Laplace transform of $f = t^2 -8t + 23$ for $t\geq 4$ and $0$ elsewhere

Question

$f$ is defined piecewise like so: \begin{cases} 0 & t < 4 \\ t^2 -8t + 23 & t \geq 4 \\ \end{cases}

So using the stepwise function notation, $u_4(t) = u(t-4)$ (as in, the stepwise function shifted to the right by $4$), I find that $f = u_4(t) \cdot (t^2 -8t + 23) $

Now I want to use the formula for Laplace transforms of functions multiplied by stepwise functions:

$$\mathcal{L} (u_a(t) \cdot f(t)) = e^{-as} \mathcal L (f(t+a))$$

First I find $\mathcal L(f(t+a)):$

$\mathcal L f(t+5)=\mathcal L ((t+5)^2-8(t+5)+23) = \mathcal L(t^2+2t-8) = \frac2{s^3} + \frac2{s^2} - \frac8s$

So the final answer should be $e^{-4s} ( \frac2{s^3} + \frac2{s^2} - \frac8s)$...but apparently this is wrong. Can anyone see why?

Answer 1

Another way to do, only with the formal definition of the Laplace transform. Then
$\mathcal{L}\lbrace f\rbrace=\displaystyle\int_{0}^{4} 0dt+\int_{4}^{\infty}e^{-st}(t^{2}-8t+23)dt$.

Answer 2

Your answer is incorrect because you wrote $f(t+a)=f(t+5)$ when the step function had a right shift of $4$. Applying

$$\mathcal{L} \left(u(t-a)f(t)\right) = e^{-as} \mathcal {L} [f(t+a)],$$

we find

$$\mathcal{L}[f(t+4)]=\mathcal L \left[(t+4)^2-8(t+4)+23\right] = \mathcal {L}[t^2+7] = \frac2{s^3} + \frac{7}{s}.$$

This gives the final answer of

$$\mathcal{L} \left(u(t-4)f(t)\right)=e^{-4s}\left(\frac{7}{s}+\frac{2}{s^3}\right).$$

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I took a slightly different approach. We are given:

$$f(t)=\begin{cases} 0, & t < 4, \\ t^2 -8t + 23, & t \geq 4, \\ \end{cases}=u(t-4)(t^2-8t+23).$$

Lets first complete the square

$$t^2-8t+23=(t-4)^2+7.$$

Then the problem becomes

$$f(t)=u(t-4)(t-4)^2+7u(t-4).$$

Applying

$$\mathcal{L}(t^2)=\frac{2}{s^{3}},\quad \mathcal{L}\big[u(t-c)\big]=\frac{e^{-cs}}{s},\quad \mathcal{L}\big[u(t-c)g(t-c)\big]=e^{-cs}\mathcal{L}[g(t)],$$

we find

$$\mathcal{L}[f(t)]=\mathcal{L}\left[u(t-4)(t-4)^2\right]+\mathcal{L}[7u(t-4)]=\frac{2e^{-4s}}{s^3}+\frac{7e^{-4s}}{s}=e^{-4s}\left(\frac{7}{s}+\frac{2}{s^3}\right).$$