The goals of both transform techniques is to convert an ordinary differential equation into an algebraic one (and a partial differential equation into an ordinary differential equation). The difference is that the domain of the differential operators for which a Fourier tranform applies is $(-\infty,\infty)$, while that for which Laplace applies is $[0,\infty)$. Thus, LT's are used in initial value problems, while FT's are used when the function is applied over the whole real line.
The Laplace transform is useful because
- it transforms an ODE into an algebraic equation in the transformed variable (or a PDE into an ODE), and
- it includes the initial conditions as part of the algebraic equation.
The usefulness of this LT of course depends on the ability to find the inverse transform of the solution to the algebraic equation. In general, this requires the evaluation of a complex integral. However, for those not versed in complex integration methods, there are tables from which one may simply write down the inverse. Thus, the LT provides a simple means of solving an ODE using algebra.
No better way to understand this than to provide a specific example. Let's solve
$$f''(t) + 4 f'(t) + 3 f(t) = t \sin{t}$$
$$f(0) = 0 \quad f'(0) = 1$$
The LT of the LHS of the equation is
$$s^2 F(s) - 1 + 4 s F(s) + 3F(s) $$
The LT of the RHS is
$$\int_0^{\infty} dt \, t \sin{t}\; e^{-s t} = \operatorname{Im}{\int_0^{\infty} dt \, t \, e^{-(s-i) t}} = \operatorname{Im}{\left [\frac1{(s-i)^2}\right ]} = \frac{2 s}{(s^2+1)^2}$$
We need only solve a simple algebraic equation to find $F(s)$:
$$F(s) = \frac{(s^2+1)^2+2 s}{(s^2+1)^2 (s+3)(s+1)} $$
We may then use tables or the residue theorem to find the inverse:
$$f(t) = \frac14 e^{-t} - \frac{47}{100} e^{-3 t} + \frac{1}{50} [(5 t+2) \sin{t} +(11-10 t) \cos{t}]$$
Best Answer
We know that
$$\mathcal{L}\{f'(t)\} = \int_0^\infty f'(t)e^{-st}dt = sF(s) - f(0)$$
Then if one takes limits on both sides we get that
$$0 = \lim_{s\to \infty} sF(s) - f(0)$$
assuming that $f'(t)$ was bounded so we could apply the dominated convergence theorem and move the limit inside of the integral.