# Show that $e^{-x}x^n$ is bounded on $[0,\infty)$ and hence prove that $\int_0^\infty e^{-x}x^n \, dx$ exists.

Show that $e^{-x}x^n$ is bounded on $[0,\infty)$ for all positive integral values of $n$. Using this result show that $\int_0^\infty e^{-x}x^n \, dx$ exists.

My work: I know $f$ is a continuous function and $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{x^n}{e^x}=\dfrac{n!}{e^x}=0$, by applying L'hospital's rule repeatedly.

But how to prove its boundedness formally? Also for existence of an integration, my book only covers for those integration where the limits are finite (Darboux's condition for integrability). So, how do I prove existence of an integration if the limits are infinite.

• The definition of $\int_0^\infty ...$ is $\lim_{M \to \infty} \int_0^M ...$; so you have to show each of the things on the right is bounded by something, and that their limit exists. Aug 10, 2017 at 2:15

In fact, this integral is a form of Gamma function: $\Gamma(n+1)$. What you want is that $\Gamma(n+1)$ is defined on $(0,+\infty)$. The domain, however, can be enlarged to $(-1,+\infty)$.

Gamma function. $$\Gamma(\alpha)=\int_0^\infty x^{\alpha-1}e^{-x}\mathrm{d}x, (\alpha>0)$$

The Gamma function is well-defined on $(0,+\infty)$, here we will show that.

Proof. It is ok to choose a large number $M$ s.t. $M>\alpha$, then we have $e^{x}>x^M/M!$ on $(0,+\infty)$ as the latter is one term in Taylor series expansion of $e^x$. Then $e^{-x}<M!x^{-M}$ on $(0,+\infty)$.

Consider the following two cases:

Case I: $\int_1^B e^{-x}x^{\alpha-1}\mathrm{d}x,$where $B>1$. \begin{align} \int_1^B e^{-x}x^{\alpha-1}\mathrm{d}x &\leq\int_1^BM!x^{-M}x^{\alpha-1}\mathrm{d}x\\ &=M!\left.\frac{x^{\alpha-M}}{\alpha-M}\right|_1^B\\ &=\frac{M!}{\alpha-M}\left(\frac{1}{B^{M-\alpha}}-1\right) \end{align}

Therefore, when $B\rightarrow\infty$, the integral is finite and well-behaved as $M-\alpha>0$.

Case II: $\int_0^1e^{-x}x^{\alpha-1}\mathrm{d}x$

$$\int_0^1e^{-x}x^{\alpha-1}\mathrm{d}x\leq\int_0^1 1\cdot x^{\alpha-1}\mathrm{d}x=\frac1\alpha$$

which is also finite for a given $\alpha$.

Therefore, we conclude that the $\Gamma(\alpha)$ is defined on $(0,+\infty)$.

You may find that $\Gamma(n+1)=n!$, if $n\in\mathbb{N}$. :)

• Seeing it as gamma function was quite awesome. Thanks a lot! Aug 10, 2017 at 8:02
• My pleasure :) @user467745 Aug 10, 2017 at 8:10

$f(x)=e^{x-1}$ is convex ($f''(x)>0$ for all real $x$), so its graph lies above the tangent at $x=1$: $e^{x-1}\ge x$ for all real $x$. This means $x\,e^{-x}\le e^{-1}$. Replacing $x$ by $x/n$ and raising to the $n$th power gives $$x^n\,e^{-x}\le n^n\,e^{-n}=C_n$$ for $x>0,$ that's the boundedness.
We have to show only the existence of the improper integral $\int^\infty_1x^n\,e^{-x}\,dx,$ since the existence of $\int^1_0x^n\,e^{-x}\,dx$ is trivial. But $$\int^b_1x^n\,e^{-x}\,dx=\int^b_1\frac{x^{n+2}\,e^{-x}}{x^2}\,dx\le \int^b_1\frac{C_{n+2}}{x^2}\,dx\le\int^\infty_1\frac{C_{n+2}}{x^2}\,dx=C_{n+2},$$ because $x^{n+2}\,e^{-x}\le C_{n+2},$ as was shown above. The LHS is monotone (the integrand is positive) and bounded, so the limit as $b\rightarrow\infty$ exists.

