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I'm doing exercises in Real Analysis of Folland and got stuck on this problem. I don't know how to calculate limit with the variable on the upper bound of the integral. Hope some one can help me solve this. I really appreciate.

Show that $\lim\limits_{k\rightarrow\infty}\int_0^kx^n(1-k^{-1}x)^k~dx=n!$

Thanks so much for your consideration.

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  • $\begingroup$ Make the substitution $u=x/k$. $\endgroup$ – Ted Shifrin Feb 22 '14 at 11:26
  • $\begingroup$ I think if you integrate by parts, keeping $x^n$ the second function, and go on recursively, $n!$ term should pop out. And then you can take limit $k \rightarrow \infty$ the remaining factor could be 1 ? Have you tried this way ?? $\endgroup$ – DiffeoR Feb 22 '14 at 11:27
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We have

$$\int_0^k x^n(1-k^{-1}x)^kdx=\int_0^\infty x^n(1-k^{-1}x)^k\chi_{(0,k)}dx$$ then since $$x^n(1-k^{-1}x)^k\chi_{(0,k)}\le x^n e^{-x},\;\forall k$$ and the function $$x\mapsto x^n e^{-x}$$ is integrable on the interval $(0,\infty)$ then by the dominated convergence theorem we have $$\lim_{k\to\infty}\int_0^k x^n(1-k^{-1}x)^kdx=\int_0^\infty x^ne^{-x}dx=\Gamma(n+1)=n!$$

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  • $\begingroup$ absolutely right. Thanks so much. $\endgroup$ – le duc quang Feb 23 '14 at 8:38
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Feb 23 '14 at 12:07
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Hereafter we'll use the Beta Function $\ds{{\rm B}\pars{x,y} = \int_{0}^{1}t^{x - 1}\pars{1 - t}^{y - 1}\,\dd t}$ with $\ds{\pars{~\Re\pars{x} > 0\,,\Re\pars{y} > 0~}}$ and its property $\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$. $\ds{\Gamma\pars{z}}$ is the Gamma Function. For $n\ \in\ {\mathbb N}$, it satisfies $\ds{\Gamma\pars{n + 1} = n!}$.

\begin{align} &\lim_{k \to \infty}\int_{0}^{k}x^{n}\pars{1 -k^{- 1}x}^{k}\,\dd x= \lim_{k \to \infty}\bracks{k^{n + 1}\int_{0}^{1}x^{n}\pars{1 - x}^{k}\,\dd x} \\[3mm]&=\lim_{k \to \infty}\bracks{k^{n + 1}{\rm B}\pars{n + 1,k + 1}} =\lim_{k \to \infty}\bracks{% k^{n + 1}\,{\Gamma\pars{n + 1}\Gamma\pars{k + 1}\over \Gamma\pars{n + k + 2}}} \\[3mm]&=n!\lim_{k \to \infty}\bracks{{k^{n + 1}k! \over \pars{n + k + 1}!}} \tag{1} \end{align}

However, \begin{align} &\lim_{k \to \infty}\bracks{k^{n + 1}k! \over \pars{n + k + 1}!} =\lim_{k \to \infty} \bracks{k^{n + 1} \over \pars{k + n + 1}\pars{k + n}\ldots\pars{k + 1}} \\[3mm]&= \lim_{k \to \infty}\bracks{% \pars{1 + {n + 1 \over k}}\pars{1 + {n \over k}}\ldots\pars{1 + {1 \over k}}}^{-1} = 1 \end{align}

By replacing this result in $\pars{1}$, we get: $$\color{#00f}{\large% \lim_{k \to \infty}\int_{0}^{k}x^{n}\pars{1 -k^{- 1}x}^{k}\,\dd x =n!} $$

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  • $\begingroup$ Yes, this is the painful approach: more complicated, more error-prone, fails as soon as one perturbs the problem, does not explain why the result holds. $\endgroup$ – Did Feb 23 '14 at 10:13
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Let $$ f_k(x)=\left\{\begin{array}{ccc} \left(1-\frac{x}{k}\right)^{k}x^n & \text{if} & x\in [0,k], \\ \\ 0 & \text{if} & x>k. \end{array} \right. $$ Then $$ 0\le f_k(x)\le \mathrm{e}^{-x}x^n, $$ for all $k$ and $x\ge 0$, and $\lim_{k\to\infty}f_k(x)=\mathrm{e}^{-x}x^n$, for all $x\ge 0$.

Lebesgue Dominated Convergence Theorem is applicable as $\mathrm{e}^{-x}x^n\in L^1(0,\infty)$, and implies that $$ \lim_{k\to\infty}\int_0^k \left(1-\frac{x}{k}\right)^k x^n\,dx= \lim_{k\to\infty}\int_0^\infty f_k(x)\,dx=\int_0^\infty \mathrm{e}^{-x}\,x^n\,dx=\Gamma(n+1)=n! $$

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  • $\begingroup$ correctly. thanks so much... $\endgroup$ – le duc quang Feb 23 '14 at 8:38

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