# What is the inverse Laplace transform of $sF''(s)$?

I need to find the inverse Laplace transform of $$sF''(s)$$ where $$F(s) = \int_0^\infty f(t)e^{-st}dt$$. This is not one of the cases in my book. What can I do to compute this?

My hunch is that the answer is $$t^2f'(t)$$ seen in the beginning of this paper.

In this paper, they inverse laplace transform

$$w^{i v}-10 z w^{\prime \prime}-10 w^{\prime}+9 z^2 w=0$$

into $$9 f^{\prime \prime}-10 t^2 f^{\prime}+\left(t^4-10 t\right) f=0$$

where $$w(z)=\int_{\mathcal{C}} f(t) e^{-z t} d t$$

The only term that I cannot compute from the Laplace transform table is the $$zw''(z)$$ term, but according to this table, it should be $$tf'(t)$$

Thank you

$$\begin{split} \mathcal{L}^{-1}\{sF''(s)\} &= \mathcal{L}^{-1}\{sF''(s) + F'(s)\} - \mathcal{L}^{-1}\{F'(s)\}\\ &= \mathcal{L}^{-1}\{\frac{d}{ds}(sF'(s))\} + tf(t)\\ &= -t\mathcal{L}^{-1}\{sF'(s)\}(t) + tf(t)\\ &= -t[\mathcal{L}^{-1}\{sF'(s) + F(s)\} - \mathcal{L}^{-1}\{F(s)\}](t) + tf(t)\\ &= -t\mathcal{L}^{-1}\{\frac{d}{ds}(sF(s))\}(t) + 2tf(t)\\ &= t^2\mathcal{L}^{-1}\{sF(s)\}(t) + 2tf(t)\\ &= t^2[\mathcal{L}^{-1}\{sF(s) - f(0)\} + \mathcal{L}^{-1}\{f(0)\}](t) + 2tf(t)\\ &= t^2f'(t) + t^2f(0)\delta(t) + 2tf(t)\\ &= t^2f'(t) + 2tf(t) \end{split}$$