Answer:
[tex]F(s)=\frac{1}{s}[/tex]
Step-by-step explanation:
Let [tex]f(t)[/tex] be defined for [tex]t\ge0[/tex].
The Laplace transform of [tex]f(t)[/tex] is an integral transform given by the Laplace integral:
[tex]\ell(f(t))=F(s)=\int\limits^{\infty}_0 {e^{-st}}\cdot f(t) \, dt[/tex], provided that this improper integral exists.
If we let [tex]f(t)=1[/tex], then
[tex]\ell(f(t))=F(s)=\int\limits^{\infty}_0 {e^{-st}}\cdot1 \, dt[/tex]
[tex]\ell(f(t))=F(s)=\lim_{a\to \infty}\int\limits^{a}_0 {e^{-st}} dt[/tex]
[tex]F(s)=\lim_{a\to \infty}-\frac{1}{s} {e^{-st}}|_{0}^{a}[/tex]
[tex]F(s)=-\frac{1}{s} {\lim_{a\to \infty}e^{-as}}--\frac{1}{s} {\lim_{a\to \infty}e^{-0\times s}}[/tex]
Recall that: [tex]\lim_{x\to \infty}e^{-x}=0[/tex]
[tex]\implies F(s)=0+\frac{1}{s}[/tex]
[tex]\implies F(s)=\frac{1}{s}[/tex]
