# 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:

$$\mathcal{L}\{ F(t) \} = \int_0^\infty e^{-st} F(t) \ dt$$

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.

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
$$\frac{d}{dy}\int_{0}^\infty f(x,y)\,dx=\int_{0}^\infty \frac{\partial}{\partial y} f(x,y)\,dx$$
• What's the $g$ in our case? May 29, 2019 at 13:23
• @GFauxPas By definition of Laplace transforms, we have that $\mid \dfrac{\partial{f}}{\partial{s}} \mid \le Me^{\alpha t}$ (in other words, the function $\dfrac{\partial{f}}{\partial{s}}$ is of exponential order). Therefore, there must, by definition, exist some function $g(t) = Me^{\alpha t} \ge \mid \dfrac{\partial{f}}{\partial{s}} \mid$. May 29, 2019 at 15:03