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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?


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You could probably use contour integration. – PEV Feb 2 '11 at 21:29
For the computation it is better to compute $n 2^n$ times your integral than to compare very small values. – Plop Feb 2 '11 at 21:41
I believe there is a nice closed formula for the integral but I have no time to investigate further. – Listing Feb 2 '11 at 21:59
Thank you very much. I actually got that far with the tan sub, but did not know about the hypergeometric. It's probably me, but shouldn't that be $$sec^{2n}(\theta)$$ in the denominator of the first integrand?. Because $$(1+tan^{2}(\theta))^{n}=(sec^{2}(\theta))^{n}$$. Anyway, thanks to both of you for your fast reply and assistance. – Cody Feb 2 '11 at 22:26
@Cody: Sorry for my mistake yes you are correct. I have edited my answer accordingly. – user17762 Feb 2 '11 at 22:51
up vote 8 down vote accepted

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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Thanks very much. – Cody Feb 4 '11 at 16:48

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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Thank you very much – Cody Feb 3 '11 at 15:31
@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. – Eric Naslund Mar 17 '11 at 16:15

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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Thank you very much. – Cody Feb 3 '11 at 15:32

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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Sorry, I don't immediately see it, what would show that it is of the same order? (using contour integration) – Jonas Teuwen Feb 2 '11 at 23:20
It would appear this problem was more complicated than I had initially thought. I am going to try and learn more about hypergeometric and how to interpret what $$hypergeom\left(\left[\frac{1}{2},\frac{3}{2}-n\right],\left[\frac{3}{2}\right]‌​,\frac{1}{2}\right)$$. This means very little without knowing how to convert it to some sort of numerical solution, given an 'n' value of course. I ran $$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}cos^{2n-2}(\theta)d{\theta}$$ through Maple and the above hyper is what it gave me. Thanks again to all. – Cody Feb 3 '11 at 15:38
@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. – Pedro Tamaroff Feb 19 '12 at 0:35

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