Basic Fourier Transform Given a real positive $n\times n$ matrix, what would the following Fourier Transform be:
$$\int_{\mathbb{R}^{n}}e^{-(Ax,x)}e^{-i\omega x}dx$$
I'm quite confused by $(Ax,x)$. I assumed it's a vector, and $Ax$ should be a $n\times 1$ vector, so $(Ax,x)$ is going to be a $2n\times 1$ vector. But I feel this might not be correct. Even if it is, I'm not particularly sure how to proceed. Everything I know about Fourier Transforms is pretty basic, so I assume this might be really trivial, but I'm honestly not sure how to compute it. Any help would be greatly appreciated, thanks.
 A: I assume that you need the Fourier transform of the multidimensional Gaussian function $e^{-x^TAx}$ with a positive-definite matrix $A$ that looks like
$$
\int_{\Bbb{R}^n}e^{-x^TAx}e^{-i\omega^Tx}\,dx=\int_{\Bbb{R}^n}e^{-x^TAx-i\omega^Tx}\,dx.
$$
Completing the squares
$$
x^TAx+i\omega^Tx=\left(x+\frac12A^{-1}\omega i\right)^TA\left(x+\frac12A^{-1}\omega i\right)+\frac14\omega^TA^{-1}\omega
$$
we get the equivalent integral
$$
e^{-\frac14\omega^TA^{-1}\omega}\int_{\Bbb{R}^n}e^{-(x+\frac12A^{-1}\omega i)^TA(x+\frac12A^{-1}\omega i)}\,dx.
$$
To calculate the last integral we notice that the matrix $A$ can be diagonalized by an appropriate unitary transformation, so with the notation $y=\frac12A^{-1}\omega$ and $\lambda_k$ being the eigenvalues of $A$ it becomes
$$
\int_{\Bbb{R}^n}e^{-(x+\frac12A^{-1}\omega i)^TA(x+\frac12A^{-1}\omega i)}\,dx=
\int_{\Bbb{R}^n}e^{-\sum_{k=1}^n\lambda_k(x_k+iy_k)^2}\,dx=\prod_{k=1}^n\int_{\Bbb{R}}e^{-\lambda_k(x_k+iy_k)^2}\,dx_k.
$$
Finally, it can be shown by considering the rectangular contour of integration with corners $(\pm N,0)$ and $(\pm N+iy_k)$ that the analytical function $e^{-\lambda_k z^2}$ vanishes on the vertical sides of the rectangular when $N\to 0$ and then 
$$
\int_{\Bbb{R}}e^{-\lambda_k(x_k+iy_k)^2}\,dx_k=\int_{\Bbb{R}}e^{-\lambda_k x_k^2}\,dx_k=\sqrt{\frac{\pi}{\lambda_k}}
$$
(the last equality is the standard Gaussian function integration). It makes the entire thing to be
$$
e^{-\frac14\omega^TA^{-1}\omega}\prod_{k=1}^n\sqrt{\frac{\pi}{\lambda_k}}=\sqrt{\frac{\pi^n}{\det(A)}}\cdot e^{-\frac14\omega^TA^{-1}\omega}.
$$

P.S. It maybe be easier to calculate the original integral by the same trick as here, but I leave it open. 
A: An other derivation with $d(Mx)=\det(M)dx$ (the volume definition of $\det$)


*

*If $$\mathcal{F}[g(x)](\xi) = \int_{\mathbb{R}^n} g(x) e^{-2i \pi \xi^Tx }dx$$
then for any invertible matrix $M \in \mathbb{R}^{n\times n}$ with  $x = My$,  $d(Mx)=\det(M)dx$  : 
$$\mathcal{F}[g(x)](M^{-T}\xi) = \int_{\mathbb{R}^n} g(x) e^{-2i \pi \xi^TM^{-1} x }dx= \int_{\mathbb{R}^n} g(My) e^{-2i \pi \xi^TM^{-1}My}d(My)$$
$$=\int_{\mathbb{R}^n} g(My) e^{-2i \pi \xi^Ty}\det(M)dy=\det(M) \mathcal{F}[g(Mx)](\xi)$$

*$\int_{-\infty}^\infty e^{-t^2}e^{-2i \pi \xi t}dt= \sqrt{\pi} e^{-\pi^2 \xi^2}$ so with $g(x) =   e^{-x^T x}, x \in \mathbb{R}^n$ : $$\mathcal{F}[g(x)](\xi) = \sqrt{\pi^n}e^{-\pi^2 \xi^T \xi}$$

*And if $A$ is positive definite then $A = B^T B$ and $g(B x) = e^{-x^T B^T B x} = e^{-x^T A x}$ so that
$$\int_{\mathbb{R}^n} e^{-x^T A x} e^{-2i \pi \xi^Tx}dx=\int_{\mathbb{R}^n} g(Bx) e^{-2i \pi \xi^Tx}dx = \mathcal{F}[g(B x)](\xi)$$
$$ = \frac{1}{\det(B)}\mathcal{F}[g(x)](B^{-T}\xi)= \frac{1}{\det(B)}\sqrt{\pi^n}e^{-\pi^2 \xi^TB^{-1}B^{-T} \xi}=\sqrt{\frac{\pi^n}{\det(A)}}e^{-\pi^2 \xi^TA^{-1} \xi}$$
