Question 4.1.3: Using Definition 4.1.1 Evaluate (a) L{e^-3t} (b) L{e^6t}...
Using Definition 4.1.1
Evaluate
(a) \mathscr{L}\left\{e^{-3 t}\right\} (b) \mathscr{L}\left\{e^{6 t}\right\}
In each case we use Definition 4.1.1.
(a) \mathscr{L}\left\{e^{-3 t}\right\}=\int_{0}^{\infty} e^{-3 t} e^{-s t} d t=\int_{0}^{\infty} e^{-(s+3) t} d t
\begin{aligned} & =\left.\frac{-e^{-(s+3) t}}{s+3}\right|_{0} ^{\infty} \\ & =\frac{1}{s+3}. \end{aligned}
The last result is valid for s>-3 because in order to have \lim _{t \rightarrow \infty} e^{-(s+3) t} \quad 0 we must require that s+3>0 or s>-3.
(b) \mathscr{L}\left\{e^{6 t}\right\} \quad \int_{0}^{\infty} e^{6 t} e^{-s t} d t \quad \int_{0}^{\infty} e^{-(s-6) t} d t
\begin{aligned} & \left.\frac{-e^{-(s-6) t}}{s-6}\right|_{0} ^{\infty} \\ & \frac{1}{s-6} . \end{aligned}
In contrast to part (a), this result is valid for s>6 because \lim _{t \rightarrow \infty} e^{-(s-6) t} \quad 0 demands s-6>0 or s>6.