Computation of the fourier transformation of a function with a matrix I want to compute the Fourier transformation of the following function:
\begin{align}
f:& \mathbb R^n \rightarrow \mathbb R \\
  & x \mapsto \exp(-\left<Ax,x\right>)
\end{align}
where $A$ is a symmetric and positive definite matrix.
My thoughts so far:
$A$ is symmetric and positive definite, hence there exists an orthogonal matrix $S$, such that $S^{T}DS$ with a diagonal matrix, which contains positive eigenvalues $\lambda_1,\ldots,\lambda_n$. Note that $\det(S)=\pm1$.Then we obtain:
\begin{align}
\widehat{f}(t)
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-\left<Ay,y\right>}e^{-i\left<y,t\right>} \, \mathrm{d}y \\
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-y^{T}Ay}e^{-iy^{T}t} \, \mathrm{d}y \\
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-(Sx)^{T}ASx}e^{-i(Sx)^{T}t} \, \mathrm{d}x\\
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-x^{T}S^{T}ASx}e^{-ix^{T}S^{T}t} \, \mathrm{d}x\\
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-x^{T}Dx}e^{-ix^{T}S^{T}t} \, \mathrm{d}x \\
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-(x_1^2\lambda_1+\cdots+x_n^2\lambda_n)}e^{-ix^{T}S^{T}t} \, \mathrm{d}x
\end{align}
I used that $S^{-1}=S^T$ , $A^T=A$ and the substitution $y=Sx$.
But I don't have any idea how I should continue. I don't think that I am finished.
 A: Can I do $\mathbb{R}^{2}$ and you do $\mathbb{R}^{n}$? Since $\mathbf{A}$ is symmetric and positive definite, there exists an orthonormal basis for $\mathbb{R}^{2}$ such that $\mathbf{A} = \begin{pmatrix} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{pmatrix}$. Rotate the coordinate system so the positive $x$-axis points in the same direction as the eigenvector corresponding to $\lambda_{1}$, and the $y$-axis points in the same direction as the eigenvector corresponding to $\lambda_{2}$. Then 
\begin{align}
\hat{f}(z_1, z_2)
&=
\frac{1}{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \exp\left(-\lambda_{1}x^{2} -\lambda_2 y^{2} -ixz_1 -iyz_2  \right) \mathrm{d}x \mathrm{d}y \\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left(-\lambda_{1}x^{2} -ixz_1 \right) \mathrm{d}x
\int_{-\infty}^{\infty} \exp\left( -\lambda_2 y^{2} -iyz_2  \right) \mathrm{d}y  
\end{align}
These integrals can be computed by completing the square, but everyone knows you just ask Wolfram alpha to compute them (their convergence is guaranteed by $\lambda_{i} > 0$). Then 
\begin{align}
\hat{f}(z_{1}, z_{2}) 
&=
\frac{1}{2\pi}\left(\frac{\sqrt{\pi}\exp(-z_{1}^{2}/4\lambda_{1}^{2})}{\sqrt{\lambda_{1}}} \right) \left(\frac{\sqrt{\pi}\exp(-z_{2}^{2}/4\lambda_{2}^{2})}{\sqrt{\lambda_{2}}}  \right)\\
&=
\frac{1}{2}\left(\frac{\exp\left(-\frac{1}{4}\left(z_{1}^{2}/\lambda_{1}^{2} + z_{2}^{2}/\lambda_{2}^{2}\right) \right)}{\sqrt{\lambda_{1}\lambda_{2}}} \right) 
\end{align}
I think your final answer will involve a $\sqrt{\det(\mathbf{A})}$, but we can't be sure yet.
Note that $(z_{1}, z_{2})$ is in the basis such that $\mathbf{A}$ is diagonal. You'll need to rotate back to the original coordinate system at the end. 
A: What about this?: 
\begin{align}
\\
&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! e^{-(x_1^2\lambda_1+\cdots+x_n^2\lambda_n)}e^{-i<t,Sx>} \, \mathrm{d}x
\\&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\int_{\mathbb R^n}^ \! \prod_{j=1}^{n}e^{-\lambda_jx_j^2}\prod_{j=1}^{n}e^{-i((S^{T}t)_jx_j)} \, \mathrm{d}x
\\&=\left(\frac{1}{2\pi}\right)^\frac{n}{2}\prod_{j=1}^{n}\int_{\mathbb R}^ \! e^{-\lambda_jx_j^2}e^{-i((S^{T}t)_jx)_j} \, \mathrm{d}x_j
\\&=\prod_{j=1}^{n}[F(e^{-\lambda_j(.)^2})]({({S^{T}t})_j})
\end{align}
Where F is the map: $F:f\rightarrow \hat{f}$ , which maps a function to its Fouriertransformation.
I am open for any other suggestions including a continuation of my way.
