Laplace of sum of shifted diracs : $\sum_{n=-\infty}^{\infty}\delta(t-nT)$

Question

I want to compute : $$\sum_{n=-\infty}^{\infty}\delta(t-nT)$$ I tried as such:
$$\cal{L} \Big( \sum_{n=-\infty}^{\infty}\delta(t-nT) \Big)= \sum_{n=-\infty}^{\infty}\cal{L}(\delta(t-nT)=\sum_{n=-\infty}^{\infty} e^{-snT} = \sum_{n=-\infty}^{0}e^{-snT} + \sum_{n=0}^{\infty}e^{-snT}$$

The second one converges (geometric series) but the first one?

Answer 1

You made a mistake after the second equality; indeed, one has $$ \sum_{n=-\infty}^\infty\mathscr{L}[\delta(t-nT)](s) = \sum_{n=\color{red}{0}}^\infty e^{-snT} $$ since $$ \mathscr{L}[\delta(t-nT)](s) = \int_{0^-}^\infty\delta(t-nT)e^{-st}\,\mathrm{d}t = \begin{cases} e^{-snT} & \mathrm{if}\quad n\ge0 \\ 0 & \mathrm{otherwise} \end{cases} $$

Answer 2

Laplace transform is usually defined on the positive half-axis. Do you want to use two-sided Laplace transform? It has rather strict convergence requirements.

I would suggest to use Fourier transform (of distributions) instead, which yields another Dirac comb after some regularization.