real analysis - Justifying differentiation under the integral sign with Dominated Convergence

Problem: Justify why you can use differentiation under the integral to differentiate:

$$I(t) := \int_{(0,\infty)}e^{-tx} \, m(dx).$$ Then use $I(t) = 1/t$ to show for all $n \in \mathbb{N}$ that $$\int_{(0,\infty)} x^ne^{-x} \,m(dx) = n!\, .$$

Attempt: We have that:

$$\dfrac{d}{dt} I(t) = \lim_{h \rightarrow 0} \dfrac{I(t+h) - I(t)}{h} = \lim_{h \rightarrow 0} \int_{(0,\infty)} e^{-tx}\left( \dfrac{e^{-hx}-1}{h}\right).$$ Since $\mathbb{R}$ is separable, it suffices to consider a sequence $\{h_n\}_{n \in \mathbb{N}}$ with $h_n \rightarrow 0$ as $n \rightarrow \infty$. Then, we consider: $$\lim_{n \rightarrow \infty} \int_{0,\infty)} e^{-tx}\left( \dfrac{e^{-h_nx}-1}{h_n}\right)$$ and presuming that $\left| e^{-tx}\left( \dfrac{e^{-h_nx}-1}{h_n}\right) \right|$ is bounded by an integrable function for all $n$, we may apply the Dominated Convergence Theorem to swap the integral sign and the limit.

Question: What can I use to bound $\left| e^{-tx}\left( \dfrac{e^{-h_nx}-1}{h_n}\right) \right|$? Certainly we can't use $e^x$ or just $e$, since neither is integrable on $(0, \infty)$. Any suggestions?

Answer

One can check that $e^u\geq 1+u$ for all $u\in\mathbb{R}$, hence $1-e^u\leq -u$. Applying this with $u=-h_nx$, we obtain

$$\Big|\frac{e^{-h_nx}-1}{h_n}\Big|=\frac{1-e^{-h_nx}}{h_n}\leq \frac{h_nx}{h_n}=x$$
if $h_n>0$, and
$$\Big|\frac{e^{-h_nx}-1}{h_n}\Big|=\frac{e^{-h_nx}-1}{-h_n}=\frac{1-e^{-h_nx}}{h_n}\leq \frac{h_nx}{h_n}=x$$

if $h_n<0$.
Therefore
$$\Big|e^{-tx}\frac{e^{-h_nx}-1}{h_n}\Big|\leq xe^{-tx}$$
which is integrable on $(0,\infty)$ if $t>0$.