a limit of the laplace transform and its derivative If $\phi(s)$ is the Laplace tranfrom of $f(t)$, then $\lim_{s\rightarrow \infty} s\phi(s) = f(0^+)$. and also $\lim_{\rightarrow \infty} s\phi'(s) = \lim_{t\rightarrow 0^+}tf(t)$ since $\phi'(s)$ is the laplace transform of $tf(t)$. These results suggest that $\lim_{s\rightarrow \infty} s\phi'(s)/\phi(s)$ is finite, and indeed it is finite for many well-known Laplace tranforms. My question is: is there a relation between $\lim_{s\rightarrow \infty} s\phi'(s)/\phi(s)$ and particular values or limits of $f(t)$ and/or its derivative?
 A: if $f(t)$ vanishes as $t^\nu$ for $t\downarrow 0$, then 
$\lim_{s\rightarrow\infty}s\phi'(s)/\phi(s)=-\nu-1$, in other words
$$\lim_{s\rightarrow\infty}s\phi'(s)/\phi(s)=-1- \lim_{t\downarrow 0}tf'(t)/f(t)$$
A: If $f(t)$ is everywhere smooth everywhere including $t=0$, then its Laplace transform will have the asymptotic expansion $\phi(s) \sim f(0)/s + f'(0)/s^2 + 2! f''(0)/s^3 + 3! f'''(0)/s^4 + \cdots$ for large $s \to \infty$. Since you can differentiate asymptotic expansions term by term, $\phi'(s)$ has the same asymptotic expansion, just shifted in terms of powers of $s$. Your limit is then $\lim_{s\to \infty} s\phi'(s)/\phi(s) = -1$ if $f(0)\ne 0$, or $-2$ if $f(0) = 0$ but $f'(0) \ne 0$, or $-(n+1)$ if $f^{(n)}(0)$ is the first non-vanishing derivative of $f(t)$ at $t=0$.
This property of the Laplace transform can be seen to derive from its expression as the analytic continuation ($is \to -s$) of the Fourier transform of $f(t)\Theta(t)$ over the whole real line, with $\Theta(t)$ the unit step function. The Fourier transform is know to transform discontinuities in the $n$-the derivatives into an asymptotic contribution of order $O(1/s^{n+1})$.
