Question

Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. Find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = t, 0 ≤ t < 1 1, t ≥ 1

Answer 1

Answer:

ℒ{f(t)}(s) = $\frac{1-e^{-s}}{s^2}$ for s>0

Step-by-step explanation:

The Laplace transform of this specific function is

$\bf \mathcal{L}[f(t)](s)=\int_{0}^{\infty}f(t)e^{-st}dt=\int_{0}^{1}te^{-st}dt+\int_{1}^{\infty}e^{-st}dt$

We can compute the first integral by parts.

By making

$\bf u=t,\;dv=e^{-st}dt\Rightarrow du=dt,\;v=\int e^{-st}dt=\frac{e^{-st}}{-s}$

and  

$\bf \int_{0}^{1}te^{-st}dt=\left [\frac{te^{-st}}{-s} \right ]_{t=0}^{t=1}+\frac{1}{s}\int_{0}^{1}e^{-st}dt=\frac{e^{-s}}{-s}+\frac{1}{s}\left [\frac{e^{-st}}{-s} \right ]_{t=0}^{t=1}=\\\\=\frac{e^{-s}}{-s}+\frac{1}{s}\left (\frac{e^{-s}}{-s}+\frac{1}{s} \right )=\frac{1}{s^2}-\frac{e^{-s}}{s^2}-\frac{e^{-s}}{s}$

the second integral can be computed as

$\bf \lim_{t \to\infty}\int_{1}^{t}e^{-sx}dx=\lim_{t \to\infty}\left [ \frac{e^{-sx}}{-s} \right ]_{x=1}^{x=t}=\lim_{t \to\infty}\left ( \frac{e^{-st}}{-s}+\frac{e^{-s}}{s} \right )$

but when s > 0

$\bf \lim_{t\to\infty}e^{-ts}=0$

hence,

$\bf \lim_{t \to\infty}\int_{1}^{t}e^{-sx}dx=\frac{e^{-s}}{s}$

and our Laplace transform is

$\bf \frac{1}{s^2}-\frac{e^{-s}}{s^2}-\frac{e^{-s}}{s}+\frac{e^{-s}}{s}=\frac{1}{s^2}-\frac{e^{-s}}{s^2}=\\\\=\boxed{\frac{1-e^{-s}}{s^2}}$