[After further searching around, I decided to modify the question slightly] Let $A$ be an invertible real matrix and $g(\mathbf{x}) = e^{ i \mathbf{x}^T A \mathbf{x} }$. Let $w$ be a real valued smooth function of compact support, in particular a Schwartz function.
I just want to establish that $$ \int_{\mathbb{R}^n} w( \mathbf{x})g( \mathbf{x}) d \mathbf{x} = (2 \pi)^{-n} \int_{\mathbb{R}^n} \hat{w}(- \boldsymbol{\xi}) \hat{g} (\boldsymbol{\xi}) d (\boldsymbol{\xi}). $$ Since $g$ is a tempered distribution that coincides with the function $ e^{ i \mathbf{x}^T A \mathbf{x} }$, $\hat{g}$ is well-defined as a function and so as the integral on the right hand side. My thought is that it is a consequence of Parseval for Schwartz functions $$ \int_{\mathbb{R}^n} \bar{a}(\mathbf{x}) {b}( \mathbf{x}) d \mathbf{x} = (2 \pi)^{-n} \int_{\mathbb{R}^n} \bar{\hat{a}}( \boldsymbol{\xi}) \hat{b} (\boldsymbol{\xi}) d \boldsymbol{\xi}, $$ for any Schwartz functions $a, b$.
My attempt: Since $g$ is not a Schwartz function I think I need to construct a sequence of Schwartz function $g_h$ such that it "converges" to $g$, so that $$ \int_{\mathbb{R}^n} w( \mathbf{x})g( \mathbf{x}) d \mathbf{x} = \lim_h \int_{\mathbb{R}^n} w( \mathbf{x})g_h( \mathbf{x}) d \mathbf{x} = \lim_h (2 \pi)^{-n} \int_{\mathbb{R}^n} \hat{w}(-\boldsymbol{\xi}) \hat{g_h} (\boldsymbol{\xi}) d (\boldsymbol{\xi}) $$ and justify that the limit and the Fourier tranform can be swapped in the end.
As mentioned in the comment, this is probably just a special case of convolution theorem for tempered distributions, but I would like to learn the details... I would appreciate if someone could explain this to me. thank you!
The issue is that there no general theorem for tempered distribution. Indeed, you cannot always define the product of tempered distributions or the convolution product of any two tempered distribution. However, it can work for a Schwarz function and a tempered distribution. It applies here because $g$ is tempered and $w$ is Schwartz.
First off, $gw$ is defined weakly by: $$ \langle gw,\phi\rangle = \langle g,w\phi\rangle $$ and defines a tempered tempered distribution since the product of two Schwartz functions is Schwartz. $\hat g,\hat w$ are respectively tempered and Schwartz. Similarly, you can define weakly the $\hat g*\hat w$ using the fact that the convolution of Schwartz functions is still Schwartz: $$ \langle \hat g*\hat w,\phi\rangle = \langle \hat g,R\hat w*\phi\rangle\\ R\hat w(x) = \hat w(-x) $$ Thus, all your quantities make sense, and you can prove a convolution theorem in this framework. You can find a proof of this for example in An Introduction to the Theory of Distributions by Barros-Neto (Theorem 4.11).
Hope this helps.