Define the Short-time (or windowed) Fourier Transform of a function $f:\mathbb{R}\rightarrow\mathbb{C}$ as follows,
$F_gf(\omega,t)=\int\limits_{\mathbb{R}}f(x)\overline{g(x-t)e^{ix\omega}}dx$.
Show that $||F_gf||^2=2\pi\cdot||f||^2\cdot||g||^2$ where all norms are $L^2$. I am having trouble manipulating things inside the integrals with the squares.
Let $\tau_t g$ denote, for a fixed $t$, the function $x \mapsto g(x-t)$. We have:
$$F_g f (\omega, t) = \int_{\Bbb R} f(x) \overline{\tau_t g(x)} e^{-i\omega x } dx = \mathcal F({f\overline{\tau_t g}})(\omega)$$
We know that $\mathcal F: L^2 \to L^2$ is an isometry up to a $2\pi$ factor, so (wrt $\omega$) $\|F_g f\|_{L^2(\Bbb R)}^2 = 2 \pi \|f \overline{\tau_t g} \|^2 $. But:
$$\|f \overline{\tau_t g} \|^2 = \int_{\Bbb R} |f(x) \overline {g(x-t)}|^2 dx = \int_{\Bbb R} |f(x)|^2|g(x-t)|^2 dx$$
We get:
$$\|F_g f\|^2_{L^2(\Bbb R \times \Bbb R)} = \int_{\Bbb R} \int_{\Bbb R} |F_gf(\omega,t)|^2 d\omega dt = \int_{\Bbb R} \|F_g f \|^2_{L^2(\Bbb R) \text{ (wrt $\omega$)}}dt \\ = 2 \pi \int_{\Bbb R} \left( \int_{\Bbb R} |f(x)|^2|g(x-t)|^2 dx \right) dt = 2 \pi \int_{\Bbb R} \left( \int_{\Bbb R} |f(x)|^2|g(x-t)|^2 dt \right) dx \\ = 2 \pi \int_{\Bbb R} |f(x)|^2\left( \int_{\Bbb R} |g(x-t)|^2 dt \right)dx$$
After a change of variable $ u = x - t$, the inner integral is just $\|g\|^2$. So, we get the desired result.