# Laplace Transforms

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.2 – Definition of the Laplace Transform

## Expected Educational Results

• Objective 19–1: I understand the Laplace transform as a map of a function onto a specific class of functions.
• Objective 19–2: I can use the definition of Laplace transform to find the Laplace Transform of a given any function.

## Laplace Transforms

Definition: Laplace Transforms

Let $f(t)$ be a function on $[0,\infty)$. The Laplace Transform of $f(t)$ is the function $F(s)$ defined as

$\displaystyle F(s)=\mathcal{L}\left\{f(t)\right\}(s)\equiv\int_0^{\infty}{e^{-st}f(t)\,dt}$

Note: You will need to know how to evaluate an improper integral; l'Hopital's Rule may be required.

Example 01: Find the Laplace transform of $f(t)={\color{gsured}t^2}$.

Solution:

$F(s)=\mathcal{L}\left\{t^2\right\}(s)$

By the definition of the Laplace Transform

$F(s)=\int_0^{\infty}{e^{-st}t^2}\,dt$

Use the definition of improper integral:

$\displaystyle\Rightarrow F(s)=\lim_{a\rightarrow\infty}{\int_0^a{e^{-st}t^2\,dt}}$

Integrate by parts:

$\displaystyle\Rightarrow F(s)=\lim_{a\rightarrow\infty}{\left.\left(-\frac{t^2e^{-st}}{s}-\frac{2te^{-st}}{s^2}-\frac{2e^{-st}}{s^3}\right)\right|_{t=0}^{t=a}}$

Rewrite rational expressions:

$\displaystyle\Rightarrow F(s)=\lim_{a\rightarrow\infty}{\left.\left(-\frac{t^2}{se^{st}}-\frac{2t}{s^2e^{st}}-\frac{2}{s^3e^{st}}\right)\right|_0^a}$

Use FTC-II:

$\displaystyle\Rightarrow F(s)=\lim_{a\rightarrow\infty}{\left[\left(-\frac{a^2}{se^{as}}-\frac{2a}{s^2e^{as}}-\frac{2}{s^3e^{as}}\right)-\left(-0-0-\frac{2}{s^3}\right)\right]}$

Use Limit Laws:

$\displaystyle\Rightarrow F(s)=-\frac{1}{s}\lim_{a\rightarrow\infty}{\left(\frac{a^2}{e^{as}}\right)}-\frac{1}{s^2}\lim_{a\rightarrow\infty}{\left(\frac{2a}{e^{as}}\right)}-\frac{1}{s^3}\lim_{a\rightarrow\infty}{\left(\frac{2}{e^{as}}\right)}-\left(-\frac{2}{s^3}\right)$

Use l’Hopital’s Rule for the first two limits:

$\displaystyle\Rightarrow F(s)=-\frac{1}{s}\lim_{a\rightarrow\infty}{\left(\frac{2}{s^2e^{as}}\right)}-\frac{1}{s^2}\lim_{a\rightarrow\infty}{\left(\frac{2}{se^{as}}\right)}-\frac{1}{s^3}\lim_{a\rightarrow\infty}{\left(\frac{2}{e^{as}}\right)}-\left(-\frac{2}{s^3}\right)$

Evaluate limits:

$\displaystyle\Rightarrow F(s)=-0-0-0+\frac{2}{s^3}=\frac{2}{s^3}$

So, $\displaystyle \mathcal{L}\left\{{\color{gsured}t^2}\right\}(s) = \frac{2}{s^3}$.

NOTE: The more that you can recognize patterns of Laplace transforms, the more helpful Laplace transforms will be.