Why a propagating pulse has no DC spectral component? I found in a text that : "Because the laser pulse represents a propagating electromagnetic wave packet,
the dc component of its spectrum vanishes. Hence the time integral over the
electric field is zero."
Why is it so? I mostly don't understand why should a pulse not have any DC component.
 A: The pulse having a DC component would imply that there is a constant electric field forever after and before the pulse arrives. Because light obeys a dispersion relation $\omega = c  k$, a field with $\omega=0$ does not propagate. Hence, the DC component is spatially and temporally uniform. In that case,  it is meaningless to attribute the DC component to the laser pulse because it will be there far after the pulse leaves and before it even arrives.
Edit: OP sounds like he is still confused. The essential point here is that if the pulse satisfies the following:
$$\lim_{t \to \pm \infty}\mathbf{E}(\mathbf{x},t)=0$$
$$\lim_{\mathbf{x} \to \pm \infty}\mathbf{E}(\mathbf{x},t)=0$$
Then there is no DC component. The proof is that at $t\to\pm\infty$ and $\mathbf{x}\to\pm\infty$ any DC component will no go to zero strictly speaking.
Provided $\mathbf{E}(\mathbf{x},t)$ satisfies these conditions (as the usual definition of a pulse should) then if you integrate $\mathbf{E}$ over all time, you average to zero.
