Show that $\mathscr{L}\{f\}(s)=\frac{1}{s+3} \quad \text{when}f(x)=e^{-3x}$

Question

Problem

Functions $f:[0,\infty[\rightarrow \mathbb{R}$ laplace transform, value at point $s > 0$ is defined with formula:

$$ \mathscr{L}\{f\}(s)=\int_{0}^{\infty}e^{-st}f(t)dt $$

Show that $$\mathscr{L}\{f\}(s)=\frac{1}{s+3} \quad \text{when }f(x)=e^{-3x}$$

Problem is I don't even understand what I am supposed to do here? If someone could give hint in the right direction that how I should begin solving this that would be highly appreciated.

Answer 1

$$\begin{array}{rcl} \mathcal{L}\{f\}(s) &=& \displaystyle \int_{0}^{\infty}e^{-st}f(t) \ \mathrm dt \\ &=& \displaystyle \int_{0}^{\infty}e^{-st}e^{-3t} \ \mathrm dt \\ &=& \displaystyle \int_{0}^{\infty}e^{-(s+3)t} \ \mathrm dt \\ &=& \displaystyle \left[\frac{-1}{s+3} e^{-(s+3)t}\right]_{0}^{\infty} \\ &=& \displaystyle \left[0 - \frac{-1}{s+3}\right] \\ &=& \displaystyle \frac{1}{s+3} \\ \end{array}$$

Answer 2

Well, solving a more general problem:

$$\text{F}_{\space\text{n}}\left(\text{s}\right):=\mathscr{L}_x\left[\exp\left(\text{n}\cdot x\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\exp\left(\text{n}\cdot x\right)\cdot e^{-\text{s} x}\space\text{d}x\tag1$$

Substitute $\text{u}:=\left(\text{n}-\text{s}\right)\cdot x$:

$$\text{F}_{\space\text{n}}\left(\text{s}\right)=\frac{1}{\text{n}-\text{s}}\cdot\lim_{\text{p}\to\infty}\int_0^{\left(\text{n}-\text{s}\right)\cdot\text{p}}\exp\left(\text{u}\right)\space\text{d}\text{u}=$$ $$\frac{1}{\text{n}-\text{s}}\cdot\lim_{\text{p}\to\infty}\left[\exp\left(\text{u}\right)\right]_0^{\left(\text{n}-\text{s}\right)\cdot\text{p}}=$$ $$\frac{1}{\text{n}-\text{s}}\cdot\lim_{\text{p}\to\infty}\left(\exp\left(\left(\text{n}-\text{s}\right)\cdot\text{p}\right)-\exp\left(0\right)\right)=$$ $$\frac{1}{\text{n}-\text{s}}\cdot\lim_{\text{p}\to\infty}\left(\exp\left(\left(\text{n}-\text{s}\right)\cdot\text{p}\right)-1\right)\tag2$$

Now, what happends to the limit when:

$$\Re\left(\text{n}\right)<\Re\left(\text{s}\right)\tag3$$

Answer 3

$$\mathscr{L}\{e^{at}\}(s)=\int\limits_{t=0}^{\infty}e^{at}e^{-st} \mathrm{d}t$$ $$\mathscr{L}\{e^{at}\}(s)=\int\limits_{t=0}^{\infty}e^{at-st} \mathrm{d}t$$ $$\mathscr{L}\{e^{at}\}(s)=\int\limits_{t=0}^{\infty}e^{(a-s)t} \mathrm{d}t$$ $$\mathscr{L}\{e^{at}\}(s)=\left[\frac{1}{a-s}e^{(a-s)t}\right]_{t=0}^{t=\infty}$$ $$\mathscr{L}\{e^{at}\}(s)=\frac{1}{a-s}\left[e^{(a-s)t}\right]_{t=0}^{t=\infty}$$ $$\mathscr{L}\{e^{at}\}(s)=\frac{1}{a-s}\left[\lim\limits_{t \to \infty} e^{(a-s)t}-e^{(a-s)0}\right]$$ $$\mathscr{L}\{e^{at}\}(s)=\frac{1}{a-s}\left[\lim\limits_{t \to \infty} e^{(a-s)t}-1\right]$$ Now we need to suppose that $a-s<0$ for the limit to exist (and to be $0$): $$\mathscr{L}\{e^{at}\}(s)=\frac{1}{a-s}\left[0-1\right]$$ $$\mathscr{L}\{e^{at}\}(s)=-\frac{1}{a-s}$$ $$\mathscr{L}\{e^{at}\}(s)=\frac{1}{s-a}$$