Please show me how to solve this problem.
I tried to make the substitution $x/n=y$ but I don't know how to go further.

(a) Prove that $$\lim_{n\to\infty}n\int_0^n\frac{\arctan\frac{x}{n}}{x(x^2+1)}\,dx=\frac{\pi}{2}$$

(b) Find the limit $$\lim_{n\to\infty}n\left(n\int_0^n\frac{\arctan\frac{x}{n}}{x(x^2+1)}\,dx-\frac{\pi}{2}\right)$$

  • $\begingroup$ May you write down your question directly in your post? Someone might be not willing to, or have internet problem with linking to the image. $\endgroup$ – Cave Johnson Sep 24 '16 at 12:55
  • $\begingroup$ I don't know how to write in Latex. $\endgroup$ – user371863 Sep 24 '16 at 12:55
  • 2
    $\begingroup$ It takes a couple of minutes to learn and quite frankly I think it's a bit rude to expect people to help you when you haven't put in effort to help them. $\endgroup$ – Zestylemonzi Sep 24 '16 at 12:59
  • $\begingroup$ Some tips here. $\endgroup$ – Cave Johnson Sep 24 '16 at 13:00
  • $\begingroup$ See math.stackexchange.com/questions/1720872/… and the rest should be easy. $\endgroup$ – abstract Sep 24 '16 at 13:09

About the first integral, we have $$ \int_{n}^{+\infty}\frac{dx}{1+x^2}=O\left(\frac{1}{n}\right) $$ and for any $x\in\mathbb{R}^+$ $$ \lim_{n\to +\infty}\frac{\arctan\frac{x}{n}}{\frac{x}{n}}=1 $$ hence by the dominated convergence theorem $$ \lim_{n\to +\infty}\int_{0}^{n}\frac{\arctan\frac{x}{n}}{\frac{x}{n}}\cdot\frac{dx}{1+x^2}=\int_{0}^{+\infty}\frac{dx}{1+x^2}=\frac{\pi}{2}.$$ In a similar way, since in a right neighbourhood of the origin we have $$ \frac{\arctan x}{x} = 1-\frac{x^2}{3}+\frac{x^4}{5}-\ldots $$ the second limit equals $$ -1+\sum_{k\geq 1}\frac{(-1)^k}{(2k-1)(2k+1)}=\color{red}{-\frac{\pi+2}{4}}.$$ Explanation: $$\begin{eqnarray*}\lim_{n\to +\infty}n\left(\int_{0}^{n}\frac{\arctan\frac{x}{n}}{\frac{x}{n}}\cdot\frac{dx}{1+x^2}-\frac{\pi}{2}\right)&=&\lim_{n\to +\infty}n\left(\int_{0}^{n}\left(\frac{\arctan\frac{x}{n}}{\frac{x}{n}}-1\right)\cdot\frac{dx}{1+x^2}-\arctan\frac{1}{n}\right)\\&=&-1+\lim_{n\to +\infty}n^2\int_{0}^{1}\left(\frac{\arctan x}{x}-1\right)\cdot\frac{dx}{1+n^2 x^2}\\&=&-1+\lim_{n\to +\infty}\int_{0}^{1}\left(\frac{\arctan x}{x}-1\right)\cdot\frac{dx}{x^2+\frac{1}{n^2}}\\(DCT)\quad &=& -1+\int_{0}^{1}\left(\frac{\arctan x}{x}-1\right)\frac{dx}{x^2}\\(Taylor)\quad&=&-1+\int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^n x^{2n-2}}{2n+1}\,dx\\&=&-1+\frac{1}{2}\sum_{k\geq 1}(-1)^k\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\\&=&-1+\frac{1}{2}\left[-\arctan(1)-\left(\arctan(1)-1\right)\right]\\&=&-\arctan(1)-\frac{1}{2}.\end{eqnarray*}$$

  • $\begingroup$ Thank you.This is the only solution? $\endgroup$ – user371863 Sep 24 '16 at 21:35
  • $\begingroup$ I guess there are plenty of equivalent alternatives, as usual, but it looks to me that the DCT provides an efficient way. $\endgroup$ – Jack D'Aurizio Sep 24 '16 at 23:05
  • $\begingroup$ Anyway, I would be glad to see the downvoter explain his downvote. $\endgroup$ – Jack D'Aurizio Sep 24 '16 at 23:06
  • $\begingroup$ Since this problem does not have an official solution yet, I do not think it is a bad idea to have a complete solution here on MSE. $\endgroup$ – Jack D'Aurizio Sep 29 '16 at 20:33

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