Problem 01 | Second Shifting Property of Laplace Transform

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-as} F(s)$
 

$F(s) = \mathcal{L} (t^2)$   and   $a = 1$

$F(s) = \dfrac{2}{s^3}$
 

Thus,
$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-s} \left( \dfrac{2}{s^3} \right)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \dfrac{2e^{-s}}{s^3}$           answer

$g(t) = \begin{cases}{f(t - 1)^2 & t > 1 \\ 0 & t 1}\end{cases}$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-st} g(t) \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^1 e^{-st} (0) \, dt + \int_1^\infty e^{-st} (t - 1)^2 \, dt$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_1^\infty e^{-st} (t - 1)^2 \, dt$
 

Let
$z = t - 1$

$t = z + 1$

$dt = dz$
 

when   $t = 1, \, z = 0$

when   $t = \infty, \, z = \infty$
 

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-s(z + 1)} z^2 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz - s} z^2 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \int_0^\infty e^{-sz} e^{-s} z^2 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-s} \int_0^\infty e^{-sz} z^2 \, dz$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-s} \mathcal{L} (z^2)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = e^{-s} \left( \dfrac{2}{s^3} \right)$

$\displaystyle \mathcal{L} \left\{ g(t) \right\} = \dfrac{2e^{-s}}{s^3}$       okay