Limit of an Integral including inverse tangent

$\begin{aligned} L & = \lim_{n \to \infty} \int_0^n \frac {x^{2017}\tan^{-1}x}{n^{2018}} dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \lim_{n \to \infty} \frac {x^{2018}\tan^{-1}x}{2018n^{2018}} \bigg|_0^n - \frac 1{2018n^{2018}}\color{#3D99F6}\int_0^n \frac {x^{2018}}{1+x^2} dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \frac \pi {4036} - \lim_{n \to \infty} \frac 1{2018n^{2018}}\color{#3D99F6}\int_0^{\tan^{-1}n} \tan^{2018} \theta\ d\theta \\ & = \frac \pi {4036} - \lim_{n \to \infty} \frac {\color{#3D99F6}I_{2018}}{2018n^{2018}} \end{aligned}$

We can solve the integral $\displaystyle I_{2018} = \int_0^{\tan^{-1}n} \tan^{2018} \theta\ d\theta$ using the reduction formula $\displaystyle \int \tan^m x = \frac {\tan^{m-1}x}{m-1} - \int \tan^{m-2} x\ dx$ or $I_m = \dfrac {\tan^{m-1}x}{m-1} - I_{m-2}$ . Therefore,

$\begin{aligned} I_{2018} & = \frac {\tan^{2017}\theta }{2017} - I_{2016} \\ & = \frac {\tan^{2017}\theta}{2017} - \frac {\tan^{2015}\theta}{2015} + I_{2014} \\ & = \sum_{k=1}^{1009} \frac {(-1)^{k+1}\tan^{2k-1}\theta}{2k-1} - \color{#3D99F6} I_0 & \small \color{#3D99F6} \text{Note that }I_0 = \int d\theta = \theta + C \end{aligned}$

$\begin{aligned} \implies \lim_{n \to \infty} \frac {I_{2018}}{2018n^{2018}} \bigg|_0^n & = \lim_{n \to \infty} \frac 1{2018n^{2018}} \sum_{k=1}^{1009} \frac {(-1)^{k+1}\tan^{2k-1} \theta} {2k-1} - \frac {\color{#3D99F6}\theta}{2018n^{2018}} \bigg|_0^{\tan^{-1} n} \\ & = \lim_{n \to \infty} \frac 1{2018n^{2018}} \sum_{k=1}^{1009} \frac {(-1)^{k+1}n^{2k-1}}{2k-1} - \frac {\tan^{-1} n}{2018n^{2018}} \\ & = \lim_{n \to \infty} \frac 1{2018} \sum_{k=1}^{1009} \frac {(-1)^{k+1}}{(2k-1)n^{2k-1}} - \frac {\frac \pi 2}{2018n^{2018}} \\ & = 0 \end{aligned}$

Therefore, $\displaystyle L = \frac \pi {4036} - \cancel{\lim_{n \to \infty} \frac {I_{2018}}{2018n^{2018}}}^0 = \boxed{\dfrac \pi {4036}}$ .

The easiest solution is via L'hopital's rule. The denominator, $n^{2018}$ , diverges and we can prove the numerator diverges because for big enough $x$ , $x^{2017} \arctan(x) > 1$ so divergence follows by comparison with $\int_0^n dx$ . So we have a $\frac{\infty}{\infty}$ case and if we apply L'hopital's we find this limit is equal to $\lim_{n \to \infty} \frac{n^{2017} \arctan(n)}{2018n^{2017}} = \lim_{n \to \infty} \frac{\arctan(n)}{2018} = \frac{\pi}{2 \times 2018} = \frac{\pi}{4036}$ .

Other ways to reach this result are by using the Laurent series for $\arctan(x)$ which tells us that $\arctan(x) = \frac{\pi}{2} + O(\frac{1}{x})$ or by using the squeeze theorem starting from the inequality $\frac{\pi}{2} - \frac{1}{x} < \arctan(x) < \frac{\pi}{2}$ .