How to compute $\int_{S^{n-1}}\|A\vec{x}\| d\vec{x}$? Given an $n\times n$ invertible matrix A, how should I compute the integral
$$\int_{S^{n-1}}\|A\vec{x}\|^t \operatorname{d}\vec{x},$$
where $t\in\mathbb{R}$, $S^{n-1}$ is the unit circle in $\mathbb{R}^n$, $\|\cdot\|$ is the usual $l_2$-norm. Is there a closed form?
 A: This is not a complete answer to your question, just a potentially useful simplification.
I think there is something to gain from writing the integrand as a quadratic form, since there are a few useful tricks for spherically symmetric integrals of quadratic forms. This gives a closed form for $t=2$, and might be slightly simpler to deal with generally.
First, let $\vec x\in\mathbb R^n$, and let $S_{n-1}$ be the surface of the unit $n$-ball, with volume element $d\sigma$. The volume of $S_{n-1}$ is $\sigma_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$. Since $\vec x\cdot\vec x=1$ everywhere in $S_{n-1}$, we can write the following intergal.
$$\sigma_{n-1}=\int_{S_{n-1}}d\sigma=\int_{S_{n-1}}\vec x\cdot\vec x\ d\sigma$$
In Euclidean coordinates, we can break this integral into parts.
$$\sigma_{n-1}=\int_{S_{n-1}}\sum_{i=1}^nx_i^2\ d\sigma=\sum_{i=1}^n\int_{S_{n-1}}x_i^2\ d\sigma$$
by symmetry, each element of this sum must be equal. We'll use this later on for the $t=2$ case.
$$\int_{S_{n-1}}x_i^2\ d\sigma=\frac{\sigma_{n-1}}n\tag{1}$$
Since $A^TA$ is symmetric, we know there exists a diagonalization $A^TA=P^TDP$ where $P$ is orthogonal and $D$ is diagonal. We can thus apply the change of coordinates $\vec y=P\vec x$, and since $P$ is a rotation, the volume element of any spherically symmetric domain of integration does not change.
$$\int_{S_{n-1}}||A\vec x||^t\ d\sigma=\int_{S_{n-1}}(\vec x^TA^TA\vec x)^{t/2}\ d\sigma=\int_{S_{n-1}}(\vec y^TD\vec y)^{t/2}\ d\sigma$$
$$\int_{S_{n-1}}||A\vec x||^t\ d\sigma=\int_{S_{n-1}}\left(\sum_{i=1}^n\lambda_iy_i^2\right)^{t/2}d\sigma\tag{2}$$
Where $\lambda_i$ are the eigenvalues of $A^TA$. The integral can be evaluated simply in the case that $t=2$.
$$\int_{S_{n-1}}||A\vec x||^2\ d\sigma=\int_{S_{n-1}}\sum_{i=1}^n\lambda_iy_i^2\ d\sigma=\sum_{i=1}^n\lambda_i\int_{S_{n-1}}y_i^2\ d\sigma$$
Now we can use $(1)$.
$$\int_{S_{n-1}}||A\vec x||^2\ d\sigma=\frac{\sigma_{n-1}}{n}\sum_{i=1}^n\lambda_i=\frac{\sigma_{n-1}}{n}\text{Tr}(A^TA)$$
For even integer (and potentially non-integer) $t$, we could look at the multinomial expansion of the integrand of $(2)$, but we would need new integrals like $\int_{S_{n-1}}x_i^p\ d\sigma$, and things would get very messy very quickly.
