TITLE: Discrepancy with the book's solution and mine of Laplace transform of a piecewise defined function
QUESTION [5 upvotes]: 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

REPLY [2 votes]: 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.