How to calculate the value of $\int_{-\infty}^{\infty} y(t)dt$?

Question

For a function $g(t)$, $\int_{-\infty}^{\infty} g(t)e^{-j\omega t}dt=\omega e^{-2\omega ^2}$ for any real value $\omega$. If $y(t)=\int_{-\infty}^{t}g(\tau)d\tau$ then how to calculate the value of $\int_{-\infty}^{\infty} y(t)dt$?

Answer 1

To compute $\int_{- \infty}^{+\infty} y(t) \, dt$ you can find its Fourier transform $Y(\omega)$ and evaluate that zero: $$Y(0) = \int_{- \infty}^{+ \infty} y(t) \, dt.$$ By Fourier transform properties we know $$Y(w) = \mathcal{F} \{ y(t) \} = \mathcal{F} \left\{ \int_{- \infty}^t g(\tau) \, d \tau \right\} = \frac{\mathcal{F}\{g(t)\}}{j \omega}.$$ What you are given is precisely the Fourier transform of $g(t)$, therefore $$Y(\omega) = \frac{\omega e^{-2 \omega^2}}{j \omega} = \frac{e^{-2 \omega^2}}{j}.$$ Finally $$Y(0) = \int_{- \infty}^{+\infty} y(t) \, dt = \frac{1}{j}.$$