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Given a limit: $$\lim_{n\to+\infty}\frac{1}{n+1}\int_{0}^{n}\arctan(x)\,dx = \alpha$$ Find the value of $\alpha$.

Well, the inner integral equals: $$\int_{0}^{n}\arctan(x)\,dx = n\cdot\arctan(n)\bigr|_{0}^{+\infty} - \frac{1}{2}\ln\left(n^2+1\right)\Bigr|_{0}^{+\infty}$$ Then, I rewrite the limit: $$\lim_{n\to+\infty}\frac{n\cdot\arctan(n)\bigr|_{0}^{+\infty} - \frac{1}{2}\ln\left(n^2+1\right)\bigr|_{0}^{+\infty}}{n+1}$$ By inserting $+\infty$ and $0$ in the limit I obtain $\alpha = 0$. However, this is not a correct answer.

How would I proceed? Thank you.

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    $\begingroup$ The inner integral is computed incorrectly. $\endgroup$
    – Ѕᴀᴀᴅ
    Jan 16, 2018 at 12:23
  • $\begingroup$ Can you detail the computation of the limit? In particular for both terms of the sum. $\endgroup$
    – mathcounterexamples.net
    Jan 16, 2018 at 12:23
  • $\begingroup$ Hint: if $f(x) \to a \ (x \to +\infty)$, then $\frac{1}{n} \int_0^n f(x) \,\mathrm{d}x \to a \ (x \to +\infty)$. $\endgroup$
    – Ѕᴀᴀᴅ
    Jan 16, 2018 at 12:25
  • $\begingroup$ @samjoe L'Hospital's rule applies for the most trivial cases only. For more general cases, L'Hospital's rule may not work. $\endgroup$
    – Ѕᴀᴀᴅ
    Jan 16, 2018 at 12:31

1 Answer 1

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Hint. There is no need to evaluate the integral. By using L'Hopital and the Fundamental Theorem of calculus, we have that $$\lim_{t\to+\infty}\frac{1}{t+1}\int_{0}^{t}\arctan(x)dx= \lim_{t\to+\infty}\arctan(t).$$

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  • $\begingroup$ Thanks for the hint. If I tried to divide both numerator and denominator by $n$, I would get the correct $pi/2$ answer. Would that be a solid approach, or just a how-to-get-a-correct-answer method? $\endgroup$
    – E.Z
    Jan 16, 2018 at 13:02
  • $\begingroup$ I meant randomly. $\endgroup$
    – E.Z
    Jan 16, 2018 at 13:34
  • $\begingroup$ I am not sure that I understood your question. In order to apply L'Hopital we use the real variable $t$. The limit along the subsequence $t=n\in\mathbb{N}$ is the same. $\endgroup$
    – Robert Z
    Jan 16, 2018 at 13:48
  • $\begingroup$ I meant if I sticked to the initial approach. After I rewrote the limit I could divide top and bottom parts by $n$ and still get $pi/2$ answer. $\endgroup$
    – E.Z
    Jan 16, 2018 at 14:01
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    $\begingroup$ Yes, your approach and the final result are correct. By using L'Hopital you avoid the integration. $\endgroup$
    – Robert Z
    Jan 16, 2018 at 14:03

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