What is the Fourier transform of $$ e^{ia\langle Ax,x \rangle/2} $$ where $a>0$ and $A$ is a symmetric invertible matrix such that $\Im(A) \geq 0$ (i.e. $\Im\langle Ax, x\rangle \geq 0$ for all $x\in\mathbb{R}^n$). I would be satisfied with a method for finding the Fourier transform in the case where $A$ is real. The answer is supposed to be $$ \frac{1}{\sqrt{\det(aA/2\pi i)}}e^{-i\langle Ax, x\rangle/2a} $$ In the above, I am not sure what the author means by the square root since the argument might not be in $\mathbb{R}_{\geq 0}$. In that case, there will be two posible values and I am not sure how to pick the right one.
Thanks in advance
I figured it out and therefore decided to answer my own question. First, note that by writing $iA = -B$ where $B = -iA$ it suffices to show that $$ \mathcal{F}\left(e^{-a\langle Bx, x\rangle/2}\right)(\xi) = \left(\frac{2\pi}{a}\right)^{n/2}\frac{1}{\sqrt{\det(B)}}e^{\frac{-1}{2a}\langle B^{-1}\xi, \xi\rangle} $$ This can be done by first considering the case where $B$ is positive definite. Then we use the result on $B+\epsilon I$ and take the limit as $\epsilon \to 0$.