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

$$\dfrac{d}{ds} \int_0^\infty e^{-st} F(t) \ dt$$


$$\int_0^\infty -te^{-st} F(t) \ dt$$

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.


1 Answer 1


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

  • $\begingroup$ What's the $g$ in our case? $\endgroup$
    – GFauxPas
    May 29, 2019 at 13:23
  • $\begingroup$ @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 $g(t) = Me^{\alpha t} \ge \mid \dfrac{\partial{f}}{\partial{s}} \mid$. $\endgroup$ May 29, 2019 at 15:03

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