# Limit of a definite integral with parameter (2)

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.

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

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

• 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?
– E.Z
Jan 16, 2018 at 13:02
• I meant randomly.
– E.Z
Jan 16, 2018 at 13:34
• 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. Jan 16, 2018 at 13:48
• 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.
– E.Z
Jan 16, 2018 at 14:01
• Yes, your approach and the final result are correct. By using L'Hopital you avoid the integration. Jan 16, 2018 at 14:03