On the relation between Laplace transform of a function and its derivative Suppose that $s>s_0$. Through integration by parts we can write $$\int_{0}^{\infty}e^{-st} f'(t)~\mathrm dt= \Big[ e^{-st} f(t)\Big]_0^\infty + s \int_{0}^{\infty}e^{-st} f(t)~\mathrm dt. $$ Book ODE by Rabenstein says that if $\int_{0}^{\infty}e^{-st} f'(t)~\mathrm dt$ and $\int_{0}^{\infty}e^{-st} f(t)~\mathrm dt$ exist then $\lim_{\ t \to +\infty} (e^{-st} f(t)) =0 $. 
My question is : What is the proof to the statement that if the integral $\int_{0}^{\infty}g(t)~\mathrm dt$ is well defined then must the limit $\lim_{r\to+\infty}g(r)=0$ ?   
 A: Through integration by parts we can actually write 
$$
\int_{0}^re^{-st} f'(t)~\mathrm dt= \Big[ e^{-st} f(t)\Big]_0^r + s \int_{0}^re^{-st} f(t)~\mathrm dt
$$
or, equivalently,
$$
\int_{0}^re^{-st} f'(t)~\mathrm dt-s \int_{0}^re^{-st} f(t)~\mathrm dt=e^{-sr} f(r)-f(0).
$$
The existence of the two improper integrals implies that one can take $r\to+\infty$ on the left and so also on the right. This shows that $\lim_{r\to+\infty}(e^{-sr} f(r))$ is well defined.
Moreover, the limit $$c=\lim_{r\to+\infty}(e^{-sr} f(r))$$ can only be zero since otherwise (and since the limit exists) the integral $\int_{0}^{\infty}e^{-st} f(t)~\mathrm dt$ would not be well defined. Indeed, if $c\ne0$, then given $\delta>0$, there exists $T>0$ such that $|e^{-st} f(t)-c|<\delta$ for $t>T$ and so $$\int_T^re^{-st} f(t)~\mathrm dt>(c-\delta)(r-T)\to+\infty$$
when $r\to+\infty$, provided that $\delta<c$ and assuming here for simplicity that all is real although we can simply consider real and imaginary parts.
A: If you write down the integration by parts formula for the integral from $0$ to $R$, then your assumption implies that the term on the left and the second summand on the right have a limit as $R \to +\infty$, which implies that $\lim\limits_{t \to +\infty} e^{-st} f(t)$ exists. If this limit was not $0$, then the improper integral from $0$ to $\infty$ of this function would diverge.
