How to Compute the Laplace Transforms of Summation

Question

Compute the Laplace Transforms of th following three unrelated functions:

$f_1(t)=\sum_{n=0}^\infty (-1)^n u(t-n)$

where $u(t-n)$ is the usual step function

$f_2(t)=\sum_{n=0}^\infty u(t-n)$

$f_3(t)=t - \left\lfloor t \right\rfloor$

where $t>0$ and $\left\lfloor t \right\rfloor$ is the floor function of $t$ and $u(t-n)$ is the usual step function.

I assume that I would just have to evaluate the sums and then take the Laplace Transform as usual, right? But if that were the case, wouldn't the two sums just converge to zero?

Answer 1

Since $f_1(t)\leqslant u(t)$ for all $t$ and $$\int_0^\infty |e^{-st}u(t)|\ \mathsf dt = \frac1s<\infty $$ for $s>0$, dominated convergence yields \begin{align} \mathcal L(f_1(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty (-1)^n u(t-n)\right) \mathsf dt\\ &= \sum_{n=0}^\infty (-1)^n\int_n^\infty e^{-st}\ \mathsf dt\\ &= \sum_{n=0}^\infty (-1)^n \frac{e^{-ns}}s\\ &= \frac1s \sum_{n=0}^\infty (-e^{-s})^n\\ &= \frac1{s(1+e^{-s})}. \end{align} Since $f_2(t)\leqslant t$ for all $t$ and $$ \int_0^\infty |e^{-st}t|\ \mathsf dt = \frac1{s^2} $$ for $s>0$, dominated convergence yields \begin{align} \mathcal L(f_2(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty u(t-n) \right)\mathsf dt\\ &= \sum_{n=0}^\infty \int_n^\infty e^{-st}\ \mathsf dt\\ &= \sum_{n=0}^\infty \frac{e^{-ns}}s\\ &= \frac1{s(1-e^{-s})}. \end{align} Since $f_3(T)\leqslant u(t)$ for all $t$ and $t\mapsto e^{-st}$ is integrable, by dominated convergence we have \begin{align} \mathcal L(f_3(t)) &= \int_0^\infty \left(e^{-st}\sum_{n=0}^\infty (t-n)(u(t-n)-u(t-n+1) \right)\mathsf dt\\ &= \sum_{n=0}^\infty \int_n^{n+1}e^{-st}(t-n)\ \mathsf dt\\ &= \sum_{n=0}^\infty \frac{\left(e^s-1-s\right) e^{-(n+1) s}}{s^2}\\ &= \frac{(e^s-1-s)e^{-s}}{s^2}\sum_{n=0}^\infty (e^{-s})^n\\ &= \frac{(e^s-1-s)e^{-s}}{s^2}\cdot \frac1{1-e^{-s}}\\ &= \frac{1+s-e^s}{s^2\left(1-e^s\right) }. \end{align}

Answer 2

$F_{1} (s)=\displaystyle \sum _{n=0}^{\infty} \dfrac{(-1)^n}{s}. e^{-ns}=\dfrac{1}{s(1+e^{-s})}$

$F_{2} (s)=\displaystyle \sum _{n=0}^{\infty}\dfrac{{e^{-sn}}}{s}=\dfrac{1}{s(1-e^{-s})}$

'

$F_{3}(s)$

=$\displaystyle \int _{0}^{\infty} (t-\lfloor{t}\rfloor)e^{-st} dt=\displaystyle \int _{0}^{1} te^{-st} dt+\displaystyle \int _{1}^{2} {(t-1)}\ e^{-st} dt+\displaystyle \int _{2}^{3} (t-2)e^{-st} dt+ \cdot {.............} +\lim{n\to\infty}\displaystyle \int _{n-1}^{n} (t-(n-1))e^{-st} dt $

$=\left(-\dfrac{e^{-s}}{s}+\dfrac{1-e^{-s}}{s^2}\right)+\left(-\dfrac{e^{-2s}}{s}+\dfrac{e^{-s}-e^{-2s}}{s^2}\right)+\left(-\dfrac{e^{-3s}}{s}+\dfrac{e^{-2s}-e^{-3s}}{s^2}\right)+.......+\left(-\dfrac{e^{-ns}}{s}+\dfrac{e^{-(n-1)s}-e^{-ns}}{s^2}\right)..$

$=-\dfrac{e^{-s}}{s}.\dfrac{1}{1-e^{-s}}+ \lim _{n \to \infty}\dfrac{1}{s^2}\left(1-e^{-s}+e^{-s}+e^{-2s}-e^{-2s}+......+e^{-(n-1)s}+e^{-ns}\right)$

$=\dfrac{1}{s^2}-\dfrac{1}{s}\dfrac{{e^{-s}}}{1-{e^{-s}}}=\dfrac{1-e^{-s}-se^{-s}}{s^2(1-e^{-s})}=\dfrac{e^s-s-1}{s^2(e^{s}- 1)}=\dfrac{1+s-e^{s}}{s^2(1-e^s)}$

Answer 3

You'd have a hard time evaluating the sums in my opinion.

No rather I'd try to evaluate $$\forall n \in \mathbb{N} \sum_{n=0}^\infty \int_{-n}^\infty |u(t)e^{-st}e^{-sn}|dt$$ (by a substitution) which exists and then since $$t \rightarrow \sum_{n=0}^\infty u(t-n)$$ converges to a function continuous on $$\mathbb{R_{+}^*}$$ you can invert integral and sum (after a substitution) and obtain the desired result. I let you verify the few other hypothesis to invert series/integral though. For the first one you can use dominated convergence theorem though.

As for $$f_3$$ you can split the integral on intervals [n,n+1] then evaluate the sum of the integral on each integral.