77 Differentiating Laplace transforms

(iii) Let $s>\beta +\delta$ for some $\delta >0$ and consider . Note that $e^{\delta x}{e}^{-sx}f(x)$ is integrable, and $x\le e^{\delta x}/\delta$, so $xf(x)$ also satisfies $(E)$. Also

$$\frac{e^{-(s+h)x}-e^{-sx}}{h}=e^{-sx}\left(\frac{e^{-hx}-1}{h}\right)\to -x{e}^{-sx}$$

as $h\to 0$. Hence

$$\frac{ℒ(f)(s+h)-ℒ(f)(s)}{h}={\int }_0^{\mathrm{\infty }}\frac{e^{-(s+h)x}-e^{-sx}}{h}f(x)𝑑x$$

$$\to -{\int }_0^{\mathrm{\infty }}{e}^{-sx}xf(x)𝑑x.$$

To make this precise, we consider