Fourier Transform of a $N$-dimensional complex Gaussian For $x \in \mathbb{R}^N$, given a complex function $f(x)\in\mathbb{C}$:
$$
f(x) = \exp \left( - \pi x^{T} (A + j B) x \right)
$$
where real matrices $A = A^{T}$ and $B = B^{T}$ are positive semi definite.
How to derive its Fourier transform in terms of $A$ and $B$? That is:
$$
F(\nu) = \int \!  f(x) \exp(-j 2\pi \nu \cdot x) \, \mathrm{d} x
$$
 A: Let $B = U^TDU$ with a diagonal matrix $D$ and an orthogonal matrix $U$. Then
$$
x^TBx + 2\nu^Tx = \sum_{k=1}^Nd_k(y_k+r_k)^2 - \sum_{k=1}^Nw_kr_k,
$$
where $y = Ux$, $w = U\nu$, and $r = D^{-1}w$. Now, set $z = y+r$. Then
$$
x^TBx + 2\nu^Tx = z^T Dz - w^Tr = |D^{1/2}z|^2 - w^Tr.
$$
As another change of variable, set $t = D^{1/2}z$. Then $x^TBx + 2\nu^Tx = |t|^2 - c$, where $c = w^Tr = |D^{-1/2}w|^2$. We get (I'm not sure whether it's $1/2$ or $-1/2$ in the exponent -- you should check it)
$$
\int f(x)e^{-2\pi i\nu^Tx}\,dx = e^{\pi ic}(\det D)^{\pm 1/2}\cdot\int e^{-\pi(D^{-1/2}t-r)^TUAU^T(D^{-1/2}t-r)}\cdot e^{-\pi i|t|^2}\,dt.
$$
Now, setting $A_1 := D^{-1/2}UAU^TD^{-1/2}$ and $\rho := D^{1/2}r$,
\begin{align}
(D^{-1/2}t-r)^TUAU^T(D^{-1/2}t-r)
&= (t-\rho)^TA_1(t-\rho)\\
&= t^TA_1t - 2\rho^TA_1t + \rho^TA_1\rho.
\end{align}
We have $A_1 = V^TEV$ with a diagonal matrix $E = diag(e_k)$ and an orthogonal matrix $V$ and we get
$$
t^TA_1t - 2\rho^TA_1t + \rho^TA_1\rho = \sum_{k=1}^Ne_k(x_k-\tau_k)^2,
$$
where $x = Vt$ and $\tau = V\rho$. Since $|t| = |x|$ (which is the main trick here), we obtain
$$
\int f(x)e^{-2\pi i\nu^Tx}\,dx = C\cdot\int e^{-\pi\sum e_k(x_k-\tau_k)^2}\cdot e^{-\pi i|x|^2}\,dx = C\prod_{k=1}^N\int e^{-\pi(e_k(s-\tau_k)^2+is^2)}\,ds.
$$
