Laplace transform of f'', L(f'')

Question

I'm looking for the Laplace Transform solution to the following problem,

Here is my attempt at a solution, however I am confused about the integration of f(t)

$\mathscr{L}${f''} = $\int_0^\infty$f''(t)e$^{-st}$dt = $s^{2}$F(s) - $sf(0) - f'(0)$

$\hspace{10mm} u = e^{-st} \hspace{19mm} dv = f''(t)$

$\hspace{10mm} du = -se^{-st} \hspace{11mm} v=f'(t)$

$e^{-st} * f'(t) - s*\int f'(t)e^{-st} dt$

$\hspace{10mm} u = u^{-st} \hspace{19mm} dv = f'(t)$

$\hspace{10mm}du = -se^{-st} \hspace{12mm} v = f(t)$

$f'(t)e^{-st} + s[e^{-st}f(t) + s\int f(t)-se^{-st} dt]$

$f'(t) e^{-st} + se ^{-st} +s^2 * \int f(t)e^{-st} dt$

evaluated from $0$ to $\infty$

Answer 1

hint: $$u=e^{-st},v'=f''$$ then repeat