How to prove Laplace transform is of an existing function

Question

How to prove this $$\frac{s}{(s^2+4)^2}$$ is the Laplace transform of a continuous exponential $f(x)$?

I understand how to solve the problem the other way around (given $f(x)$ you prove that the convolution of $ e^{-st} f(x) $ converges) but how do you approach the problem when given the Laplace transform and have to prove it is in fact an existing function that produced it?

Answer 1

If you compare it with tables, you can discover that $$f(t)=\frac{t \sin (2 t)}{4} =\mathcal{L}^{-1}\!\left[\frac{s}{\left(s^2+4\right)^{\!2}}\right]. $$ To finish, take the Laplace transform of this $f(t)$ and show that you get the RHS.

Answer 2

You can calculate the inverse Laplace Transform:

$$\mathcal {L} \left\{f(t)\right \}= \left \{\frac{s}{(s^2+4)^2}\right \}$$ $$\mathcal {L} \left \{f(t)\right \}=\frac 1 2 \left \{\frac{2s}{(s^2+4)^2}\right \}$$ $$\mathcal {L} \left \{f(t)\right \}=-\frac 1 2\left \{\frac d {ds}\frac{1}{(s^2+4)}\right \}$$ $$\mathcal {L} \left \{f(t)\right \}=-\frac 1 4\left \{\frac d {ds}\frac{2}{(s^2+4)}\right \}$$ $$\mathcal {L} \left \{f(t)\right \}=-\frac 1 4\left \{\frac d {ds}\mathcal {L} \left \{\sin(2t) \right \}\right \}$$ $$\boxed {f(t)=\frac 1 4{t\sin(2t)}}$$