Adding power spectrum of real time-integrated kernel
@@ -639,14 +639,42 @@ The integrated kernel is
\begin{proof}{celerite-integral}
k_\Delta(\tau) = \frac{2}{\left(c\,\Delta\right)^2}\,k(\tau)\,
\begin{cases}
- \cosh(c\,\Delta) - 1 & \tau > \Delta \\
- c(\Delta - \tau) e^{c\tau} + \cosh{(c\tau)} e^{-c(\Delta - \tau)} -1 & \tau \le \Delta
+ \cosh(c\,\Delta) - 1 + c\,\Delta-\sinh(c\,\Delta) & \tau = 0 \\
+ c(\Delta - \tau) e^{c\tau} + \cosh{(c\tau)} e^{-c(\Delta - \tau)} -1 & 0 < \tau \le \Delta \\
+ \cosh(c\,\Delta) - 1 & \Delta < \tau
\end{cases}
%k_\Delta(\tau) = \frac{2}{\left(c\,\Delta\right)^2}\,\left[
% \cosh(c\,\Delta) - 1
%\right]\,k(\tau)
\end{proof}
-where $\Delta$ is the integration time.
+where $\Delta$ is the integration time. Note that in the standard
+use case of a single light curve with a single telescope and a
+uniform exposure time, the times of each cadence will be separated
+by greater than the exposure time, and so the kernel will be
+semi-separable. However, there is a slight difference in
+the implementation as the $\tau = 0$ term, which corresponds to
+the diagonal of the covariance matrix, has an extra term,
+$\frac{2}{(c\Delta)^2}(c\Delta-\sinh(c\Delta))$, times $k(\tau)$.
+This additional term is negative, but can simply be added to the white
+noise component of the covariance matrix when implementing the \celerite\
+kernel. Given that the kernel is positive, the matrix will still
+be positive definite.
+
+Although this expression appears complicated, the Fourier transform
+is less so, and is given by:
+\begin{equation}
+S(\omega) = \frac{ac}{\sqrt{2\pi}(c^2 + \omega^2)} \frac{\sin^2(\tfrac{1}{2}\omega \Delta)}{(\tfrac{1}{2}\omega\Delta)^2}.
+\end{equation}
+Note that this is the Fourier transform of the kernel, $k(\tau)$,
+multiplied by the square of a sinc function, $\mathrm{sinc}{(\omega\Delta/2)}$,
+which results from the convolution of the kernel with a top-hat
+time-integration of half-width $\Delta/2$. Convolution in Fourier
+space turns into multiplication, and the square of the sinc function
+results from taking the square of the Fourier transform to obtain
+the power spectrum. Also note that the since function goes to zero
+when $\omega\Delta = 2\pi$, which corresponds to the twice the Nyquist frequency,
+$f = \frac{2\pi}{\omega} = \Delta^{-1}$.
+
\paragraph{Complex term}
