I'm trying to solve the following problem:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{F}(\text{s})+\text{G}(\text{s})}{1-\exp\left(-\frac{\text{s}}{4x}\right)}\right]_{\left(t\right)}\tag1$$
Where:
- $$\text{F}(\text{s})=\int_0^\text{n}\left(\epsilon\cdot\exp\left(-\frac{t}{\text{r}}\right)\right)\cdot e^{-\text{s}t}\space\text{d}t\tag2$$
- $$\text{G}(\text{s})=\int_\text{n}^\frac{1}{4x}\left(\left|\text{k}\cdot\sin\left(2\pi\cdot x\cdot t\right)\right|-\text{p}\right)\cdot e^{-\text{s}t}\space\text{d}t\tag3$$
My work:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{F}(\text{s})+\text{G}(\text{s})}{1-\exp\left(-\frac{\text{s}}{4x}\right)}\right]_{\left(t\right)}=\int_0^t\mathcal{L}_\text{s}^{-1}\left[\text{F}(\text{s})+\text{G}(\text{s})\right]_{\left(\tau\right)}\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1-\exp\left(-\frac{\text{s}}{4x}\right)}\right]_{\left(t-\tau\right)}\space\text{d}\tau=$$ $$\int_0^t\left(\mathcal{L}_\text{s}^{-1}\left[\text{F}(\text{s})\right]_{\left(\tau\right)}+\mathcal{L}_\text{s}^{-1}\left[\text{G}(\text{s})\right]_{\left(\tau\right)}\right)\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1-\exp\left(-\frac{\text{s}}{4x}\right)}\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag4$$
For the integral:
$$\text{F}(\text{s})=\int_0^\text{n}\left(\epsilon\cdot\exp\left(-\frac{t}{\text{r}}\right)\right)\cdot e^{-\text{s}t}\space\text{d}t=\frac{\epsilon\text{r}}{1+\text{rs}}\cdot\left(1-\exp\left(-\text{n}\left(\frac{1}{\text{r}}+\text{s}\right)\right)\right)\tag5$$
Now, we also know that (when $\exp\left(\Re\left(x\right)\right)>1$):
$$\frac{1}{1-\exp\left(-x\right)}=\sum_{\text{m}=0}^\infty\exp\left(-\text{m}\cdot x\right)\tag6$$
So, we get:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{1-\exp\left(-\frac{\text{s}}{4x}\right)}\right]_{\left(t-\tau\right)}=\mathcal{L}_\text{s}^{-1}\left[\sum_{\text{m}=0}^\infty\exp\left(-\text{m}\cdot\frac{\text{s}}{4x}\right)\right]_{\left(t-\tau\right)}=$$ $$\sum_{\text{m}=0}^\infty\mathcal{L}_\text{s}^{-1}\left[\exp\left(-\text{m}\cdot\frac{\text{s}}{4x}\right)\right]_{\left(t-\tau\right)}=4\cdot\sum_{\text{m}=0}^\infty\delta\left(\frac{\text{m}}{x}-4\cdot\left(t+\tau\right)\right)\tag7$$
And:
$$\text{f}\left(\tau\right)=\mathcal{L}_\text{s}^{-1}\left[\text{F}\left(\text{s}\right)\right]_{\left(\tau\right)}=\epsilon\cdot\exp\left(-\frac{\tau}{\text{r}}\right)\cdot\left(1-\theta\left(\tau-\text{n}\right)\right)\tag8$$
So, for now we have:
$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{F}(\text{s})+\text{G}(\text{s})}{1-\exp\left(-\frac{\text{s}}{4x}\right)}\right]_{\left(t\right)}=$$ $$\int_0^t\left(\epsilon\cdot\exp\left(-\frac{\tau}{\text{r}}\right)\cdot\left(1-\theta\left(\tau-\text{n}\right)\right)+\mathcal{L}_\text{s}^{-1}\left[\text{G}(\text{s})\right]_{\left(\tau\right)}\right)\cdot\left(4\cdot\sum_{\text{m}=0}^\infty\delta\left(\frac{\text{m}}{x}-4\cdot\left(t+\tau\right)\right)\right)\space\text{d}\tau\tag9$$
Now, I've no idea how to proceed further.
EDIT:
I also found that:
$$\text{g}\left(\tau\right)=\mathcal{L}_\text{s}^{-1}\left[\text{G}\left(\text{s}\right)\right]_{\left(\tau\right)}=$$ $$\theta\left(\tau-\frac{1}{4x}\right)\cdot\left(\text{p}-\text{k}\sin\left(2\pi x\tau\right)\right)+\theta\left(\tau-\text{n}\right)\cdot\left(\text{k}\left|\sin\left(2\pi x\tau\right)\right|-\text{p}\right)\tag{10}$$
To finish off your proof, just use the sifting property of the delta function: $\int_0^\infty h(t)\delta(t-a)dt = h(a)$. I think a sign error has also crept in, but I'm not totally sure where.
However, if you want to do it a little more cleanly, it's easier to use the time shift identity for the Laplace transform: $\mathcal{L}[h(t-a)\theta(t-a)](s) = \exp(-as)\mathcal{L}[h(t)](s)$ \begin{multline} \mathcal{L}^{-1}\left[\frac{F(s)+G(s)}{1-\exp\left(-\frac{s}{4x}\right)}\right](t) = \sum_{m=0}^\infty \mathcal{L}^{-1}\left[\left(F(s)+G(s)\right)\exp\left(-\frac{ms}{4x}\right)\right](t) \\= \sum_{m=0}^\infty\theta\left(t-\frac{m}{4x}\right)\mathcal{L}^{-1}\left[F(s) + G(s)\right]\left(t - \frac{m}{4x}\right) \end{multline} You've already found $\mathcal{L}^{-1}[F(s)](t)$ and $\mathcal{L}^{-1}[G(s)](t)$, so you should be able to go from here.