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I need to find de Laplace transform of $$f(x)=\frac{|x-a|}{x-a}$$ for $a>0$. So, I proposed $f(x)$ such that $$f(x)= \left\{ \begin{matrix} -1, & \mbox{$0<x<a$} \\ 1, & \mbox{$x>a$} \end{matrix} \right. $$

then,

\begin{equation} \begin{split} \mathcal{L}\left\{ f(x) \right\} & = \int_{0}^{a} -e^{-sx} dx + \int_{a}^{\infty} e^{-sx}dx \\ & = \frac{1}{s}e^{-sx} \Biggr|_{0}^{a} + \lim_{b \to \infty} \left[ -\frac{1}{s}e^{-sx} \Biggr|_{a}^{b} \right] \\ & = \frac{1}{s}e^{-sa} - \frac{1}{s} + \lim_{b \to \infty} \left[ -\frac{1}{s}e^{-sb} + \frac{1}{s}e^{-sa}\right] \\ & = \frac{2}{s}e^{-sa} - \frac{1}{s}+ \lim_{b \to \infty} \left[ -\frac{1}{s}e^{-sb} \right] \\ & = \frac{2}{s}e^{-sa} - \frac{1}{s} \end{split} \end{equation}

with $s>0$. Is this right?

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  • $\begingroup$ It looks fine to me. $\endgroup$
    – DonAntonio
    Jul 9, 2020 at 4:41

1 Answer 1

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I will give you another way of looking at this function, and you can cross-check yourself whether it's right. The function $\frac{|x-a|}{x-a}$ is essentially the sum of two step functions at $x=a$

Denoting a step function (or heavyside function) as $u_a(x)$, we have

$$f(x) = u_a(x) + (u_{a}(x)-1) = 2u_a(x) - 1$$

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