Answer:
ℒ{f(t)}(s) = [tex]\frac{1-e^{-s}}{s^2}[/tex] for s>0
Step-by-step explanation:
The Laplace transform of this specific function is
[tex] \bf \mathcal{L}[f(t)](s)=\int_{0}^{\infty}f(t)e^{-st}dt=\int_{0}^{1}te^{-st}dt+\int_{1}^{\infty}e^{-st}dt[/tex]
We can compute the first integral by parts.
By making
[tex] \bf u=t,\;dv=e^{-st}dt\Rightarrow du=dt,\;v=\int e^{-st}dt=\frac{e^{-st}}{-s}[/tex]
and
[tex] \bf \int_{0}^{1}te^{-st}dt=\left [\frac{te^{-st}}{-s} \right ]_{t=0}^{t=1}+\frac{1}{s}\int_{0}^{1}e^{-st}dt=\frac{e^{-s}}{-s}+\frac{1}{s}\left [\frac{e^{-st}}{-s} \right ]_{t=0}^{t=1}=\\\\=\frac{e^{-s}}{-s}+\frac{1}{s}\left (\frac{e^{-s}}{-s}+\frac{1}{s} \right )=\frac{1}{s^2}-\frac{e^{-s}}{s^2}-\frac{e^{-s}}{s}[/tex]
the second integral can be computed as
[tex] \bf \lim_{t \to\infty}\int_{1}^{t}e^{-sx}dx=\lim_{t \to\infty}\left [ \frac{e^{-sx}}{-s} \right ]_{x=1}^{x=t}=\lim_{t \to\infty}\left ( \frac{e^{-st}}{-s}+\frac{e^{-s}}{s} \right )[/tex]
but when s > 0
[tex] \bf \lim_{t\to\infty}e^{-ts}=0[/tex]
hence,
[tex] \bf \lim_{t \to\infty}\int_{1}^{t}e^{-sx}dx=\frac{e^{-s}}{s}[/tex]
and our Laplace transform is
[tex] \bf \frac{1}{s^2}-\frac{e^{-s}}{s^2}-\frac{e^{-s}}{s}+\frac{e^{-s}}{s}=\frac{1}{s^2}-\frac{e^{-s}}{s^2}=\\\\=\boxed{\frac{1-e^{-s}}{s^2}}[/tex]
