 + \epsilon}$ for some positive $\epsilon$.

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    $\begingroup$ have you checked for a simple case, say $n=2$, whether the two integrals indeed coincide? a priori I see no reason they should... $\endgroup$ Nov 19, 2015 at 11:03
  • $\begingroup$ @JeanDuchon, thank you for the corrections. Yes you are right! $\endgroup$
    – Cupitor
    Nov 19, 2015 at 17:11

1 Answer 1


the two integrals are not the same, here is a simple example for $n=2$:

$$U=\begin{pmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{pmatrix},\;\;d\mu(U)=d\alpha$$

$$M=\frac{1}{2}\begin{pmatrix} i+\sqrt 2 & 1\\ -1 & -i+\sqrt 2 \end{pmatrix},\;\;b={1\choose 2}$$

$$\int g(UMb)d\mu(U)=5.80664$$

$$\int f(Ub)d\mu(U)=\infty$$

  • $\begingroup$ Thank you for the example. Can you please elaborate on this a bit. Do you have any intuition that why this is the case? And also how did you calculate the first integral? $\endgroup$
    – Cupitor
    Nov 19, 2015 at 16:59
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    $\begingroup$ well, the first integral is just some definite integral over $\alpha$ which I evaluated numerically; its value does not matter, what matters is that it is a finite number, while the second integral diverges; and for intuition: it is the other way around, my intuition tells me the two integrals should be different because the invariant measure is only invariant under orthogonal transformations, not under arbitrary unitary transformations; so I would not have expected any identity of the two integrals to begin with. $\endgroup$ Nov 19, 2015 at 17:05
  • $\begingroup$ Very reasonable. Thank you. In my original problem I indeed have a $\epsilon$ in the denominator, but as far as I can see that doesn't matter! $\endgroup$
    – Cupitor
    Nov 19, 2015 at 17:09

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