## Calculus: Early Transcendentals 8th Edition

The Laplace transforms of the functions f are: (a) $$F(s)=\frac{1}{s}$$ and the domain of $F$ is ${s: s\gt 0}$. (b) $$F(s)=\frac{1}{s-1}$$ and the domain of $F$ is ${s: s\gt 1}$. (c) $$F(s)=\frac{1}{s^{2}}$$ and the domain of $F$ is ${s: s\gt 0}$
To find the Laplace transforms of the following functions: (a) $f(t)=1$ the Laplace transform of $f$ is the function $F$ defend by $$F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$$ so $$\begin{split} F(s)&=\int_{0}^{\infty} f(t) e^{-s t} d t \\ &=\int_{0}^{\infty} e^{-s t} d t \\ &=\lim _{n \rightarrow \infty}\left[-\frac{e^{-s t}}{s}\right]_{0}^{n} \\ &=\lim _{n \rightarrow \infty}\left(\frac{e^{-s n}}{-s}+\frac{1}{s}\right) \quad\left[\begin{array}{c}{ \text {only if } s \gt 0 }\end{array}\right] \\ &=0+\frac{1}{s}\\ &=\frac{1}{s} \end{split}$$ Therefore $F(s)=\frac{1}{s}$ with domain ${s: s\gt 0}$ (b) $f(t)=e^{t}$ the Laplace transform of $f$ is the function $F$ defend by $$F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$$ so $$\begin{split} F(s)&=\int_{0}^{\infty} f(t) e^{-s t} d t \\ &=\int_{0}^{\infty} e^{ t} e^{-s t} d t \\ &=\int_{0}^{\infty} e^{-(s-1) t} d t \\ &=\lim _{n \rightarrow \infty}\left[\frac{e^{-(s-1)t}}{1-s}\right]_{0}^{n} \\ &=\lim _{n \rightarrow \infty}\left(\frac{e^{-(s-1)n}}{1-s}-\frac{1}{1-s}\right) \quad\left[\begin{array}{c}{ \text {only if } (s-1) \gt 0 }\end{array}\right] \\ &=0+\frac{1}{s-1}\\ &=\frac{1}{s-1} \end{split}$$ Therefore $F(s)=\frac{1}{s-1}$ with domain ${s: s\gt 1}$ (c) $f(t)=t$ the Laplace transform of $f$ is the function $F$ defend by $$F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$$ so $$\begin{split} F(s)&=\int_{0}^{\infty} f(t) e^{-s t} d t \\ &=\int_{0}^{\infty} t e^{-s t} d t \\ &=\lim _{n \rightarrow \infty} \int_{0}^{n} t e^{-s t} d t \\ & \quad\quad\quad\left[\text { use integration by parts with } \right] \\ &\quad\quad\quad \left[\begin{array}{c}{u=t, \quad\quad dv= e^{-st}dt } \\ {d u= dt, \quad\quad v= \frac{1}{-s}e^{-st} }\end{array}\right] , \text { then }\\ &=\lim _{n \rightarrow \infty}\left(\frac{-te^{-st}}{s}]_0^{n}+\frac{1}{s}\int_{0}^{n} e^{-s t} d t \right) \\ &=\lim _{n \rightarrow \infty}\left[\frac{-te^{-st}}{s}-\frac{1}{s^{2}} e^{-s t} \right] _{0}^{n} \\ &=\lim _{n \rightarrow \infty}\left[\frac{-ne^{-sn}}{s}+\frac{1}{s^{2}} e^{-s n} +0+\frac{1}{s^{2}}\right] \quad\left[\begin{array}{c}{ \text {only if } s \gt 0 }\end{array}\right] \\ \\ &=\frac{1}{s^{2}} \end{split}$$ Therefore $F(s)=\frac{1}{s^{2}}$ with domain ${s: s\gt 0}$