$\displaystyle F=\int_0^\infty e^{-st}t^2 dt$

My unsuccessful attempt:

Integration by parts: $\displaystyle \int fg'=fg-\int f'g$

Using "liate":

$\displaystyle f=t^2 f'=2t$

$\displaystyle g=- \frac{1}{s} e^{-st} g'=e^{-st}$

$\displaystyle F=- \frac{1}{s} e^{-st} \cdot t^2 |^\infty_0 - \int _0^\infty 2t\cdot - \frac{1}{s} e^{-st} dt$

$\displaystyle F= \lim_{A \to \infty} (-\frac{1}{s} e^{-sA} \cdot A^2) - \int _0^\infty 2t\cdot - \frac{1}{s} e^{-st} dt$

Using L'Hopital on the limit I get that the limit is equal to 0. So...

$\displaystyle F= \frac{2}{s} \int _0^\infty t\cdot e^{-st} dt$

I use integration by parts again and I get...

$\displaystyle F= \frac{2}{s} [-\frac{1}{s}e^{-st}\cdot t|^\infty _0 -\int _0^\infty e^{-st}dt ]$

$\displaystyle F= \frac{2}{s} [\lim_{A\to\infty} (0) + \frac{1}{s} \lim_{A\to\infty} (-1)]$

$\displaystyle F= \frac{-2}{s^2}$

The answer is supposed to be $\displaystyle F=\frac{2!}{s^3}$

Where did I go wrong?