Evaluating laplace transform of {x}

Question

I came upon a question on Laplace transforms in my exams. $${\scr L}[\{x\}]$$ where $\{x\}=x - [x]$ denotes the fractional part of $x$.
This is how I approached the problem. $$f(x)=\{x\}=\sum_{n=-\infty}^\infty{(x-n)(u_n(x)-u_{n+1}(x))}$$ where $u_a(x)$ denotes the Heaviside function shifted to $x=a$. $$\implies \scr L[f(x)]=\sum_{n=-\infty}^\infty{e^{-ns}\scr L[x-n]} - e^{-(n+1)s}\scr L[x-n]$$ $$=\sum_{n=-\infty}^\infty{\frac{e^{-ns}-e^{-(n+1)s}}{s^2}-\frac{n(e^{-ns}-e^{-(n+1)s})}{s}}$$ $$=\sum_{n=-\infty}^\infty{\frac{e^{-ns}-e^{-(n+1)s}}{s}\left(\frac{1}{s}-n\right)}$$

Is the way I approached the problem correct? I noticed that the last summation is a telescopic series which results in a $e^{\infty s}$ term which is making me doubtful. I've checked the procedure multiple times, and the vanishing property seems to hold too, but still, I'm not sure what to make of the telescopic series term.

So, in short, I need an experienced look on my method and tell me if I went wrong somewhere. Thanks in advance.

Answer 1

You almost have it correct, but it appears that you are trying to take the bilateral Laplace Transform of $\{x\}$, which doesn't exist.

If you take the unilateral Laplace Transform, then the summation extends form $n=0$ to $\infty$.

Also, note that for $f_n(x)=(x-\lfloor x\rfloor)(u_n(x)-u_{n+1}(x))=(x-n)(u_n(x)-u_{n+1}(x))$, we have

$$\begin{align} \sum_{n=0}^\infty\mathscr{L}\{ f_n\}(s)&=\sum_{n=0}^\infty \left(e^{-ns}\mathscr{L}\{x\}(s)-e^{-(n+1)s}\mathscr{L}\{x+1\}(s)\right)\\\\ &=\sum_{n=0}^\infty\left(e^{-ns}\mathscr{L}\{x\}(s)-e^{-(n+1)s}\mathscr{L}\{x+1\}(s) \right)\\\\ &=\sum_{n=0}^\infty\left(\frac{e^{-ns}-e^{-(n+1)s}}{s^2}-\frac{e^{-(n+1)s}}{s}\right) \end{align}$$

which is not quite what you have.

---

Here is how I would suggest to proceed. Let $f(x)=\{x\}$. Then, the Laplace Transform of $f$ is given by

$$\begin{align} \mathscr{L}\{f\}(s)&=\int_0^\infty \{x\}e^{-sx}\,dx\\\\ &=\int_0^\infty \left(x-\lfloor x\rfloor\right)e^{-sx}\,dx\\\\ &=\frac1{s^2}-\int_0^\infty \lfloor x\rfloor e^{-sx}\,dx\\\\ &=\frac1{s^2}-\sum_{n=0}^\infty \int_{n}^{n+1}\lfloor x\rfloor e^{-sx}\,dx\\\\ &=\frac1{s^2}-\sum_{n=0}^\infty ne^{-ns}\int_{0}^{1} e^{-sx}\,dx\\\\ &=\frac1{s^2}-\left(\frac{1-e^{-s}}{s}\right)\sum_{n=0}^\infty ne^{-ns}\\\\ &=\frac1{s^2}-\left(\frac{1-e^{-s}}s\right)\frac{e^{-s}}{(1-e^{-s})^2}\\\\ &=\frac1{s^2}-\frac{e^{-s}}{s(1-e^{-s})}\\\\ &=\frac1{s^2}-\frac{e^{-s/2}}{s(e^{s/2}-e^{-s/2})}\\\\ &=\frac1{s^2}+\frac1{2s}-\frac{\coth(s/2)}{2s} \end{align}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ There is a serie representation of $\ds{\braces{x}}$. Namely, $$ \braces{x} = {1 \over 2} - {1 \over \pi}\sum_{k = 1}^{\infty} {\sin\pars{2\pi kx} \over k} $$ Then, \begin{align} \int_{0}^{\infty}\braces{x}\expo{-xs}\,\dd x & = {1 \over 2s} - {1 \over \pi}\sum_{k = 1}^{\infty}{1 \over k}\ \overbrace{\int_{0}^{\infty}\sin\pars{2\pi k x}\expo{-sx}\,\dd x} ^{\ds{2\pi k \over 4\pi^{2}k^{2} + s^{2}}} \\[5mm] & = {1 \over 2s} - {1 \over 2\pi^{2}}\sum_{k = 1}^{\infty} {1 \over k^{2} + \bracks{s/\pars{2\pi}}^{2}} \\[5mm] & = {1 \over 2s} - {1 \over 2\pi^{2}}\sum_{k = 0}^{\infty} {1 \over \bracks{k + 1 - \ic s/\pars{2\pi}}\bracks{k + 1 + \ic s/\pars{2\pi}}} \\[5mm] & = {1 \over 2s} - {1 \over 2\pi^{2}} {\Psi\pars{1 + \ic s/\bracks{2\pi}} - \Psi\pars{1 - \ic s/\bracks{2\pi}} \over \ic s/\pi} \\[5mm] & = {1 \over 2s} - {1 \over \pi s}\,\Im\,\Psi\pars{1 + \ic\,{s \over 2\pi}} \\[5mm] & = {1 \over 2s} - {1 \over \pi s} \bracks{-\,{1 \over 2}\,{1 \over s/\pars{2\pi}} + {1 \over 2}\,\pi\coth\pars{\pi\,{s \over 2\pi}}} \\[5mm] & = \bbox[15px,#ffd,border:1px solid navy]{{1 \over 2s} + {1 \over s^{2}} - {1 \over 2s}\coth\pars{s \over 2}} \end{align}