Discrepancy with the book's solution and mine of Laplace transform of a piecewise defined function

Question

Determine the Laplace transform of $f(t)$ below: $$ f(t)= \begin{cases} 0, & \text{if } t < 2 \\ (t-2)^2, & \text{if } t \geqslant 2 \end{cases} $$

So my answer is

$$ 2e^{-2s}/s^3 $$

But the actual answer [the answer in the book] is

$$ 2e^{-s}/s^3 $$

Why? What am i doing wrong? I raised the boundary to the negative e just as how its supposed to be done. Thanks

Answer 1

Given the Laplace transform as \begin{align} f(s) = \int_{0}^{\infty} e^{-st} f(t) \, dt \end{align} then for $$ f(t)= \begin{cases} 0, & \text{if } t < 2 \\ (t-2)^2, & \text{if } t \geqslant 2 \end{cases} $$ it is seen that \begin{align} f(s) &= \int_{0}^{2} e^{-st} \, (0) \, dt + \int_{2}^{\infty} e^{-st} (t-2)^{2} \, dt \\ &= 0 + \int_{0}^{\infty} e^{-s(u+2)} u^{2} \, du \\ &= \frac{2 e^{-2s}}{s^{3}}. \end{align}

If the answer is stated in a book then the most likely difference is due to a typo.

Answer 2

Why do you say the actual answer is $2e^{-s}/s^3$? With problems of this form, we can simplify the integration by recognizing the derivative. \begin{align} \int_2^{\infty}(t-2)^2e^{-st}dt &=\int_2^{\infty}t^2e^{-st}dt-4\int_2^{\infty}te^{-st}dt+4\int_2^{\infty}e^{-st}dt\\ &=\frac{\partial^2}{\partial s^2}\int_2^{\infty}e^{-st}dt+4\frac{\partial}{\partial s}\int_2^{\infty}e^{-st}dt+4\int_2^{\infty}e^{-st}dt\\ &=\frac{\partial^2}{\partial s^2}\frac{e^{-2s}}{s}+\frac{\partial}{\partial s}\frac{4e^{-2s}}{s}+\frac{4e^{-2s}}{s} \end{align} From this, we obtain the answer you arrived at not the books.

Answer 3

Given the function $f(t) = (t-2)^2 u(t-2)$, where $u$ is the unit step function, the Laplace Transform $F(s)$ of $f$ is given by

$$F(s)= \int_0^{\infty} (t-2)^2 u(t-2) e^{-st}dt$$

$$=\int_2^{\infty} (t-2)^2 e^{-st}dt$$ Now, substituting $u=t-2$, so that $du=dt$, and the limits of integration on $u$ go from $0$ to $\infty$, reveals

$$F(s)= \int_0^{\infty} u^2 e^{-s(u+2)}du=e^{-2s}\int_0^{\infty} u^2 e^{-st}dt$$

Now, we may write

$$F(s) = e^{-2s}\frac{d^2}{ds^2} \left(\int_0^{\infty} e^{-st}dt\right)$$ $$e^{-2s}\frac{d^2}{ds^2} \left(\frac{1}{s}\right)$$ $$=e^{-2s}\frac{2}{s^3}$$which recovers the desired result! Note that uniform convergence justified the interchange of differentiation and the improper integral.