How to prove limit of integral is $\frac{1}{n2^n}$?

Question

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

Answer 1

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 $n \epsilon_n^2 \rightarrow 0$

Answer 2

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 $$ \frac{{2 - x}}{2} \le \frac{1}{{x^2 + 1}} \le \frac{1}{{2x}}. $$

Answer 3

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.

Answer 4

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.

Answer 5

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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': $$ {1 \over \pars{2n - 1}2^{n}} \sim {1 \over n2^{n + 1}} \sim \color{#00f}{\large{1 \over n\,2^{n}}} $$