Calculate the laplace transform...

Question

Calculate the laplace transform of $$t^2u(t-2)$$

I don't know how to manipulate t^2 in order for it to meet the form of the product between a function and a heaviside function. Number (27) on http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

Answer 1

Use #28 from your table instead of #27. Then $c=2$ and $$g(t)=t^2\implies g(t+c)=g(t+2)=(t+2)^2=t^2+4t+4$$ so $$\mathscr{L}\{g(t+c)\}=\mathscr{L}\{t^2+4t+4\}={2\over s^3}+{4\over s^2}+{4\over s}.$$

Thus, from #28 in your table, $$ \mathscr{L}\{t^2u(t-2)\}=e^{-2s}\mathscr{L}\{g(t+2)\}=e^{-2s}\left({2\over s^3}+{4\over s^2}+{4\over s}\right).$$

Answer 2

The Heaviside function effective changes the lower limit of integration so the LT is

$$\int_2^{\infty} dt \, t^2 \, e^{-s t} = \frac{d^2}{d s^2} \int_2^{\infty} dt \, e^{-s t} = \frac{d^2}{d s^2} \frac{e^{-2 s}}{s} $$

Taking the derivative, the LT takes the form

$$ -\frac{d}{ds} \left [ \left ( \frac{2}{s} + \frac1{s^2} \right ) e^{-2 s} \right ] = \left ( \frac{4}{s} + \frac{4}{s^2} + \frac{2}{s^3} \right ) e^{-2 s} $$