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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!

Adrian Keister
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UnsinkableSam
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  • You can just compute the Riemann integral: $$\int_0^\infty e^{-x}x^n,dx=n!$$ If it's Riemann integrable, it's Lebesgue integrable. – Adrian Keister Feb 12 '20 at 21:47
  • I would just argue that $f$ is dominated by some integrable function outside a compact set. For example, by $g(x) = x^{-2}$. – David Kraemer Feb 12 '20 at 21:48
  • @AdrianKeister Thank you! However, how would I compute its Riemann Integral though? What would be a suitable partition to use? – UnsinkableSam Feb 12 '20 at 22:04
  • You don't have to do that. The function is continuous on the domain, and if you simply integrate by parts enough times (or use mathematical induction), you find out that the integral is finite. Therefore, it's Riemann integrable. – Adrian Keister Feb 12 '20 at 22:05
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    @AdrianKeister Even Would that work even with our interval being unbounded? I think I had this proposition saying that if $f:[a,b]\to \mathbb{R}$ is riemann integrable if and only if $f$ is bounded and continuous and I wondering if $[0, \infty]$ would qualify for this one? Thanks! – UnsinkableSam Feb 12 '20 at 22:16
  • Oh, see your point. The Riemann integral isn't technically defined on unbounded intervals. You could try some limiting process. Or you could try David's suggestion, which is a good one. – Adrian Keister Feb 12 '20 at 22:22
  • Well, the improper integral is $\lim_{R\to \infty} \int_0^R e^{-x} x^n,dx$ and the result does apply for integrals over $[0,R]$. Then, the monotone convergence theorem on $\chi_{[0,N]}(x) e^{-x} x^n$ will give Lebesgue integrability. – Daniel Schepler Feb 13 '20 at 00:47

2 Answers2

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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 $[0,\infty)$.

RRL
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  • This avoids using the dominated convergence theorem explicitly with the obvious but unjustified assertion that $x^{-2}$ is Lebesgue integrable. – RRL Feb 13 '20 at 03:31
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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

$\int_{a(n)}^{\infty}x^ne^{-x}dx\le \int_{a(n)}^{\infty}e^{\frac{1}{2}x}e^{-x}dx=\int_{a(n)}^{\infty}e^{-\frac{1}{2}x}dx\le \infty.\tag2.$

Matematleta
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