What is the the Laplace transform of $\sum^{\infty}_{n=0}(-1)^nu(t-n)$?

Question

What is the the Laplace transform of $\sum^{\infty}_{n=0}(-1)^nu(t-n)$?

u(t-n) is a step function. $$ u(t-n) = \left\{ \begin{array}{ll} 0 & \quad x < n \\ 1 & \quad x \geq n \end{array} \right. $$

My professor gives this question on a homework. I am baffled by how to deal with the varying $n$ in the step function and the changing sign. Can I turn this into an integral?

Answer 1

i think there is no closed form, the Laplace tranfsorm of the step function given by $ u(t-n)$ is equal to

$$ \int_{0}^{\infty}dt u(t-n)exp(-st) =\frac{e^{-sn}}{s} $$

Answer 2

\begin{align} \mathcal L \{f(t)\}& = \sum_{n=0}^\infty \int_0^\infty (-1)^n e^{-st} u(t-n) \\ & = \sum_{n=0}^\infty \frac{(-1)^n e^{-ns}}{s} = \frac{1}{s}\frac{1}{1+e^{-s}} \end{align}