i need to compute the inverse laplace transform of $\frac{e^s}{1 + e^s}$, but i'm kind of lost.
i know that $\frac{e^s}{1 + e^s} = \frac{1}{1 - e^{-s}}$, so this can be expressed as a geometric series, i.e., $\frac{1}{1 - e^{-s}} = 1 + e^{-s} + e^{-2s} + e^{-3s} + ... + e^{-ks} + ...$.
i also know that the laplace transform of each element of this series expansion is $\delta(t - k)$, i.e., $L(1) + L(e^{-s}) + L(e^{-2s}) + L(e^{-3s}) + ... + L(e^{-ks}) + ... = \delta(t) + \delta(t - 1) + \delta(t - 2) + \delta(t - 3) + ... + \delta(t - k) + ...$.
so this suggests to me that $L(\frac{1}{1 - e^{-s}}) = L(\frac{e^s}{1 + e^s}) = \sum_{i = 0}^k \delta(t - k)$.
and i think this last result implies that $L^{-1}(\sum_{i = 0}^k \delta(t - k)) = \frac{e^s}{1 + e^s}$, but i have also been told that another possible result is $\delta(sin(\pi \cdot x))$.
any advice? thank you in advance.