# How can the Laplace transform be used to solve piecewise functions?

For example, suppose we have the following two problems that we'd like to find the Laplace transform of:

1. $f(t) = \begin{cases} 1, & t \lt 2 \\ 0, & t \geq 2 \end{cases}$

2. $f(t) = \begin{cases} 1-0.5t, & t \lt 2 \\ 5, & t \geq 2 \end{cases}$

Is there a general method to solve these sorts of problems? How can I approach them and make solving them more intuitive?

Thanks for the help!

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The definition of the Laplace transform is given as

$F(s)= \mathcal{L} \left\{f\right\}(s) =\int_{0}^{\infty} \mathrm{e}^{-st} f(t)\,\mathrm{d}t, \qquad s\in\mathbb{C}.$

So if we take example 1), we get

\begin{align*}F(s) = \mathcal{L} \left\{f\right\}(s) =\int_{0}^{\infty} \mathrm{e}^{-st} f(t)\,\mathrm{d}t &= \int_{0}^{2} \mathrm{e}^{-st} f(t)\,\mathrm{d}t+ \int_{2}^{\infty} \mathrm{e}^{-st} f(t)\,\mathrm{d}t \\ &= \int_{0}^{2} \mathrm{e}^{-st} \cdot 1\,\mathrm{d}t+ \int_{2}^{\infty} \mathrm{e}^{-st} \cdot 0\,\mathrm{d}t \\ &=\int_{0}^{2} \mathrm{e}^{-st} \,\mathrm{d}t \\ &= \Bigl[-\frac{1}{s} e^{-st} \Bigr]_0^ 2 = -\frac{1}{s} e^{-2s} +\frac{1}{s} \end{align*}

So the general approach is to split the integral at that points, at which $f$ changes.

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Here is the first one

$$\int_{0}^{\infty}f(t)e^{-st}dt = \int_{0}^{2}1.e^{-st}dt+ \int_{2}^{\infty}0.e^{-st}dt$$

You can do the other function with the same technique.

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