# Finding the Laplace Transform of $\frac{|x-a|}{x-a}$

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 $$f(x)= \left\{ \begin{matrix} -1, & \mbox{0a} \end{matrix} \right.$$

then,

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

with $$s>0$$. Is this right?

• It looks fine to me. Jul 9, 2020 at 4:41

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