Dtyjlui

I would like to understand how to fourier transform the function $f:mathbb R^ntomathbb R$,

$f(x):=e^{-lt x,Axgt}$

with $A$ being a positive definite, symmetric matrix.

I understand that $lt x,Axgt$ will end up being something like $sum_{i=1}^nsum_{j=1}^n a_{ij}x_ix_j$ with a couple of summarizable summands due to the symmetry of $A$:

$frac 1{{(2pi)}^{frac 1 n}}int_{mathbb R^n} e^{-(a_{11}x_1^2+a_{22}x_2^2+...+a_{nn}x_n^2+2a_{12}x_1x_2+...+2a_{23}x_2x_3+...+2a_{n,n-1}x_nx_{n-1})}e^{-i lt t,xgt}d^nx $

(Is that even correct?..)

But I am having trouble actually calculating the integral, as I can't see anything I already know I could substitute in this, not even for $n=1$.

I'd appreciate any help with this.

You can simplify the Fourier transform expression if you express everything in the basis of eigenvectors of $A$.

Remember that, by the spectral theorem, $A$ can be diagonalized in an orthonormal basis. That means there is a unitary transform $U$ and a diagonal matrix $Lambda$ such that
$$A=U^TLambda U$$
And because $A$ is positive definite, $Lambda$'s diagonal elements are all strictly positive. Also, because $U$ is unitary, for any $t$ and $x$ in $mathbb R^n$, $langle t, xrangle = langle Ut, Ux rangle$.

With that, the Fourier transform of $f$ now becomes
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt Ux, Lambda Ux gt}e^{-i lt Ut,Uxgt}d^nx\
end{split}$$
By changing the variable $y=Ux$, we get
$$begin{split}
hat{f}(t) &=
frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt Ut,ygt}|det U|d^nx\
end{split}$$
Noting that $|det U|=1$ and $U^TU=I$
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} e^{-lt y, Lambda y gt}e^{-i lt t,ygt}d^ny$$
If $lambda_1,...,lambda_n$ denote the diagonal elements of $Lambda $ (all strictly positive),
$$hat{f}(U^Tt)=frac 1{(2pi)^n}int_{mathbb R^n} prod_{k=1}^ne^{-lambda_ky_k^2}e^{-i t_ky_k}d^ny=prod_{k=1}^n frac 1 {2pi}int_{mathbb R}e^{-lambda_ky_k^2}e^{-i t_ky_k}dy_k$$
In other words, if you rotate the axes to align with the eigenvectors of $A$, the $n$-dimensional Fourier transform is the product of $1$-dimensional Fourier transforms of the univariate Gaussian function. And you know the expression for the Fourier transform of a $1$-dimensional Gaussian function.