Poynting vector for a whole spectrum Considering an electric and magnetic plane wave, $E(t)=E_p(\omega)e^{i\omega t}$ and $H(t)=H_p(\omega)e^{i\omega t}$, where $E_p(\omega)$ and $H_p(\omega)$ represent the frequency dependent peak amplitudes of the wave. The time-averaged Poynting vector is defined as $<S_x> = \frac{1}{2}\text{Re}\{E_p(\omega)H_p^*(\omega)\}$ with unit $\left[\frac{V}{m}\frac{A}{m} = \frac{W}{m^2}\right]$ (see: https://en.wikipedia.org/wiki/Poynting_vector#Time-averaged_Poynting_vector).
How to determine the total power if there is a whole radiation spectrum? For example, if you want to know the power carried by the solar spectrum? I would expect something like
$$ S_{tot} = \frac{1}{2} \text{Re}\{ \int\limits_0^\infty E_p(\omega) H_p^*(\omega) d\omega \} \,.$$
However, this expression has a wrong unit $\left[\frac{V}{m}\frac{A}{m}\frac{1}{s} = \frac{W}{m^2}\frac{1}{s}\right]$...
My question: what is the correct expression of the total power in function of the reciprocal electric and magnetic field?
 A: Confusion is caused by misleading notation. $E_p (\omega), H_p (\omega)$ is the first and the second case, are slightly different things.
In the case of monochromatic wave - they are the whole electric and magnetic field.
Because there is a wave with only one frequency I would drop the $\omega$ index, and denote them as $E, H$.
And in the second case - it is a density of electric(magnetic) field for a given frequency.
To see the connection between the first and the second case you may write:
$$
E = \int E_p (\omega) d \omega^{'} = \int E \ \delta (\omega - \omega^{'}) d \omega^{'}
$$
So the units of $E_p (\omega)$ are $\left[\frac{V s}{m}  \right]$. Now perform the integration ($T \rightarrow \infty$):
$$
S = \frac{1}{2T} \text{Re} \left[\int_{-T}^{T} dt \int d \omega \int d \omega^{'} E_p (\omega) H_p^{*} (\omega^{'}) e^{i (\omega - \omega^{'}) t  / T} \right] = 
\frac{1}{2} \text{Re} \left[ \int d \omega  \ E_p (\omega) H_p^{*} (\omega) \right]
 $$
Checking the dimensions of S:
$$
S = \left[\frac{V s}{m} \frac{A s}{m} \frac{1}{s} \frac{1}{s} \frac{s}{s}\right] = \left[\frac{W}{m^2} \right] 
$$
