I have this problem to solve, but I couldn't get the solution:
$F(s) = \frac{s-2}{s^{2}-1}$
$s>1$
$\int_{0}^{+\infty}\left ( \frac{f(t)}{e^{2t}} \right )dt$
$L\left [ e^{-2t}f(t)) \right ]=F\left ( s-a \right )=\frac{s}{(s+2)^{2}-1}$
and
$L\left [ \int_{0}^{\infty}\left ( \frac{f(t))}{e^{2t}} \right ) \right ] = \frac{F(s))}{s}=\frac{1}{(s+2)^{2}-1}$
so
$\int_{0}^{\infty}\left ( \frac{f(t))}{e^{2t}} \right ) = L^{-1}\left [ \frac{1}{(s+2)^{2}-1} \right ]$
$\int_{0}^{\infty}\left ( \frac{f(t))}{e^{2t}} \right ) = \frac{1}{2}\left ( e^{-t} + e^{-3t} \right )$
I am struggling to solve this with Laplace properties. I have to evaluate the integral without explicitly calculate f(t). F(s) is the Laplace transform of f(t). I suppose that my resolution is wrong since the superior limit is infinity. The solution must be achieved without performing f(t) at the beginning. The method must use F(s) and not f(t) directly on the integral. Can anyone help me?
$$ F(s)=\frac{s-2}{s^{2}-1},\quad{s}\gt{1}. \\ L^{-1}\left[F(s)\right]=L^{-1}\left[\frac{s-2}{s^{2}-1}\right]=L^{-1}\left[\frac{s}{s^{2}-1}\right]+L^{-1}\left[\frac{-2}{s^{2}-1}\right]= \\ =\left(\cosh(t)-2\sinh(t)\right)=\frac{e^{t}+e^{-t}-2e^{t}+2e^{-t}}{2}=\frac{-e^{t}+3e^{-t}}{2},\implies \\ \implies{f(t)}=\frac{-e^{t}+3e^{-t}}{2}. $$ $$ \int{\frac{(-e^{t}+3e^{-t})e^{-2t}}{2}}dt=\left(\int{\frac{-e^{-t}}{2}}dt+\int{\frac{3e^{-3t}}{2}}dt\right)=0.5e^{-t}-1.5e^{-3t}. $$