In my textbook I have the following theorem about the integration of the frequency (F(s)):
Let the Laplace transform of a function $f(t)$ be $\mathscr{L}\{f(t)\}=F(s)$.
If $\dfrac{f(t)}{t}$ is the original, then
$\mathscr{L}\{\dfrac{f(t)}{t}\}=\int_s^\infty F(s)ds$.
Proof:
Let $\Phi(s)$ be the Laplace transform of function $\dfrac{f(t)}{t}$. For every Laplace transform holds $\lim_{s \to \infty} \Phi(s) = 0$. Using the theorem about the integration of the Laplace transform
$\mathscr{L}\{t\dfrac{f(t)}{t}\}=-\Phi'(s)$
so $\Phi'(s) = -F(s)$
Because of that, $\Phi(s)$ can be obtained by the definite integral:
$\Phi(s)=-\int_{s_{0}}^{s} F(s)ds+C=\int_{s}^{s_{0}} F(s)ds+C$
If we put $s=s_{0}$, we obtain that $C=\Phi(s_{0})$. We now let that $s_{0}$ goes to infinity:
$\Phi(s)=\lim_{s_{0} \to \infty}\int_{s}^{s_{0}} F(s)ds + \lim_{s_{0} \to \infty}\Phi(s_{0})=\int_{s}^{\infty} F(s)ds$
I don't understand how can that last equality be correct. Shouldn't the left hand side be equal to: $\lim_{s_{0} \to \infty}\Phi(s)$ instead only $\Phi(s)$, since we have taken the limit of the whole expression?
Best Answer
You are correct.
However, $\Phi(s)$ is just a constant, and does not depend on $s_0$, so $$\lim_{s_{0} \to \infty}\Phi(s)=\Phi(s).$$