Question

Consider the signal x(t) =$\{\begin{array}{c}{e}^{-t},t\ge 0\\ \\ 0,t<0\end{array}$
Let $x(\omega$) denote the Fourier transform of this signal.
The integral $\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}X\left(\omega \right)d\omega$ is

Answer 1

$x\left(t\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}X\left(\omega \right)e^{-j\omega t}d\omega$

putting t = 0 in the above expression we get,

$x\left(0\right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}X\left(\omega \right)e^{0}d\omega$

$\therefore \frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}X\left(\omega \right)d\omega =x\left(0\right)=e^{-0}=1$