For a function g(t), it is given that $$\int\limits_{ – \infty }^{ + \infty } {g\left( t \right){e^{ – j\omega t}}dt = \omega {e^{ – 2{\omega ^2}}}} $$ for any real value $$\omega $$ . If $$y\left( t \right) = \int\limits_{ – \infty }^t {g\left( \tau \right)} d\tau ,\,{\rm{then}}\,\int\limits_{ – \infty }^{ + \infty } {y\left( t \right)} dt$$ is. . . . . . . .

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.