Solution
= Solution
Use the <Fourier transform> convention $\widetilde a(\omega)=\int_{-\infty}^{\infty}a(t)e^{-i\omega t}\,dt$, with inverse factor $1/(2\pi)$. Direct <integration> over the pulse gives
$$
\boxed{\widetilde a(\omega)=\int_{-1}^1e^{-i\omega t}\,dt=\frac{2\sin\omega}{\omega},\qquad\widetilde a(0)=2.}
$$
The apparent singularity at zero is removable. Changing the values of the pulse at its two endpoints does not affect the transform.