Question

In Problems 3 - 24 fill in the blanks or answer true or false.

$\mathbf{21}\mathbf{.}\mathit{L}\left\{e^{-5t}\right\}\mathit{e}\mathit{x}\mathit{i}\mathit{s}\mathit{t}\mathit{s}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{s}\mathbf{>}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

$s>-5$

Step-by-step solution

The Laplace equations

The formula for Laplace transform

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)\text{d}t$

We have that$f\left(t\right)=e^{-5t}$, so use the above formula.

$L\left\{e^{-5t}\right\}$$={\int }_0^{\infty }{e}^{-st}{e}^{-5t}dt$

$={\int }_0^{\infty }{e}^{-(s+5)t}dt$

$=-{\left.\frac{e^{-(s+5)t}}{s+5}\right|}_0^{\infty }$

$=\frac{1}{s+5}$

In order to have$\underset{t\to \infty }{Lim}{e}^{-(s+5)t}=0$we need to require that . Thus, the last result exists for $s>-5$.

Final proof

$s>-5$