The correct answer is $\boxed{{j \over 2}}$.
To solve this, we can use the following property of the Fourier transform:
$$\int_{-\infty}^{\infty} y(t) e^{-j\omega t} dt = \pi \left[ \text{Re} \left( \hat{y}(\omega) \right) + j \text{Im} \left( \hat{y}(\omega) \right) \right]$$
where $\hat{y}(\omega)$ is the Fourier transform of $y(t)$.
In this case, we are given that
$$\hat{g}(\omega) = \omega e^{-2\omega^2}$$
Therefore, the Fourier transform of $y(t)$ is
$$\hat{y}(\omega) = \pi \omega e^{-2\omega^2}$$
The inverse Fourier transform of $\hat{y}(\omega)$ is
$$y(t) = \frac{\pi}{2} \int_{-\infty}^{\infty} \omega e^{-2\omega^2} e^{j\omega t} d\omega = \frac{j}{2}$$
Therefore,
$$\int_{-\infty}^{\infty} y(t) dt = \frac{j}{2}$$
Here is a brief explanation of each option:
- Option A: $0$. This is the integral of an even function over all real numbers. Since $y(t)$ is an even function, its integral over all real numbers is $0$.
- Option B: $-j$. This is the integral of an odd function over all real numbers. Since $y(t)$ is an odd function, its integral over all real numbers is $-j$.
- Option C: $-\frac{j}{2}$. This is the integral of a function that is neither even nor odd over all real numbers. Since $y(t)$ is a function that is neither even nor odd, its integral over all real numbers is $\frac{j}{2}$.
- Option D: $\frac{j}{2}$. This is the correct answer.