Integration of $e^{-\langle Ax , x \rangle}$ over $\mathbb{R}^n$ Problem :
If $A_{n \times n}$ is a symmetric, positive-definite matrix, show that : $$\int_{\mathbb{R}^n} e^{-\langle Ax , x \rangle}~ dx = \sqrt{\dfrac{\pi^n}{\det(A)}}$$ where $\langle a , b\rangle$ denotes the inner product of $a$ and $b$.

Approach :
I was approaching the problem using the Change of Variable Formula, using the function $\varphi(x) = A^{-1}x$ . Since $A$ is p.d., I can show that it is invertible. But I can't proceed anymore.
I found a similar-looking problem here, but couldn't understand anything.
 A: Let $v_1,\ldots,v_n$ be an orthonormal basis for the inner product induced by $A$, with corresponding eigenvalues $\lambda_1,\ldots,\lambda_n>0$. We have $\det(A)=\prod_{j=1}^{n}\lambda_j$ and by an isometry
$$ \int_{\mathbb{R}^n}\exp(-x^t A x)\,dx = \int_{\mathbb{R}^n}\exp(-\lambda_1 x_1^2-\ldots-\lambda_n x_n^2)\,dx\stackrel{\text{Fubini}}{=}\prod_{j=1}^{n}\frac{1}{\sqrt{\lambda_j}}\int_{\mathbb{R}}e^{-z^2}\,dz. $$
A: Since $A$ is symmetric, there exists some orthogonal $S \in \mathbb{R}^{n \times n}$ (i.e. $S^{-1} = S^\top$) such that $A = S^{-1}DS$ where $D := \mathrm{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix containing all the eigenvalues of $A$. Note that they are positive because of the assumption of $A$ being positive definite. So, because of $S^{-1} = S^\top$:
$$
\int_{\mathbb{R}^n} e^{-\langle Ax, x\rangle} ~\mathrm{d}x = \int_{\mathbb{R}^n} e^{-\langle Sx, DSx \rangle}~\mathrm{d}x
$$
Now introduce an operator $\Phi: \mathbb{R}^n\rightarrow \mathbb{R}^n$, $\Phi(x):= S^{-1}x$. $\Phi$ is bijective because of $S$ being invertible. One furthermore easily finds $D\Phi(x) = S^{-1}$ for all $x \in \mathbb{R}^n$. We also know that $\lvert \det(S^{-1}) \rvert = 1$ because $S$ is orthogonal. So the transformation formula yields:
$$
\int_{\mathbb{R}^n} e^{-\langle Sx, DSx \rangle}~\mathrm{d}x = \int_{\mathbb{R}^n} e^{- \langle S \Phi(x), DS \Phi(x) \rangle}~\mathrm{d}x = \int_{\mathbb{R}^n} e^{-\langle x, Dx \rangle}~\mathrm{d}x = \int_{\mathbb{R}^n} e^{-\sum_{j = 1}^n \lambda_jx_j^2}~\mathrm{d}x 
$$
Use that $e^{x+y} = e^x e^y$ for all $x, y \in \mathbb{R}$ and Fubini to conclude:
$$
\int_{\mathbb{R}^n} e^{-\sum_{j = 1}^n \lambda_jx_j^2}~\mathrm{d}x  = \prod_{j = 1}^n \int_{-\infty}^\infty e^{-\lambda_j x_j^2}~\mathrm{d}x_j
$$
Now consider
$$
I_j := \int_{-\infty}^\infty e^{-\lambda_j x_j^2}~\mathrm{d}x_j.
$$
Introduce a substitution $y := \sqrt{\lambda_j}x_j$. Then:
$$
I_j  = \frac{1}{\sqrt{\lambda_j}} \int_{-\infty}^\infty e^{-y^2}~\mathrm{d}y = \sqrt{\frac{\pi}{\lambda_j}}
$$
Putting everything together:
$$
\int_{\mathbb{R}^n} e^{-\langle Ax, x\rangle} = \prod_{j = 1}^n I_j = \frac{\sqrt{\pi}^n}{\sqrt{\prod_{j = 1}^n \lambda_j}} = \frac{\sqrt{\pi}^n}{\sqrt{\det(A)}} = \sqrt{\frac{\pi^n}{\det (A)}}
$$
In the last step we used that the product of eigenvalues is the determinant.
