How can I evaluate $\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$?

Question

I'm here wondering if this integral that our math teacher gave us (students) is even possible to evaluate? I just started to study real analysis so I find this very disturbing. Here you go, and if anyone has any idea I will be very grateful.

$$\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$$

Answer 1

The change of variables $x\mapsto nx$ gives $$ \begin{align} \lim_{n\to\infty}\int_n^\infty\frac{n^2\arctan\left(\frac1x\right)}{x^2+n^2}\,\mathrm{d}x &=\lim_{n\to\infty}\int_1^\infty\frac{n\arctan\left(\frac1{nx}\right)}{x^2+1}\,\mathrm{d}x\tag1\\ &=\int_1^\infty\frac{\frac1x}{x^2+1}\,\mathrm{d}x\tag2\\ &=\int_1^\infty\left(\frac1x-\frac{x}{x^2+1}\right)\,\mathrm{d}x\tag3\\ &=\lim_{n\to\infty}\left(\int_1^n\frac1x\,\mathrm{d}x-\int_1^n\frac x{x^2+1}\,\mathrm{d}x\right)\tag4\\ &=\lim_{n\to\infty}\left(\log(n)-\frac12\log\left(\frac{n^2+1}2\right)\right)\tag5\\[3pt] &=\frac12\log(2)\tag6 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto nx$
$(2)$: $0\le\frac1x-n\arctan\left(\frac1{nx}\right)\le\frac1{3n^2x^3}$; therefore, $0\le(2)-(1)\le\frac1{12n^2}$
$(3)$: partial fractions
$(4)$: write the improper integral as a limit
$(5)$: evaluate the integrals
$(6)$: evaluate the limit

Answer 2

Sorry for overlooking some constant, the following is for the integral $\displaystyle\int_{n}^{\infty}\dfrac{n\tan^{-1}(1/x)}{x^{2}+n^{2}}dx$, the integral in question will be addressed in the second part.

By change of variable $u=x/n$, then the integral is \begin{align*} \int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}du. \end{align*} But \begin{align*} \int_{1}^{\infty}\dfrac{1}{1+u^{2}}du<\infty \end{align*} and \begin{align*} \dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}\leq\dfrac{\tan^{-1}1}{1+u^{2}} \end{align*} for all $u\geq 1$, $n\geq 1$, use Lebesgue Dominated Convergence Theorem to conclude that the limit is zero.

It seems that there is an elementary answer.

\begin{align*} &\lim_{n\rightarrow\infty}\int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}du\\ &=\lim_{n\rightarrow\infty}\tan^{-1}u\tan^{-1}(1/(nu))\bigg|_{u=1}^{u=\infty}+\lim_{n\rightarrow\infty}\int_{1}^{\infty}\dfrac{\tan^{-1}u}{1+(1/(nu))^{2}}\dfrac{1}{nu^{2}}du\\ &=\lim_{n\rightarrow\infty}n\int_{1}^{\infty}\dfrac{\tan^{-1}u}{n^{2}u^{2}+1}du, \end{align*} but \begin{align*} n\int_{1}^{\infty}\dfrac{1}{n^{2}u^{2}+1}du&=\int_{n}^{\infty}\dfrac{1}{v^{2}+1}dv\\ &=\dfrac{\pi}{2}-\tan^{-1}n\\ &\rightarrow 0. \end{align*} However, \begin{align*} n\int_{1}^{\infty}\dfrac{\tan^{-1}u}{n^{2}u^{2}+1}du\leq\dfrac{\pi}{2}\cdot n\int_{1}^{\infty}\dfrac{1}{n^{2}u^{2}+1}du. \end{align*}

For the integral in question:

So the integral is $\displaystyle\int_{1}^{\infty}\dfrac{n\tan^{-1}(1/(nu))}{1+u^{2}}$, and we have $n\tan^{-1}(1/(nu))\rightarrow(1/u)$ by L'Hopital rule or some sort, and note that $\tan^{-1}v$ behaves like $v$ for small $|v|$.

And in fact, one deduces using Taylor formula or some sort that $n\tan^{-1}(1/(nu))\leq 1/u$, and that \begin{align*} \int_{1}^{\infty}\dfrac{1}{u(1+u^{2})}\leq\int_{1}^{\infty}\dfrac{1}{1+u^{2}}du<\infty, \end{align*} so Lebesgue Dominated Convergence Theorem implies that the limit in question is then \begin{align*} \int_{1}^{\infty}\dfrac{1}{u(1+u^{2})}du&=\int_{1}^{\infty}\left(\dfrac{1}{u}-\dfrac{1}{1+u^{2}}\right)du\\ &=\lim_{M\rightarrow\infty}\left(\log u\bigg|_{u=1}^{u=M}-\dfrac{1}{2}\log(1+u^{2})\bigg|_{u=1}^{u=M}\right)\\ &=\lim_{M\rightarrow\infty}\left(\dfrac{1}{2}\log 2+\log\dfrac{M}{\sqrt{1+M^{2}}}\right)\\ &=\dfrac{1}{2}\log 2+\log 1\\ &=\dfrac{1}{2}\log 2. \end{align*}

Answer 3

EDIT: This was intended to finish a previous answer by user284331, which contained an error I failed to spot. I'll keep this because it demonstrates a useful technique, but this is solving the wrong integral. Please refer to the other answers instead.

