The Laplace transform of $f(x)$ is $$\int^{\infty}_0f(x)e^{-xs}dx$$
Obviously, the result would be a function of $s$.
The $f(x)$ in most cases is so complicated that the result most likely could not be written in terms of elementary functions(or even special functions like $Li(x), \Gamma(x)$). However, in my study, I am only interested in the result in the form of Taylor series(if exist).
So, would there be any shortcut such that we can expand the integrand into an infinite sum, then do an easy integration, and then obtain a Taylor series in terms of $s$?
For instance, with $f(x)=e^{-x}u(x)$($u(x)$ is the Heaviside step function), is there any easier way(although the standard method is already very easy) to obtain the result of its Laplace transform $$\sum_{k=0}^{\infty}(-s)^k$$?
It is not necessary to find the general term. The firsr few terms will do the job.
Any idea?
You may find few terms of the Taylor series of $f(x)e^{-sx}$ by formally multiplying the first few terms of Taylor series of $f(x)$ and Taylor series of $e^{-sx}.$
Then you can integrate the result.