$\left[ – \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\} \right]_s^\infty$ to $\mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}\bigg\vert_s$

Question

I'm currently studying the textbook An Introduction to Laplace Transforms and Fourier Series, second edition, by Phil Dyke. Chapter 2.2 Derivative Property of the Laplace Transform gives the following theorem and proof:

Theorem 2.3 If $\mathcal{L}(F(t)) = f(s)$ then $\mathcal{L}\left\{ \dfrac{F(t)}{t} \right\} = \int_s^\infty f(u) \ du$, assuming that $\mathcal{L}\left\{ \dfrac{F(t)}{t} \right\} \to 0$ as $s \to \infty$.

Proof Let $G(t)$ be the function $F(t)/t$, so that $F(t) = t G(t)$. Using the property
$$\mathcal{L}\{ t G(t) \} = – \dfrac{d}{ds}\mathcal{L}\{ G(t) \}$$
we deduce that
$$f(s) = \mathcal{L} \{ F(t) \} = – \dfrac{d}{ds} \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}.$$
Integrating both sides of this with respect to $s$ from $s$ to $\infty$ gives
$$\int_s^\infty f(u) \ du = \left[ – \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\} \right]_s^\infty = \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}\bigg\vert_s = \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}$$
since
$$\mathcal{L} \left\{ \dfrac{F(t)}{t} \right\} \to 0 \ \ \ \text{as} \ \ \ s \to \infty$$
which completes the proof.

I'm confused about this part:

$$\int_s^\infty f(u) \ du = \left[ – \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\} \right]_s^\infty = \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}\bigg\vert_s = \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}$$

How do we get from $\left[ – \mathcal{L} \left\{ \dfrac{F(t)}{t} \right\} \right]_s^\infty$ to $\mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}\bigg\vert_s$, and then from $\mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}\bigg\vert_s$ to $\mathcal{L} \left\{ \dfrac{F(t)}{t} \right\}$?

Answer 1

This seems like mostly a confusion about notation. The notation

$$\left[f(x)\right]_s^\infty$$

is shorthand for

$$\lim_{b\to\infty} f(b)-f(s)\ .$$

So

$$\left[- \mathcal{L} \left\{ \dfrac{F(t)}{t}\right\}\right]_s^{\infty} = -\lim_{b\to\infty}\mathcal{L}\left\{\frac{F(t)}{t}\right\}\bigg\vert_b + \mathcal{L}\left\{\frac{F(t)}{t}\right\}\bigg\vert_s$$

Because of the assumption that the Laplace transform approaches $0$ as the input approaches $\infty$ the first term vanishes. The second is just the value of the Laplace transform of $\frac{F(t)}{t}$ evaluated at $s$. Both

$$\mathcal{L}\left.\left\{\frac{F(t)}{t}\right\}\right\vert_s$$

and

$$\mathcal{L}\left\{\frac{F(t)}{t}\right\}$$

are ways to denote this. In the second case the dependence on $s$ has been suppressed.