Since Laplace transforms of the type: $$\mathcal{L} \left\{g\left(t\right)\right\}\left(s\right) = sF\left(s\right)$$ can be simplified as by finding $f\left(t\right)$ such that: $$f\left(t\right)=\mathcal{L}^{-1}\left\{F\left(s\right)\right\}\left(t\right)$$ , and $f\left(0\right) = 0$, then: $$g(t)=\frac{d}{dt}f\left(t\right)$$
This process simplifies the Inverse laplace process for $sF\left(s\right)$, by only needing to find the Inverse Laplace of $F\left(s\right)$. I was wondering if there's also a method for solving $\mathcal{L}^{-1}\left\{\left(-\frac{1}{s}\right)F\left(-\frac{1}{s}\right)\right\}(t)$ by using $\mathcal{L}^{-1}\left\{F\left(s\right)\right\}$ ?
To summarize the question:
Given: $$\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)=F\left(s\right)$$ and $$\mathcal{L}\left\{h\left(t\right)\right\}\left(s\right)=\left(-\frac{1}{s}\right)F\left(-\frac{1}{s}\right)$$
Can we find a general formula for the relation between $f\left(t\right)$ and $h\left(t\right)$, however complicated the relation may be?
Thank you.