Evaluating $\int_{\mathbb R^n}e^{-\langle Ax,x \rangle}dx$ where $A$ is symmetric and positive definite

Question

I was asked the following question, and while I think I made little progress, I'd like a push in the right direction.

Let $A$ be $n$x$n$ positive definite symmetric matrix with real entries. then for all $x \in \mathbb R^n$, evaluate the integral:

$$\int_{\mathbb R^n}e^{-\langle Ax,x \rangle}dx$$

Hint: Start by assuming $A$ is diagonal and remember that any symmetric matrix is also diagonlizable.

What I did:

I think the hint is good. Assume $A$ is diagonal with the entries $\lambda_1 ,\lambda_2 ,...\lambda_n$, then $\langle Ax,x \rangle=\sum_{i=1}^{n} \lambda_ix_i^2$

So now the integral becomes $$\int_{\mathbb R^n}e^{\sum_{i=1}^{n}\lambda_ix_i^2}dx$$

But what now? can we split integral and integrate by $x_i$ each time?

Answer 1

Yes. You're exactly right.

$$\int_{\mathbb{R}^n} e^{-\sum_{i=1}^n \lambda_i x_i^2} d\mathbf{x} = \int_{\mathbb{R}^n} \prod_{i=1}^n e^{-\lambda_i x_i^2} d\mathbf{x}$$ then we integrate with respect to each dimension separately

$$\int_{\mathbb{R}}\cdots\int_{\mathbb{R}} \,\,\prod_{i=1}^n e^{-\lambda_i x_i^2} \,\,dx_1\ldots dx_n$$

Giving the answer as

$$\left(\sqrt{\pi}\right)^n \prod_{i=1}^n \lambda_i^{-\frac{1}{2}}$$

Where we have used $\int_{\mathbb{R}} e^{-\lambda x^2}dx = \sqrt{\pi/\lambda}$.