STFT calculation with Gaussian Window
so MJ, my approach to using the Gaussian window for the STFT is first to use the "ordinary frequency" definition of the continuous-time Fourier Transform:

$$ \mathscr{F}\Big\{x(t)\Big\} \triangleq X(f) = \int\limits_{-\infty}^{\infty}x(t)\, e^{-j 2 \pi f t} \,\mathrm{d}t$$

$$ \mathscr{F}^{-1}\Big\{X(f)\Big\} \triangleq x(t) = \int\limits_{-\infty}^{\infty}X(f)\, e^{j 2 \pi f t} \,\mathrm{d}f$$

and start with this really cool isomorph of the Fourier Transform:

$$ \mathscr{F}\Big\{e^{-\pi t^2}\Big\} = e^{-\pi f^2} $$

and use the well known time-scaling and translation or frequency-scaling and translation theorems of the Fourier Transform to get you

$$ \mathscr{F} \Big\{ e^{a t^2 + b t + c} \Big\} = e^{A f^2 + B f + C} $$

where the constants $A$, $B$, and $C$ can be explicitly mapped from $a$, $b$, and $c$. It appears to me that the mapping is:

$$\begin{align}
A &amp;= \frac{\pi^2}{a} \\
\\
B &amp;= j \frac{\pi b}{a} \\
\\
C &amp;= c - \frac{b^2}{4a} - \tfrac{1}{2}\log\left(-\frac{a}{\pi}\right) \\
\end{align}$$

and the inverse mapping (which should be self-similar) is:

$$\begin{align}
a &amp;= \frac{\pi^2}{A} \\
\\
b &amp;= -j \frac{\pi B}{A} \\
\\
c &amp;= C - \frac{B^2}{4A} - \tfrac{1}{2}\log\left(-\frac{A}{\pi}\right) \\
\end{align}$$

Looks like $\Re\{a\}&lt;0$ and $\Re\{A\}&lt;0$ for the integrals to converge and for the $\log(\cdot)$ to be real and finite in the mapping.