I'm working through Keener and Sneyd's Mathematical Physiology I: Cellular Physiology and had trouble solving a derivation. To summarize, we have a 1D Markov chain (illustration in screenshot below). I'm assuming that the parameter $k_{i}$ is a constant $k_{0} = k_{1} = ... \equiv k $ to make things easier.
\begin{align} \frac{dP_{0}}{dt} &= - k P_{0}\\ \frac{dP_{1}}{dt} &= k (P_{0} - P_{1})\\ \vdots\quad &=\quad \vdots\\ \frac{dP_{n-1}}{dt}&= k (P_{n-2} - P_{n-1}) \end{align}
with boundary conditions $P_{0}=1$ and $P_{n}=0$ for $n>0$. I'm trying to obtain the function $P_{n-1}(t)$ as the result of a Fourier transform, but I'm not familiar with how to apply Fourier transforms to Markov chains. Once I obtain a function $P_{n-1}(t)$, I should be able to obtain the mean $\mu$ as
$$\mu = k \int_{0}^{\infty} t\cdot P_{n-1}(t)dt$$
The first equation is easy to solve it gives you:
$$P_0(t) = P_0(0)e^{-kt} = e^{-kt}.$$
I suggest using Laplace transform since the domain is $(0, \infty)$ you will have that
$$\mathcal L\left[P_0\right](\omega) = \int_{0}^\infty e^{-\omega t}P_0(t)\mathrm d t = \int_{0}^\infty e^{-(\omega + k)t}\mathrm d t = \frac{1}{k+\omega}$$
and for $\ell=1,\ldots,n-1$,
$$\mathcal L\left[\frac{\mathrm d}{\mathrm dt}{P}_\ell\right](\omega) = k \left(\mathcal L\left[P_{\ell-1}\right] - \mathcal L\left[P_{\ell}\right]\right)$$
Since $$\mathcal L\left[\frac{\mathrm d}{\mathrm dt}{P}_\ell\right](\omega) = -\omega \mathcal L\left[{P}_\ell\right](\omega) + P_\ell(0)$$
then you will have:
$$\mathcal L\left[P_\ell\right] = \frac{k}{k-\omega}\mathcal L\left[P_{\ell-1}\right]$$
and by recursive $$\mathcal L\left[P_\ell\right] = \left(\frac{k}{k-\omega}\right)^\ell \frac1{k+\omega}$$
Now if you want $P_\ell$ apply the inverse Laplace transform. But if you only need to compute $\mu$, you can use the fact that,
$$\mu = k\int_{0}^\infty tP_{n-1}(t)\mathrm d t = -k\frac{\mathrm d}{\mathrm d\omega} \mathcal L\left[P_{n-1}\right]_{\Big| \omega = 0} = k\left[k^{n-1}\left(\frac{n-1}{\left(k-\omega\right)^{n}}\frac1{k+\omega} - \frac1{(k-\omega)^{n-1}}\frac1{(k+\omega)^2} \right)\right]_{\Big| \omega = 0} = \frac{n - 2}{k^2}.$$