Encontrar la transformada de fourier de la siguiente señal:

$$ x(t) = [u(t + 4)-u(t - 4)] $$

\(x(t) = 1\) para \( t \in [-4, 4] \)
Reemplazamos en la formula de la transformada de Fourier: \[ X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j \omega t} \, dt \] \[ X(\omega) = \int_{-4}^{4} 1 e^{-j \omega t} \, dt \] \[ X(\omega) = \int_{-4}^{4} e^{-j \omega t} \, dt \] \(e^v\)
\(v = -j \omega t\)
\(v' = -j \omega\)
\[ X(\omega) = \int_{-4}^{4} e^{-j \omega t} \frac{-j \omega}{-j \omega} \, dt \] \[ X(\omega) = \frac{1}{-j \omega}\int_{-4}^{4} e^{-j \omega t} (-j \omega) \, dt \] \[ X(\omega) = \frac{e^{-j \omega t}}{-j \omega} \Big|^{4}_{-4} \] \[ X(\omega) = \frac{e^{-j \omega 4}}{-j \omega} - \frac{e^{-j \omega (-4)}}{-j \omega} \] \[ \color{lime} \boxed{ \color{black}X(\omega) = \frac{e^{-j \omega 4}}{-j \omega} + \frac{e^{j \omega 4}}{j \omega} } \]