\subsection{General form}

The procedure here is exactly the same as the previous section.
The integral normalizing $G_0$ just become a little more complicated\ldots

Fourier transforming eq. \ref{eq.DHO} and applying \ref{eq.four_deriv} we have
\begin{align}
  (-m\omega^2 + i \gamma \omega + k) x(\omega) &= F(\omega)
                                              \label{eq.DHO_freq} \\
  (\omega_0^2-\omega^2 + i \beta \omega) x(\omega) &= \frac{F(\omega)}{m} \\
  |x(\omega)|^2 &= \frac{|F(\omega)|^2/m^2}
                        {(\omega_0^2-\omega^2)^2 + \beta^2\omega^2}
                                              \label{eq.DHO_xmag},
\end{align}
where $\omega_0 \equiv \sqrt{k/m}$ is the resonant angular frequency
 and $\beta \equiv \gamma / m$ is the drag-acceleration coefficient.

We compute the \PSD\ by plugging eq. \ref{eq.DHO_xmag} into \ref{eq.psd_def}
\begin{equation}
  \PSD(x, \omega)
        = \normLimT \frac{2 |F(\omega)|^2/m^2}
                         {(\omega_0^2-\omega^2)^2 + \beta^2\omega^2}.
                                               \label{eq.DHO_psd}
\end{equation}

Plugging eq. \ref {eq.GOdef} into \ref{eq.DHO_psd} we have
\begin{equation}
  \PSD(x, \omega) = \frac{G_0/m^2}{(\omega_0^2-\omega^2)^2 + \beta^2\omega^2}.
                                       \label{eq.model_psd}
\end{equation}
Integrating over positive $\omega$ to find the total power per unit time yields
\begin{align}
  \iOInf{\omega}{\PSD(x, \omega)}
     &= \frac{G_0}{2m^2}
      \iInfInf{\omega}{\frac{1}{(\omega_0^2-\omega^2)^2 + \beta^2\omega^2}} \\
     &= \frac{G_0}{2m^2} \cdot \frac{\pi}{\beta\omega_0^2} \\
     &= \frac{G_0 \pi}{2m^2\beta\omega_0^2} \\
     &= \frac{G_0 \pi}{2m^2\beta \frac{k}{m}} \\
     &= \frac{G_0 \pi}{2m \beta k}
\end{align}
The integration is detailed in Section \ref{sec.integrals}.
By our corollary to Parseval's theorem (eq. \ref{eq.parseval_var}), we have
\begin{equation}
  \avg{x(t)^2} = \frac{G_0 \pi}{2m^2\beta\omega_0^2} \label{eq.DHO_var}
\end{equation}

Plugging eq. \ref{eq.DHO_var} into the equipartition theorem
(eqn. \ref{eq.equipart}) we have
\begin{align}
  k \frac{G_0 \pi}{2m \beta k} &= k_BT \\
  G_0 &= \frac{2}{\pi} k_BT m \beta.  \label{eq.GO}
\end{align}

So we expect $x(t)$ to have a power spectral density per unit time given by
\begin{equation}
  \PSD(x, \omega) = \frac{2 k_BT \beta}
                         {\pi m \left[
                                (\omega_0^2-\omega^2)^2 + \beta^2\omega^2
                                \right] }
\end{equation}
