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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Orthogonal_matrix) this would have been true (due to properties of Haar measure), yet here $M$ is [unitary](https://en.wikipedia.org/wiki/Unitary_matrix).


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 $\epsilon$.



Change of variable for integration with respect to Haar measure
the two integrals are not the same, here is a simple example for $n=2$:


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


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


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


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


