2
$\begingroup$

If $A$ is a $n\times n$ symmetric positive definite matrix, I would like to compute this integral:

$\int_{\mathbb R^n} \|x\|^2 e^{-x\cdot Ax} dx$.

I have tried to use a Cholesky decomposition for $A = L^TL$. Then, applying the change of variables formula I get that it is the same as:

$\frac{1}{\sqrt{det(A)}} \int_{\mathbb R^n} \|Lu\|^2 e^{-\|u\|^2}du$.

It seems to be a simpler form, but I don't know what to do from here. Any help would be appreciated. Thank you!

1 Answers 1

0

You're on the right track. Continuing we find $$\begin{align*} \int_{\mathbb R^n} \|x\|^2 e^{-x^T A x} dx &= \det L^{-1} \int_{\mathbb R^n} \|L^{-1} u \|^2 e^{-\|u\|^2} du \\ &= \frac{1}{\sqrt{\det A}} \left({L^{-1}}^T L^{-1}\right)_{ii} \int_{\mathbb R^n} u_i^2 e^{-\|u\|^2} du \\ &= \frac{1}{\sqrt{\det A}} \left(A^{-1}\right)_{ii} \int u_i^2 e^{-u_i^2} d u_i \prod_{j\ne i} \int e^{-u_j^2} du_j \\ &= \frac{1}{\sqrt{\det A}} \left(A^{-1}\right)_{ii} \frac{\sqrt{\pi}}{2} \prod_{j\ne i} \sqrt{\pi} \\ &= \frac{\pi^{n/2}}{2} \frac{\mathrm{Tr}\, A^{-1}}{\sqrt{\det A}}. \end{align*}$$ Note that integrals of the form $\int u_i u_j e^{-\|u\|^2}du$ with $i\ne j$ vanish identically.

  • 0
    Thanks! I think there is a $\sum_i$ missing in the second line. So you will get that the final answer is $n\frac{\pi^{n/2}}{2}\frac{TrA^{-1}}{\sqrt{\detA}}$.2012-12-09
  • 0
    @George: Glad to help. I'm using Einstein's summation convention so summation is implied. Thus, $(A^{-1})_{ii} = \sum_i (A^{-1})_{ii} = \mathrm{Tr}\,A^{-1}$. There is no factor of $n$.2012-12-09
  • 0
    (Calculate the integral when $A$ is the identity matrix to convince yourself of this fact.)2012-12-09
  • 0
    Yes Sorry! Thanks again!2012-12-09
  • 0
    @George: No problem. Cheers!2012-12-09