Check my answer - simple laplace transform of piecewise continuous function.

Question

I'd just like to check that I got the idea right, first exercise im doing in laplace transforms and am a bit clueless.

We are given $f(t)=0$ if $0<t<2$ and $f(t)=t$ if $t>2$. We are asked to find the laplace transform of $f(t)$.

My answer:

$F(s)=L(f(t))=\int_{0}^{\infty}e^{-st}f(t)dt=\int_{0}^{2}0*e^{-st}dt+\int_2^\infty te^{-st}dt=\int_2^\infty te^{-st}dt$

Using integration by parts I got

$$\int_2^\infty te^{-st}dt=-\frac{e^{-st}(st+1)}{s^2}|_{t=2}^{t=\infty}=\lim_{a \to \infty} -\frac{e^{-sa}(sa+1)}{s^2}+\frac{e^{-2s}(2s+1)}{s^2}=\frac{e^{-2s}(2s+1)}{s^2}$$

So my final answer is $$F(s)=L(f(t))=\frac{e^{-2s}(2s+1)}{s^2}$$

Is this result correct?

Answer 1

Yes, your result is correct.

I would also recommend solving it using the Heaviside Unit Step Function.

We can write your piecewise function (see my response Using laplace transforms to solve a piecewise defined function initial value problem) as:

$$f(t) = 0 - 0 u(t-2) + t u(t-2) = t u(t-2)$$

The Laplace Transform (see item 27) is:

$$\mathscr{L} (t~ u(t-2)) = \dfrac{e^{-2 s} (2 s+1)}{s^2}$$