Let $A \in \mathbb{R}^{n \times n}$ be symmetric and invertible. $K=\{ x \in \mathbb{R}^n : \|Ax\|_2 \le 1 \}$. Now I have to calculate:

$$\int_K \langle A^2x, x \rangle \mathrm d x $$

It exists a orthogonal matrix $S \in \mathbb{R}^{n \times n}$ with $$SAS^T=D \text{ and } SA^2S^T = D^2 \ (D=\text{diag}(\lambda_1,\ldots,\lambda_n))$$

With $\varphi(x)=Sx$ and $B=\{ x \in \mathbb{R}^n : \|Dx\|_2\le 1 \}$ we get $\varphi(B)=K$. Because $|\det(\varphi')|=1$ we get

$$ \int_{\varphi(B)} \langle A^2x, x \rangle \mathrm dx = \int_{B} \langle D^2x,x \rangle \mathrm d x = \int_B \lambda_1^2 x_1^2+\ldots +\lambda_n^2 x_n^2 \mathrm d x $$

At this point I stuck. Is there any further step to simplify this integral?

  • $\begingroup$ How would you compute volume of the $n$-dimensional ball $\{x\in \Bbb R^n: \|x\|_2\le 1\}$? $\endgroup$ – Quang Hoang Aug 18 '14 at 16:15
  • $\begingroup$ $n$-dimensional polar coordinates. Seems very messy to do so. $\endgroup$ – DerJFK Aug 18 '14 at 16:23
  • $\begingroup$ This is the exact integral, but you have $r^{n+1}dr$ instead of $r^{n-1}dr$. I don't think there's a better way. $\endgroup$ – Quang Hoang Aug 18 '14 at 16:39
  • $\begingroup$ Why you guys ask the same question? here $\endgroup$ – Troy Woo Aug 18 '14 at 16:40
  • $\begingroup$ Didn't see the other post. $\endgroup$ – DerJFK Aug 18 '14 at 16:43

Let $\phi(x) = A^{-1} x$, and $f(x) = \|Ax\|^2$.

Note that $K=\{x | \|Ax\| \le 1 \} = \{ A^{-1}y | \|y \| \le 1 \} = \phi(B(0,1))$.

$\int_K \|Ax\|^2 dx = \int_{\phi(B(0,1))} f = \int_{B(0,1)} f \circ \phi |J_\phi| = {1 \over \det A} \int_{B(0,1)} \|x\|^2 dx$.


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