I am working on the Stockwell Transform and want to show that time averaging of the Stockwell Transform yields the spectrum of the function $u(t)$.
$G(t'-t,\omega)$ being a Gaussian Window Function, with $t'$ being the center time.
$$ G(t'-t,\omega) = ae^{-b(t'-t)^2} $$
satisfying the condition
$$ \int G(t_\alpha,\omega) dt_\alpha = 1 $$
The Stockwell Transform is defined as
$$ S(t,\omega)=\int{u(t')} G(t'-t,\omega) e^{-i \omega t'} dt' $$
It is claimed that by the integral properties of the Gaussian function, the relation between $S(t'-t,\omega)$ and $u(\omega)$ (the Fourier transform of $u(t)$) is
$$ \int S(t,\omega)\; dt = u(\omega) $$
Attempt at a proof
$$ \rm{To\; show\; that\;} \int S(t,\omega)\; dt = u(\omega) $$
Have done time averaging as $\int S(t,\omega) dt$ assuming the condition $\int G(t'-t,\omega) dt' = 1$
$$ \int S(t,\omega) dt = \int \int{u(t')}\; [G(t'-t,\omega) e^{-i \omega t'} dt']\;dt $$
$$ = \int \bigg[\int{u(t')} e^{-i \omega t'} dt'\bigg]\; G(t'-t,\omega)\; dt $$
Because $G(t'-t,\omega)$ is dependent on $t'$, I hit a wall.
The specific expression for $G(t'-t,\omega)$ is a Gaussian Function
$$ G(t'-t,\omega) = a\; e^{-b(t'-t)^2} $$