I've been trying for the last few hours to solve the following problem, which looks like this :

Find the limit $l$ : $$l=\mathop {\lim }\limits_{n \to \infty } \,\,\,n\int\limits_0^n {\frac{{\arctan (\frac{x}{n})}}{{x(x^2 + 1)}}} \,dx$$

Use the result to compute: $$ \mathop {\lim }\limits_{n \to \infty } \,\,\,n\,(\,n\int\limits_0^n {\frac{{\arctan (\frac{x}{n})}}{{x(x^2 + 1)}}} \,dx - \frac{\pi }{2}) $$ I've tried Taylor expansion, partial fraction decomposition, but I can't really find anything useful. Some help would be really appreciated. I'm much more interested in the method than in the actual result.

  • $\begingroup$ Let $~t=\dfrac xn$ $\endgroup$ – Lucian Dec 3 '14 at 0:57

Hint/suggestion: do the change of variables $u=\frac{x}{n}$, then use Lebesgue's dominated convergence theorem to compute the limit of the resulting integral.

  • $\begingroup$ this is actually an approach I discarded right at the beginning, never thought it'd be this simple heh...thank you a lot ! $\endgroup$ – browning Dec 3 '14 at 0:59
  • $\begingroup$ It's not going to be this straightforward, though -- the limit is most likely $\ell=\frac{\pi}{2}$, which is a clue that a direct application of the TCD won't cut it (if you could apply it, the limit would be 0, since $g(n,u) \xrightarrow[n\to\infty]{} 0$). $\endgroup$ – Clement C. Dec 3 '14 at 1:01
  • $\begingroup$ My answer sheet actually says $l = \frac{\pi }{2}$ $\endgroup$ – browning Dec 3 '14 at 1:04
  • $\begingroup$ Yes (you can guess hit from the second part, which at first glance asks to compare the rate of convergence to $1/n$). $\endgroup$ – Clement C. Dec 3 '14 at 1:05
  • $\begingroup$ $ \int\limits_0^1 {\frac{{\arctan (t)}}{{t(n^2 t^2 + 1)}}} \,dt$ . The integral looks like this upon using the suggested substitution, but I'm not really sure where to go from here. $\endgroup$ – browning Dec 3 '14 at 1:11

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