Use of the Leibniz integral rule in Laplace transform proof My Laplace transform textbook presents the following theorem:

If $\mathcal{L}\{ F(t) \} = f(s)$, then $\mathcal{L}\{ t F(t) \} = - \dfrac{d}{ds}f(s)$ and in general $\mathcal{L}\{ t^n F(t) \} = (-1)^n \dfrac{d^n}{ds^n} f(s)$.

The proof then begins as follows:

Proof Let us start with the definition of Laplace transform
$$\mathcal{L}\{ F(t) \} = \int_0^\infty e^{-st} F(t) \ dt$$
and differentiate this with respect to $s$ to give
$$\begin{align} \dfrac{df}{ds} &= \dfrac{d}{ds} \int_0^\infty e^{-st} F(t) \ dt \\ &= \int_0^\infty -te^{-st} F(t) \ dt \end{align}$$
...

My understanding is that the author went from
$$\dfrac{d}{ds} \int_0^\infty e^{-st} F(t) \ dt$$
to
$$\int_0^\infty -te^{-st} F(t) \ dt$$
by using the Leibniz integral rule to change the ordinary derivative to a partial derivative.
However, as you can see from the Wikipedia page, the Leibniz integral rule is only valid for $\int_{a(x)}^{b(x)}, b(x) < \infty$, whereas the Laplace transform has $b(x) = \infty$. Doesn't this mean that the Leibniz rule is invalid?
I would greatly appreciate it if people could please take the time to clarify this.


The Leibniz Rule for an infinite region
If there is a positive function $g(x, y)$ that is integrable, with respect to $x$, on $[0,∞)$, for each $y$, and such that $|\frac{∂f}{∂y} (x, y)| ≤ g(x, y)$ for all $(x, y)$, then 
$$\frac{d}{dy}\int_{0}^\infty f(x,y)\,dx=\int_{0}^\infty \frac{\partial}{\partial y} f(x,y)\,dx$$


Ref.: https://math.hawaii.edu/~rharron/teaching/MAT203/LeibnizRule.pdf
