I must analyze the Heisenberg box of $$f(t)=e^{-ax^2},\: a>0$$
Namely, I must calculate $\sigma_t$ and $\sigma_w$ and then verify $\sigma_t\sigma_w=1/2$.
The mean of wave function is $$u=\frac{1}{||f||_2^2}\int_{-\infty}^{\infty}t|f(t)|^2\: dt$$
The average momentum is $$\varepsilon = \frac{1}{2\pi ||f||_2^2}\int_{-\infty}^{\infty} w|\widehat{f}(w)|^2\: dw$$
The variances are defined as $$\sigma_t=\frac{1}{||f||_2^2}\int_{-\infty}^{\infty}(t-\mu)^2|f(t)|^2\: dt$$ $$\sigma_w=\frac{1}{2\pi||f||_2^2}\int_{-\infty}^{\infty}(w-\varepsilon)^2|\widehat{f}(w)|^2\: dw$$
I know that the energy of the function is $$||f||_2^2=\frac{\pi}{2a}$$
The mean therefore is $$\mu=0$$
Right now I'm stumbling on $\sigma_t$ because it involves the integral $$\int_0^{\infty}t^2e^{-2at^2}\: dt$$ I've tried looking up Laplace transform tables to find anything applicable, but no luck so far.
That integral is Feynman trickable
$$\int_0^\infty t^2 e^{-2at^2}\:dt = -\frac{1}{2}\frac{d}{da}\int_0^\infty e^{-2at^2}\:dt$$
$$= -\frac{1}{2}\frac{d}{da}\left(\frac{1}{2}\sqrt{\frac{\pi}{2a}}\right) = \sqrt{\frac{\pi}{128a^3}}$$