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I'll finish off the other answer using theorems involving the Riemann integral only for completeness - Lebesgue's dominated convergence may not be in the scope of your course (but is the natural way to solve this problem).

We have the sequence $$\int_1^\infty \frac{\arctan \frac{1}{nu}}{1+u^2} \, du$$ and observe the functions $f_n(u) = \frac{\arctan \frac{1}{nu}}{1+u^2}$ converge uniformly to $0$. We would like to show convergence of the integrals to $0$. You will likely come across the theorem that for a sequence of uniformly convergent functions on a bounded interval, the limit of the integrals is the integral of the limit. But here, the issue is that the functions are not defined on a bounded interval.

So, instead split into two regions. We have that $f_n \leq \frac{\arctan 1}{1+u^2}$ for all $n$, and this is an integrable function. So, $\forall \epsilon > 0$, there is some $c$ such that $$\int_c^\infty \frac{\arctan 1}{1+u^2} \, du < \frac{\epsilon}{2}$$ but then $$\begin{align}\int_0^\infty f_n(u)\, du &= \int_0^c f_n(u)\, du + \int_c^\infty f_n(u)\, du \\ &\leq c \cdot \frac{\epsilon}{2c} + \frac{\epsilon}{2} = \epsilon\end{align}$$ by choosing $n$ such that $\|f_n(u)\| < \frac{\epsilon}{2c}$, which exists by uniform convergence.

So, the limit is 0.

Answer 4

I may be totally wrong but I have the feeling that the limit is not $0$.

If you look here, you will find (Theorem II.1) very nice bounds for $\tan^{-1}(y)$ $$\frac{y}{\frac{4}{\pi ^2}+\sqrt{\left(1-\frac{4}{\pi ^2}\right)^2+\frac{4 y^2}{\pi ^2}}} < \tan^{-1}(x) <\frac{y}{1-\frac{6}{\pi ^2}+\sqrt{\left(\frac{6}{\pi ^2}\right)^2+\frac{4 y^2}{\pi ^2}}}$$ Using $y=\frac 1x$, we face, for each bound, the problem of $$I=\int \frac{n^2}{\left(n^2+x^2\right) \left(a x+\sqrt{b^2 x^2+c}\right)}\,dx$$ which is tedious but doable (I shall not write the expressions here, at least for the time being).

Computing from $I$ $$J=\int_n^\infty \frac{n^2}{\left(n^2+x^2\right) \left(a x+\sqrt{b^2 x^2+c}\right)}\,dx$$ and developing as Taylor series for infinitely large values of $n$, we get for the left bound $$J=\frac{\log (2)}{2}+\frac{\log (2)-1}{\left(\pi ^2-4\right) n^2}+O\left(\frac{1}{n^4}\right)$$ and for the right bound $$J=\frac{\log (2)}{2}+\frac{\log (2)-1}{6 n^2}+O\left(\frac{1}{n^4}\right)$$

So, using the squeeze theorem, the limit looks to be $\frac{1}{2}\log (2)$.

For a sanity check, let $n=10^k$ and numerical integration of the given function leads to $$\left( \begin{array}{cc} k & \text{result} \\ 1 & 0.346064090433349 \\ 2 & 0.346568476259452 \\ 3 & 0.346573539137855 \\ 4 & 0.346573589768551 \\ 5 & 0.346573590274858 \\ 6 & 0.346573590279922 \\ 7 & 0.346573590279972 \\ 8 & 0.346573590279973 \\ 9 & 0.346573590279973 \end{array} \right)$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\int_{n}^{\infty}{n^{2}\arctan\pars{1/x} \over x^{2} + n^{2}} \,\dd x & \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \lim_{n \to \infty}\int_{0}^{1/n}{\arctan\pars{x} \over x^{2} + 1/n^{2}} \,\dd x \\[5mm] & = \lim_{n \to \infty}\bracks{% \int_{0}^{1/n}{x \over x^{2} + 1/n^{2}}\,\dd x + \int_{0}^{1/n}{\arctan\pars{x} - x \over x^{2} + 1/n^{2}}\,\dd x} \end{align}

With $\ds{0 < x < 1/n}$, note that $\ds{\exists\ \xi \mid 0 < \xi < x}$ which satisfies

$\ds{\verts{\arctan\pars{x} - x} = x\,{\xi^{2} \over 1 + \xi^{2}} < {1 \over n^{3}}}$ such that

$$ 0 < \verts{\int_{0}^{1/n}{\arctan\pars{x} - x \over x^{2} + 1/n^{2}}\,\dd x} < {1 \over n^{3}}\verts{\int_{0}^{1/n}{\dd x \over x^{2} + 1/n^{2}}} = {1 \over n^{3}}\,{n\pi \over 4} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0} $$

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Then \begin{align} \lim_{n \to \infty}\int_{n}^{\infty}{n^{2}\arctan\pars{1/x} \over x^{2} + n^{2}} \,\dd x & = \lim_{n \to \infty}\int_{0}^{1/n}{x \over x^{2} + 1/n^{2}}\,\dd x = \lim_{n \to \infty} \bracks{{1 \over 2}\ln\pars{1/n^{2} + 1/n^{2} \over 1/n^{2}}} \\[5mm] = \bbx{{1 \over 2}\,\ln\pars{2}} \approx 0.3466 \end{align}