Given $\int_{-\infty}^{\infty} g(t)\cdot e^{-jwt} dt = we^{-2w^2}$, what the value $\int_{-\infty}^{\infty}y(t) dt$

Question

For a function $g(t)$, it is given that $\int_{-\infty}^{\infty} g(t)\cdot e^{-jwt} dt = w\cdot e^{-2w^2}$

if $y(t) = \int_{-\infty}^t g(\tau) d\tau$, then

$\int_{-\infty}^{\infty}y(t)dt = ?$

Any hint for that?

I'd tried to related Fourier transform with some derivative proprieties such as

$G(w) = w\cdot e^{-2w^2} \Rightarrow jwG(w) = w\cdot e^{-2w^2}+w(-2\cdot 2w)\cdot e^{-2w^2}$ but, i think its not a good strategy.