Problem. Consider the function $$F(t) = \int_{0}^{\infty}x^{2}e^{-tx}dx,\quad t>0.$$ Show, using the Dominated Convergence Theorem, that $F$ is differentiable for every $t > 0$, with $$F'(t) = - \int_{0}^{\infty}x^{3}e^{-tx},\quad \forall t>0.$$
This problem is a particular case of Differentiating under the integral sign, however, this result has not been proven so far. The only result I can use is the Dominated Convergence Theorem (and those before him, of course).
We are considering the Lebesgue measure.
My approach. Let $(t_{n})$ be any sequence such that $t_{n} \to t$ and $t_{n} \neq t$ for each $t$. $$\frac{F(t_{n}) - F(t)}{t_{n} - t} = \int_{0}^{\infty}\underbrace{\frac{x^{2}e^{-t_{n}x} - x^{2}e^{-tx}}{t_{n} - t}}_{\varphi_{n}(x)}dt.$$
For each $x$ we have $$\lim_{n \to \infty}\varphi_{n}(x) = \lim_{n \to \infty}\frac{x^{2}e^{-t_{n}x} - x^{2}e^{-tx}}{t_{n} - t} = -x^{3}e^{-tx}.$$
If $g$ is a integrable function such that $|x^{3}e^{-tx}| \leq g(x)$ for each $x$, by the Means Value Theorem $$\varphi_{n}(x) = \frac{x^{2}e^{-t_{n}x} - x^{2}e^{-tx}}{t_{n} - t} = -x^{3}e^{-cx}$$ for some $c \in (t,t_{n})$. So, $|\varphi_{n}(x)| \leq g(x)$. Thus, by the Dominated Convergence Theorem, $$\lim_{n \to \infty}\int_{0}^{\infty}\varphi_{n}(x)dt = \int_{0}^{\infty}-x^{3}e^{-tx}dt = F'(t)$$
Is this correct? Moreover, how can I find $g$?
In order to find a suitable function $g$, we can use the fact that the sequence $(t_n)$ consists of positive numbers and converges to a positive number hence there exists a positive $\delta_0$ such that for all $n$, $\delta_0\leqslant t_n$. This guarantees that for all $x \gt 0$, $e^{-cx}\leqslant e^{-\delta_0x}$ hence we can define $$ g(x):=x^3e^{-\delta_0x}. $$ This work since this function is integrable and $\delta_0$ depends only on the sequence $(t_n)_n$.