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.
 A: 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 $n\in\mathbb{N}$. :)

A: 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.
A: $f(x)=e^{x-1}$ 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.
A: 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 $n!$.
