Inverse Laplace Transform of $\frac{1}{e^s-1}$ and $\frac{s}{e^s-1}$

Question

Question is as in the title. My complex analysis isn't where it used to be and these ILTs have come up in a problem I'm working on. They arise from taking the Laplace transform of a sum of delta functions, which may be why they're so nasty to calculate.

Answer 1

It's not hard to check that

$$\int_0^{\infty}\sum_{n=1}^{\infty}\delta(t-n)e^{-st}=\sum_{n=1}^{\infty}e^{-ns}=\frac{1}{e^s-1}$$

and

$$\int_0^{\infty}\sum_{n=1}^{\infty}\delta'(t-n)e^{-st}=-\sum_{n=1}^{\infty}s e^{-ns}=-\frac{s}{e^s-1}$$

and hence from the Laplace uniqueness theorem $$\mathcal{L}^{-1}\left[\frac{1}{e^s-1}\right](t)=\sum_{n=1}^{\infty}\delta(t-n)$$

$$\mathcal{L}^{-1}\left[\frac{s}{e^s-1}\right](t)=-\sum_{n=1}^{\infty}\delta'(t-n)$$

which is your original train of delta functions. If you are seeking a simpler form of these Laplace transforms , I doubt you would find it, since evaluation with complex analysis (if possible at all without introduction of smearing) should yield the same result.