I am interested in the laplace transform of $f(t)f'(t)$. If it is known that there is no solution to this then there will be a pretty short answer however I would like to know if one exists and if so what it is.
I tried to find a solution using integration by parts: $$\begin{equation}\begin{aligned} \mathscr{L}\{f(t)f'(t)\}&=\int_0^{\infty}e^{-st}f(t)f'(t)\ dt \\ &=\big[f(t)e^{-st}f(t)\big]_0^{\infty}-\int_0^{\infty}f(t)(-se^{-st}f(t)+e^{-st}f'(t))\ dt \\ &=\big[f(t)e^{-st}f(t)\big]_0^{\infty}-\int_0^{\infty}e^{-st}f(t)f'(t)\ dt+\int_0^{\infty}se^{-st}f(t)^2\ dt \\ 2\mathscr{L}\{f(t)f'(t)\}&=-f(0)^2+\int_0^{\infty}e^{-st}f(t)^2\ dt \\ \end{aligned}\end{equation}$$
From this I evaluated $\mathscr{L}\{f(t)^2\}$ $$\begin{equation}\begin{aligned} \mathscr{L}\{f(t)^2\}&=\int_0^{\infty}e^{-st}f(t)^2\ dt \\ &=\bigg[-\frac{1}{s}e^{-st}f(t)^2\bigg]_0^{\infty}-\int_0^{\infty}-\frac{1}{s}e^{-st}\times 2f(t)f'(t)\ dt \\ &=\bigg[-\frac{1}{s}e^{-st}f(t)^2\bigg]_0^{\infty}+\frac{2}{s}\int_0^{\infty}e^{-st}f(t)f'(t)\ dt \\ &=\frac{f(0)^2}{s}+\frac{2}{s}\int_0^{\infty}e^{-st}f(t)f'(t)\ dt \\ \end{aligned}\end{equation}$$
I then substituted this back into the expression for $\mathscr{L}\{f(t)f'(t)\}$ to get $$\begin{equation}\begin{aligned} 2\mathscr{L}\{f(t)f'(t)\}&=-f(0)^2+\frac{f(0)^2}{s}+\frac{2}{s}\mathscr{L}\{f(t)f'(t)\} \\ \bigg(2-\frac{2}{s}\bigg)\mathscr{L}\{f(t)f'(t)\}&=-f(0)^2+\frac{f(0)^2}{s} \\ \mathscr{L}\{f(t)f'(t)\}&=\frac{f(0)^2}{2(s-1)}-\frac{sf(0)^2}{2(s-1)} \\ \end{aligned}\end{equation}$$
This seems odd as a solution to me as it doesn't have any functions of the form $\mathscr{L}\{f(t)\}$. I also am aware that it is wrong due to it not working for basic functions such as $e^x$ etc.
Essentially my question is:
What was wrong about my method, and is there a known expression for $\mathscr{L}\{f(t)f'(t)\}?$