Chapter 8, Problem 1RE

To determine

The Laplace transform for the function $f\left(t\right)=1-t$ by using the definition of the Laplace transform.

Answer to Problem 1RE

The Laplace transform for the function $f\left(t\right)=1-t$ is $\underset{\_}{\frac{s-1}{s^2}}$.

Explanation of Solution

Procedure used:

If $f\left(t\right)$ is a function defined in the interval $\left(0,\infty \right)$. Then, the Laplace transform of the function $f\left(t\right)$ is:

$F\left(s\right)=ℒ\left\{f\left(t\right)\right\}={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

The formula for evaluating improper integrals:

${\displaystyle \underset{0}{\overset{\infty }{\int }}f\left(t\right)dt=\underset{M\to \infty }{\lim }}{\displaystyle \underset{0}{\overset{M}{\int }}f\left(t\right)dt}$

Calculation:

Consider the function,

$f\left(t\right)=1-t$

Apply the Laplace transform.

$\begin{array}{c}F\left(s\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}\left(1-t\right)dt}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}dt-{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}tdt}}\end{array}$

Apply the formula for evaluation improper integrals:

$\begin{array}{c}F\left(s\right)=\underset{M\to \infty }{\lim }{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt-\underset{M\to \infty }{\lim }{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}tdt}}\\ =\underset{M\to \infty }{\lim }{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt}-\left\{\underset{M\to \infty }{\lim }{\left[-\frac{1}{s}t{e}^{-st}\right]}_0^M+\underset{M\to \infty }{\lim }\frac{1}{s}{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt}\right\}\\ =\underset{M\to \infty }{\lim }{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt}-\underset{M\to \infty }{\lim }{\left[-\frac{1}{s}t{e}^{-st}\right]}_0^M-\underset{M\to \infty }{\lim }\frac{1}{s}{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt}\\ =\left(1-\frac{1}{s}\right){\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt}-\underset{M\to \infty }{\lim }{\left[-\frac{1}{s}t{e}^{-st}\right]}_0^M\end{array}$

Simplify the above equation.

$\begin{array}{c}\left(1-\frac{1}{s}\right)\underset{M\to \infty }{\lim }{\displaystyle \underset{0}{\overset{M}{\int }}{e}^{-st}dt}-\underset{M\to \infty }{\lim }{\left[-\frac{1}{s}t{e}^{-st}\right]}_0^M=\left(1-\frac{1}{s}\right)\underset{M\to \infty }{\lim }{\left[-\frac{1}{s}{e}^{-st}\right]}_0^M-0\\ =\left(1-\frac{1}{s}\right)\underset{M\to \infty }{\lim }\left[-\frac{1}{s}{e}^{-sM}+\frac{1}{s}\right]\end{array}$

If $s>0$, then $\underset{M\to \infty }{\lim }{e}^{-sM}=0$.

$\begin{array}{c}\left(1-\frac{1}{s}\right)\underset{M\to \infty }{\lim }\left[-\frac{1}{s}{e}^{-sM}+\frac{1}{s}\right]=\left(1-\frac{1}{s}\right)\frac{1}{s}\\ =\frac{s-1}{s^2}\end{array}$

Therefore, the Laplace transform for the function $f\left(t\right)=1-t$ is $\underset{\_}{\frac{s-1}{s^2}}$.