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I was wondering how to prove that $\lim_{n\to \infty}\int_{1}^{n}\frac{1}{(x^{2}+1)^{n}}dx\sim \frac{1}{n\cdot 2^{n}}?.$

This appears to be asymptotic to $\frac{1}{n2^{n}}$, but how to prove it?.

I checked with larger and larger values of n, and it does get closer and closer to $\frac{1}{n2^{n}}.$

i.e $\int_{1}^{10}\frac{1}{(x^{2}+1)^{10}}dx\approx .00009843725636$ and $\frac{1}{10\cdot 2^{10}}\approx .00009765625.$

The larger $n$, the closer they get.

I tried using parts to no avail. I also thought $\sum_{k=0}^{\infty}\binom{-n}{k}x^{2k}=\frac{1}{(1+x^{2})^{n}}$ may be useful in some manner.

Does anyone have a good idea as to how to prove this?

Thanks

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    Your notation is a bit sloppy. The *limit* of the integral is zero. It is the integral itself that is asymptotic to $1/n2^n$.2011-02-03

5 Answers 5

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} I_{n} \equiv \int_{1}^{n}{\dd x \over \pars{x^{2} + 1}^{n}}&= {n \over \pars{n^{2} + 1}^{n}} - {1 \over 2^{n}} + 2n\int_{1}^{n}{x^{2} \over \pars{x^{2} + 1}^{n + 1}}\,\dd x \\[3mm]&= {n \over \pars{n^{2} + 1}^{n}} - {1 \over 2^{n}} + 2nI_{n} - 2n\int_{1}^{n}{\dd x \over \pars{x^{2} + 1}^{n + 1}} \end{align}

\begin{align} I_{n} &= {n \over \pars{1 - 2n}\pars{n^{2} + 1}^{n}} + \color{#00f}{{1 \over \pars{2n - 1}2^{n}}} + {2n \over 2n - 1}\,\bracks{% I_{n + 1} - \int_{n}^{n + 1}{\dd x \over \pars{x^{2} + 1}^{n + 1}} } \end{align} The $\color{#00f}{\large blue}$ one is 'the leading term': $ {1 \over \pars{2n - 1}2^{n}} \sim {1 \over n2^{n + 1}} \sim \color{#00f}{\large{1 \over n\,2^{n}}} $

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Divide $\int_{1}^{n}\frac{2^n}{(x^{2}+1)^{n}}dx$ between $1$ and $1+n^{-2/3}$ and between $1+n^{-2/3}$ and $n$. The second part is less than $n \left( \frac{2}{(1+n^{-2/3})^2+1}\right)^n$ which is equivalent to $n \exp (n(-n^{-2/3}+o(n^{-2/3}))$ which can be neglected considering the equivalent we are going to find for the other part.

For the first part, we use $\frac{2}{(1+u)^2+1}=1-u+O(u^2)$ as $u \rightarrow 0$ to evaluate $\int_0^{n^{-2/3}}\left(\frac{2}{(1+u)^2+1}\right)^n du$. This way we get that $e^{-nu-Cnu^2} \leq \left(\frac{2}{(1+u)^2+1}\right)^n \leq e^{-nu+Cnu^2}$ for some constant $C \gt 0$ and for all $u \in [0,1]$. So the first part falls between $e^{-Cn^{-1/3}} \int_0^{n^{-2/3}} e^{-nu} du$ and $e^{Cn^{-1/3}} \int_0^{n^{-2/3}} e^{-nu} du$, so is equivalent to $\int_0^{n^{-2/3}} e^{-nu} du = \frac{1}{n}(1-e^{-n^{1/3}}) \sim 1/n$.

$n^{-2/3}$ could be replaced by any $\epsilon_n$ such that $n \epsilon_n \rightarrow + \infty$ and $n \epsilon_n^2 \rightarrow 0$

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    @Cody: Might I suggest accepting answers you find useful? For each answer there is a check mark button next to the upvote spot. If one answer was particularly helpful, you can accept it to tell the author you thought so, and also so everyone knows the question was answered.2011-03-17
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Since $\int_2^n {\frac{1}{{(x^2 + 1)^n }}\,{\rm d}x} \le \frac{{n - 2}}{{5^n }}$, it suffices to show that $ \frac{1}{{(n + 1)2^n }} \le \int_1^2 {\frac{1}{{(x^2 + 1)^n}}\,{\rm d}x} \le \frac{1}{{(n - 1)2^n }}. $ This follows straight from $ \frac{{2 - x}}{2} \le \frac{1}{{x^2 + 1}} \le \frac{1}{{2x}}. $

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Motivated by the simplicity of integrating from $0$, rather than $1$, substituting $x = 1+u$ gives

$2^n\int_1^n{\frac{dx}{(1+x^2)^n}} = \int_0^{n-1}{\frac{du}{(1+u+u^2/2)^n}}.$

We seek estimates--an upper bound and a lower bound for the integrand--that will be simple to integrate. Dropping the "complicated" $u^2/2$ term gives one obviously integrable estimate; continuing the pattern $1+u+u^2/2 + \cdots + u^k/k! + \cdots = e^u$ leads to another. The resulting inequalities $1 + u < 1 + u + u^2/2 < e^u$ yield (for $n \ge 1$)

$\eqalign{ \frac{1}{n}\left(1 - e^{-n(n-1)}\right) &= \int_0^{n-1}{e^{-n u}du} \cr &\lt \int_0^{n-1}{\frac{du}{(1+u+u^2/2)^n}} \cr &\lt \int_0^{n-1}{\frac{du}{(1+u)^n}} = \frac{1}{n-1}\left(1 - \frac{1}{n^{n-1}}\right) \cr &\lt \frac{1}{n-1}. }$

Both estimates are asymptotically $1/n$, QED.

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This one should be a comment.

As PEV said you could use contour integration to show that $\int_{0}^{1} \frac{1}{(1+x^2)^n} dx$ is of the same order as $\int_{1}^{n} \frac{1}{(1+x^2)^n} dx$

To evaluate it directly you could try the following.

Plug in $x = \tan(\theta)$.

The integral becomes $\displaystyle \int_{\pi/4}^{\tan^{-1}n} \frac{\sec^2 (\theta)}{\sec^{2n} (\theta)} d \theta = \int_{\pi/4}^{\tan^{-1}n} \cos^{2n-2}(\theta) d \theta \approx \int_{\pi/4}^{\pi/2} \cos^{2n-2}(\theta) d \theta = \frac{\text{(Hypergeometric function)}}{2^{n-1}(2n-1)}$.

The approximation tends to better and better asymptotically since $\cos(\theta)$ is bounded.

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    @Sivaram I was trying to solve the problem this way and you beat me to it. Bah! Nevertheless, I guess I wouldn't have known how to come up with the hypergeometric function (which is rather Alien to me), but rather solve it in terms of factorials.2012-02-19