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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?

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  • $\begingroup$ I don't know what level of rigor you are looking to achieve, but note that $\Phi(s)$ does not depend on $s_0$ (since $\Phi(s)$ is a constant), since $s_0$ is just a book-keeping parameter that we use to manipulate the expression $\int_{s}^{s_{0}} F(s)ds+\Phi(s_0)$. So while it's technically correct to write $\lim_{s_{0} \to \infty}\Phi(s)$ in the last step, it's sort of redundant, as it just simplifies to $\Phi(s)$. Does that answer your question? $\endgroup$ Dec 6, 2014 at 17:54
  • $\begingroup$ Also, your integration property equation is incorrect, it should be properly written as $$\mathscr{L}\left\{\frac{f(t)}{t}\right\}(s)=\int_s^\infty F(\sigma)d\sigma,$$ not $\mathscr{L}\{\dfrac{f(t)}{t}\}=\int_s^\infty F(s)ds$ (you used the same variable $s$ for both the integration limit and the dummy variable in the integral). $\endgroup$ Dec 6, 2014 at 17:55

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You are correct.

However, $\Phi(s)$ is just a constant, and does not depend on $s_0$, so $$\lim_{s_{0} \to \infty}\Phi(s)=\Phi(s).$$

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