Find Laplace Transform of the following function

Question

How do I find the Laplace transform for the function:

$f(t)=t, 0 \leq t \leq 1$ and $2-t, t \geq 1$

I tried looking up the process online, but it remains unclear to me.

Thanks in advance!

Answer 1

You must break the integral up on the discontinuities to take the Laplace Transform of a discontinuous function: $$\begin{align} \mathcal{Lf}(s) &= \int_0^\infty f(t)e^{-st}dt \\ &=\int_0^1\underbrace{t}_{\text{f(t) for t < 1}}e^{-st}dt + \int_1^\infty\underbrace{(2-t)}_{\text{f(t) for t > 1}}e^{-st}\;dt\\ &=\cdots \end{align}$$

Please let me know if this needs more explaining.

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More explanation:

The function $f$ is defined by: $$f(t) = \begin{cases}\color{red}{t} & \text{if }0\le t \le1\\ \color{blue}{2-t} & \text{if } 1\le t\end{cases}$$

To compute the Laplace transform, we compute the integral: $$\int_0^\infty f(t)e^{-st}dt$$

Based on the property of integrals that says $\int_a^b f(x)\;dx + \int_b^cf(x)\;dx = \int_a^c f(x)\;dx$, we can write: $$\int_0^\infty f(t)e^{-st}dt = \int_0^1f(t)e^{-st}\;dt + \int_1^\infty f(t)e^{-st}\;dt$$

We now replace $f(t)$ based on the definition: $$\int_0^1\color{red}{f(t)}e^{-st}\;dt + \int_1^\infty \color{blue}{f(t)}e^{-st}\;dt = \int_0^1\color{red}{t}e^{-st}\;dt + \int_1^\infty \color{blue}{(2-t)}e^{-st}\;dt$$

The rest is simple integration. (The first you do by-parts, the second you distribute, then a $u$-sub and by-parts again.)