I want to show that $f(x)=e^{-x}x^n$ on $[0, \infty]$ is Lebesgue integrable for some positive integer $n.$
For the start, I am aware that the function is bounded and I thought about dividing the interval up, but I couldn't think of any intervals that are particularly useful or any functions that I could perhaps use the comparison test.
Any help is appreciated!
On
The integrand is continuous and, therefore, Riemann integrable on $[0,k]$ for every $k \in \mathbb{N}$. Riemann and Lebesgue integrals coincide on finite intervals, whence, with $f(x) = e^{-x}x^n$ we have
$$\int_{[0,\infty)}f_k =\int_{[0,\infty)}f\chi_{[0,k]} = \int_0^k f(x) \, dx$$
Since $\lim_{x \to \infty} f(x)x^{2} = 0$, there exists $x_0 > 0$ such that $f(x) < x^{-2}$ for all $x \geqslant x_0$. We also have (using Riemann integrals), $\int_{x_0}^k f(x) \, dx \leqslant \int_{x_o}^k x^{-2} \, dx = 1/x_0 - 1/k < 1/x_0$, and, consequently, the following conditions are met:
(1) The sequence $(f_k)$ is a monotone increasing sequence of nonnegative, measureable functions converging to $f$; and (2) the integrals $\int_{[0,\infty)} f_k$ are bounded for all $k \in \mathbb{N}$.
When both of these conditions hold, the monotone convergence theorem guarantees that $f$ is Lebesgue integrable on $[0,\infty)$.
You don't need to consider improper integrals. That's one of the advantages of the Lebesgue theory. In fact, your functions are Lebesgue integrable for $every$ fixed integer $n$. There is an $a(n)\in \mathbb R^+$ such that $e^{\frac{1}{2}x}\ge x^n$ whenever $x\ge a(n).$ Then,
$\int_0^{\infty}x^ne^{-x}dx=\int_0^{a(n)}x^ne^{-x}dx+\int_{a(n)}^{\infty}x^ne^{-x}dx.\tag1$
The first term on the right-hand-side is clearly Lebesgue integrable and the second is as well because
$\int_{a(n)}^{\infty}x^ne^{-x}dx\le \int_{a(n)}^{\infty}e^{\frac{1}{2}x}e^{-x}dx=\int_{a(n)}^{\infty}e^{-\frac{1}{2}x}dx\le \infty.\tag2.$