• So, if $\int_a^\infty f dx < K$, where $K$ is any constant or a bounded function, then $\int_a^\infty f dx$ exists? How do we know this? Is it some standard theorem? Aug 10, 2017 at 7:19
• @user467745 I hope it's clearer, now.
– user436658
Aug 10, 2017 at 7:31
• Yes, thank you. Quick question, what do you mean by trivial? I tried integrating that integral by parts. It gives me a summation series. Aug 10, 2017 at 7:40
• @user467745 Existence is trivial, it's a continuous integrand on a finite interval. Why would you want to calculate its value?
– user436658
Aug 10, 2017 at 7:43
• You are right, I wouldn't. Thank you so much for your time. This was one of the simplest solutions I came across. Aug 10, 2017 at 7:49

For the boundedness, yoou can take the derivative and show that it is negative on some interval $[a,\infty)$ and thus show the function is bounded on $[a,\infty)$. It is bounded on $[0,a]$ since it's continuous.

The way you show the integral on the infinite interval exists is to remember the definition of an improper integral existing: $$\lim_{b\to\infty}\int_0^b x^n e^{-x} dx$$ exists and is finite. It should be said that bounded on $[0,\infty)$ does not imply that the integral exists (take $\frac{1}{x+1}$ for instance), so that part of the question seems a bit misleading. However, it's true here and a simple way to prove it is by induction. You can show the integral exists for $n=0$ straightforwardly, and then for the induction step, use integration by parts.

• $f$ can be shown to be bounded if $f'(x)$ is negative in that interval? It seems logical. But is that an accepted way of proving boundedness? Aug 10, 2017 at 2:26
• I think the question expects to prove the existence using the fact that $f$ is bounded. So could you please elaborate on what those other crucial steps are? I will also try by induction. Thanks! Aug 10, 2017 at 2:37
• Well you need to use the amount of rigor that's appropriate, and in fact I've been sloppy since this is only true since $f(x)>0.$ Can you show if something on $[a,\infty)$ is decreasing, continuous and bounded from below that it is bounded? Regarding the integral, for instance, can you show that it's not only bounded but bounded by a function that is integrable on $[0,\infty)$? Aug 10, 2017 at 2:48
• Actually, I rescind my comment about 'with some other crucial steps'. I think a more important message is that boundedness does not imply convergence of the integral. The 'other crucial steps' are everything. Aug 10, 2017 at 2:58
• Do you mean that if $f$ is bounded by an integrable and bounded function, say $g$, then $\int_0^\infty f dx$ exists? Aug 10, 2017 at 2:59

It's probably simplest to use the fact that $e^x$ is analytic: $$e^x = \sum_{k\ge 0}\frac{x^k}{k!} \ge \sum_{k=0}^{n+2}\frac{x^k}{k!} \ge \frac{x^{n+2}}{(n+2)!}. \tag{1}$$ By $(1)$, the function $f(x)=e^{-x}x^n$ is no more than $\dfrac{(n+2)!x^n}{x^{n+2}} = \dfrac{(n+2)!}{x^2}$. For each $x\in (1,\infty)$, the inequality $f(x)< (n+2)!$ holds. Since $f$ is continuous, it is bounded on the compact set $[0,1]$, say by $M>0$. Thus, for every $x\in[0,\infty)$, the inequality $f(x)< \max\{M,(n+2)!\}$ holds, so $f$ is bounded on $[0,\infty)$.

On the other hand, if you insist on using L'Hospital's rule, you can still get boundedness. Apply L'Hospital's rule $n$ times to deduce that $\lim_{x\to\infty}f(x) = 0$. It follows that there exists a compact set $K\subset[0,\infty)$ such that for all $x\in [0,\infty)\smallsetminus K$, the inequality $|f(x)| < 1$ holds. Since $K$ is compact, and $f$ is continuous, $f$ is bounded by a constant $B>0$ for all $x\in K$. Thus, for any given $x\in[0,\infty)$, $f(x) < \max\{1,B\}$, so $f$ is bounded on $[0,\infty)$.

Boundedness alone isn't enough to prove that $\int_0^\infty f(x)\,dx$ exists. (Consider $f(x) = 1$. This is clearly bounded, but $\int_0^\infty 1\,dx = \infty$.) We have to show that the limit $\lim_{b\to\infty}\int_0^bf(x)\,dx$ exists (and is finite).

We have that $\int_0^\infty f(x)\,dx= \int_0^1f(x)\,dx + \int_1^\infty f(x)\,dx$, provided both these latter integrals exist. The integral $\int_0^1 f(x)\,dx$ certainly exists since $f$ is continuous, so it suffices to show that $\int_1^\infty f(x)\,dx$ exists. We can do this by using our estimate from $(1)$: $$\int_1^bf(x)\,dx \le \int_1^b\frac{(n+2)!}{x^2} = (n+2)!\left(1 - \frac{1}{b}\right).$$ Letting $b\to\infty$, we obtain $\int_1^\infty f(x)\,dx \le (n+2)!$, so the integral certainly exists. Fun fact: the integral $\int_0^\infty f(x)\,dx$ is actually equal to $n!$.