Calculating power in frequency band using power spectral density In a text I am using it states: 


  The fourier transform $F(\omega)$ has both positive and negative
  frequencies and $F(\omega) = F^*(-\omega)$. The power spectral
  density $S(\omega)$ is similarly defined to have the positive and
  negative frequencies, and the power spectral density is similarly
  equally divided between the positive and negative frequencies. Thus
  power spectrum $S(\omega)$ is symmetric about $\omega = 0$. So the
  contribution to the total power that is contained in the band
  $[\omega_1, \omega_2]$ is $$P_{[\omega_1, \omega_2]} = \frac{1}{2 \pi} \int_{\omega_1}^{\omega_2} [S(\omega) + S(-\omega)]d \omega.$$


Question: If $S(\omega)$ is the power per frequency $\omega$, even though $\omega$ takes both positive and negative values, I still don't see why the power contained in the band $[\omega_1, \omega_2]$ is not simply given by $$\frac{1}{2 \pi}\int_{\omega_1}^{\omega_2} S(\omega) d \omega?$$ I am only beginning to learn about these things, hence there might be something trivial that I am missing. Thanks.
It's less about power spectrum and more about Fourier transform of real signals. Let's take an easy signal, $g(x) = \cos(10\pi x)$.

Then:
$$ F(\omega) = \frac{\delta(\omega -10) + \delta (\omega +10)}2.$$

When you ask "what is the energy in the band $ \omega = [9,11]$ you ask what is the energy of that cosine (of course this is an example).

If you look only at the positive side of the frequency you lose half the energy.