Laplace Transform of Final integral Function

Question

I want to prove that if signal $u(t)$ satisfies:

$$ \lim_{t \to \infty} \int_0^t u(r) \, dr=c<\infty $$

then the Laplace Transform of the signal the following:

$$ \lim_{t \to \infty} \int_0^t \text{e}^{-sr}u(r) \, dr=d(s)<\infty $$

P.S. I don't know even this conjecture is true or not.

Answer 1

I think the question is crystal clear.

If some function $f:\mathbb{R}^+\to\mathbb{R}$ is weakly Riemann-integrable over $\mathbb{R}^+$, meaning that $$ \lim_{t\to +\infty} \int_{0}^{t}f(u)\,du = C<+\infty, $$ is it true that for any $s>0$ $$ (\mathcal{L} f)(s) = \lim_{t\to +\infty}\int_{0}^{t} e^{-us}f(u)\,du $$ is finite?

The answer is yes and the proof goes by integration by parts. Since the first limit is finite, $$ M = \sup_{t\in\mathbb{R}^+}\left|\int_{0}^{t}f(u)\,du\right| $$ is finite as well. For any $s,t>0$, by denoting as $F(t)=\int_{0}^{t}f(u)\,du$, we have:

$$ \int_{0}^{t}e^{-su}f(u)\,du = \left[e^{-su}F(u)\right]_{0}^{t}+s\int_{0}^{t}e^{-su}F(u)\,du $$ where the absolute value of the RHS is bounded by $$ 2M+Ms\int_{0}^{t}e^{-su}\,du \leq 2M+Ms\int_{0}^{+\infty}e^{-su}\,du=3M $$ no matter what the value of $t$ is. By considering the limit as $t\to +\infty$, the claim follows.