Let I be an open subinterval of $\mathbb R$ and let $f:\mathbb R \to \mathbb R$ be a Borel measurable function such that $x \to e^{tx}f(x)$ is Lebesgue integrable for each t in I. Define $h: I \to \mathbb R$ by $h(t)= \int_\mathbb R e^{tx}f(x)\lambda(dx)$. Show that h is differentiable, with derivative given by $h'(t)=\int_\mathbb R xe^{tx}f(x)\lambda(dx)$, at each t in I. Of course, it is part of your task to shat that $x \to xe^{tx}f(x)$ is integrable for each t in I.
My idea is to show using the MacLaurin expansion of $e^u$ that $\lvert e^u-1 \rvert \lt \lvert u \rvert e^{\lvert u \rvert}$ holds for each u in $\mathbb R$ and after use the Monotone Convergence Theorem but I'm a bit stuck. Can you help me?
Let $(t_n)$ be a sequence converging to a point $t \in I = (a,b)$, and define
$$g_n(x,t) = \frac{e^{t_n x}f(x)-e^{tx}f(x)}{t_n-t}.$$
The sequence converges to the partial derivative with respect to t, which as a limit of a sequence of measurable functions is measurable:
$$\lim_{n \to \infty}g_n(x,t) = xe^{tx}f(x).$$
By the mean value theorem, there is a point $\xi_n$ between $t$ and $t_n$ such that
$$|g_n(x,t)| = e^{\xi_n x}f(x) \leqslant \max(e^{ax},e^{bx})f(x) = g(x).$$
Since $g$ is integrable we can apply the Dominated Convergence Theorem to show
$$h'(t) = \lim_{n \to \infty} \frac{h(t_n)-h(t)}{t_n-t} \\ = \lim_{n \to \infty}\int g_n(x,t) \lambda(dx)\\ = \int \lim_{n \to \infty} g_n(x,t) \lambda(dx) \\= \int xe^{tx}f(x) \lambda(dx)$$