Integrating over unit sphere, using Haar measure

Question

I'm trying to compute the following integral for a general matrix $n \times n$ matrix $A$ $$\int_{\mathbb{S}^{n-1}} \|A x\|^4 d\mathcal{H}^{n-1}(x)$$ where $\mathcal{H}$ is the $(n-1)$ dimensional Hausforff measure. In the case that our integral is instead $$\int_{\mathbb{S}^{n-1}} \|A x\|^2 d\mathcal{H}^{n-1}(x)$$ we can use this result to conclude that $$\int_{\mathbb{S}^{n-1}} \|A x\|^2 d\mathcal{H}^{n-1}(x) = \frac{\text{tr}({A^T A})}{n} \text{area}(\mathbb{S}^{n-1}) = \frac{\||A\||^2}{n} \text{area}(\mathbb{S}^{n-1})$$ where I've used the Frobenius norm for the matrix in the final equation. Can I generalise this result to the other integral? For the $|Ax|^2$ you can actually just use the divergence theorem, but I can't see this in the case of $|Ax|^4$. The integral is no longer linear in $A$ but it is still invariant under conjugation with an orthogonal matrix. Can you use the Haar measure here? This is not a subject I am familiar with and trying to solve this with ordinary calculus is not getting me anywhere. Any help appreciated.

EDIT: Based on Mittens' input, I came up with the following solution which makes it easier to keep the solution explicitly in terms of $A$.

$\|A \theta \|^4 = (\theta ^T A^T A \theta)^2 = (\theta ^T P^T D P \theta)^2$ for some diagonal matrix $D$ and orthogonal matrix $P$ since $A^T A$ is clearly symmetric. Therefore by invariance under orthogonal transformations, $$\int_{\mathbb{S}^{n-1}} \|A \theta \|^4 d\mathcal{H}^{n-1}(\theta) = \int_{\mathbb{S}^{n-1}}(\theta ^T D \theta)^2 d\mathcal{H}^{n-1}(\theta). $$ Now define a vector field on the closed unit ball by $v(x) = D x x^T D x$. Then using the divergence theorem $$\int_{\mathbb{S}^{n-1}}(\theta ^T D \theta)^2 d\mathcal{H}^{n-1}(\theta) = \int_{\mathbb{S}^{n-1}}\langle v(\theta), \theta \rangle d\mathcal{H}^{n-1}(\theta) =\int_{\mathbb{B}^n} \text{div } v(x) dx.$$ Letting the elements of the diagonal of $D$ be $\lambda_1, \dots \lambda_n$ and computing, \begin{align*} \text{div } v(x) &= \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \bigg(\sum_{j=1}^{n} (Dx x^T D)_{ij} x_j \bigg)=\sum_{i=1}^{n} \frac{\partial}{\partial x_i} \bigg(\sum_{j=1}^{n} (Dx \otimes Dx)_{ij} x_j \bigg) \\ &= \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \bigg(\sum_{j=1}^{n} (Dx)_i (Dx)_j x_j \bigg) = \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \bigg(\sum_{j=1}^{n} \lambda_i \lambda_j x_i x_j^2\bigg) \\ &= \sum_{i=1}^{n} \bigg(\sum_{j=1}^{n} \lambda_i \lambda_j x_{j}^{2} + 2\lambda_{i}^{2} x_{i}^{2} \bigg) = \text{tr}(D) \langle Dx, x \rangle + \langle Dx, Dx \rangle . \end{align*} Therefore $$\int_{\mathbb{S}^{n-1}}(\theta ^T D \theta)^2 d\mathcal{H}^{n-1}(\theta) =\int_{\mathbb{B}^n} \text{tr}(D) \langle Dx, x \rangle + \langle D^2 x, x \rangle dx.$$ Taking each in turn \begin{align*} \int_{\mathbb{B}^n}\langle Dx, x \rangle dx &= \int_{\mathbb{B}^n}\sum_{i=1}^{n} \lambda_i x_{i}^{2}dx = \sum_{i=1}^{n} \lambda_i \int_{\mathbb{B}^n} x_{i}^{2}dx = \frac{\text{tr}(D)}{n} \int_{\mathbb{B}^n} |x|^2 dx \\&= \omega_n \text{tr}(D) \int_{0}^{1} r^2 r^{n-1} dr = \frac{\omega_n \text{tr}(D)}{n+2}, \end{align*} and so by the same reasoning $$\int_{\mathbb{B}^n}\langle D^2 x, x \rangle dx = \frac{\omega_n \text{tr}(D^2)}{n+2}.$$ Finally, $\text{tr}(D)$ is precisely the sum of the eigenvalues of $A^T A$, which is $\text{tr}(A^T A) = \|A\|^2$ where again this is the Frobenius norm, and identically $\text{tr}(D^2)^2 = \|A^T A\|^2$, so putting it all together $$\int_{\mathbb{S}^{n-1}} \|A \theta \|^4 d\mathcal{H}^{n-1}(\theta) = \frac{\omega_n}{n+2} \bigg(\|A\|^4 + \|A^T A\|^2 \bigg).$$

Answer 1

Notice that

$$|A\boldsymbol{x}|^4=\big( \boldsymbol{x}^\intercal A^\intercal A\boldsymbol x\big)^2=\big( \boldsymbol{x}^\intercal P^\intercal\Lambda P\boldsymbol x\big)^2$$ where $P$ is an orthogonal matrix and $\Lambda$ is a diagonal matrix with nonnegative entries. Since $\sigma_{n-1}$, the Lebesgue measure on the unit sphere $S_{n-1}=\{x\in\mathbb{R}^n:|\boldsymbol{x}|=1\}$, is invariant under orthogonal transformations

\begin{align} \int_{S_{n-1}}|A\boldsymbol{x}|^4\,\sigma_{n-1}(d\boldsymbol{x})&=\int_{S_{n-1}}\big(\sum^n_{k=1}\lambda_kx^2_k\big)^2\,\sigma_{n-1}(d\boldsymbol{x})\\ &=\sum^n_{k=1}\lambda^2_k\int_{S_{n-1}}x^4_k\sigma_{n-1}(d\mathbf{x})+2\sum_{1\leq i<j\leq n}\lambda_i\lambda_j\int_{S_{n-1}}x^2_ix^2_j\,\sigma_{n-1}(d\boldsymbol{x})\\ &=\Big(\int_{S_{n-1}}x^4_1\,\sigma_{n-1}(d\boldsymbol{x})\Big)\sum^n_{k=1}\lambda^2_k + 2\Big(\int_{S_{n-1}}x^2_1x^2_2\,\sigma_{n-1}(d\boldsymbol{x})\Big)\sum_{1\leq i<j\leq n}\lambda^2_i\lambda^2_j \end{align} where the last equation follows from the invariance of $\sigma_{n-1}$ under orthogonal transformations. I leave the estimation of the integrals $\int_{S_{n-1}}x^4_1\,\sigma_{n-1}(d\boldsymbol{x})$ and $\int_{S_{n-1}}x^2_1x^2_2\,\sigma_{n-1}(d\boldsymbol{x})$ the the OP. One may need a specific parametrization of spherical coordinates and some change of variables