Question Am I correctly resuming the series to invert this Laplace Transform? Specific points of interest are bullet pointed.
The Laplace Transform of a function, $C(t)$, is given by,
$$ F(\beta) = \mathcal{L}C(t) = \int_0^\infty C(t)e^{-\beta t}\,dt $$ where, in this case, $$ F(\beta) = \frac{\mu e^{-\beta \tau}}{1-\mu e^{-\beta \tau}} = \sum^\infty_{j=1} \mu^je^{-jt\beta} \quad \text{for}\quad |\mu|<1 $$ and I expanded the geometric sum on the RHS. This is valid as in my case, $\mu=\cos^2\tau$
Aside...
To invert the Laplace Transformed form back to $C(t)$ I note that $\mathcal{L}\delta(t) = \int_0^{\infty}\delta(t)e^{-t\beta}dt=1$ and more generally that, $$ \mathcal{L}\sum^\infty_{j=0}\mu^j\delta(t-j\tau) = \sum^\infty_{j=0}\mu^je^{-j\tau\beta}=\frac{1}{1-\mu e^{-\beta \tau}} $$ ...End Aside
Is it valid to extend the argument to the summation starting at $j=1$? $$ \mathcal{L}\sum^\infty_{j=1}\mu^j\delta(t-j\tau) \stackrel{?}{=} F(\beta) = \sum^\infty_{j=1} \mu^je^{-jt\beta} $$ and hence, $$ C(t) = \sum^\infty_{j=1}\mu^j\delta(t-j\tau) $$
Should this be regarded similarly to an integral only accepting values $j=t/\tau$ for $t/\tau\ge1$?
If so this would give, $$ C(t) = \mu^{t/\tau} \quad \text{for}\quad t/\tau \in \mathbb{Z}^+ $$ by integrating over the delta function. As per the comments I would treat this as a numerical approximation for small $\tau$.