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