Calculating a limit of integral

Since $\arctan$ is increasing we get $$\frac{1}{ n^2}\int_{n\pi}^{2n\pi }\frac{t}{\arctan 2\pi n}dt \leq I \leq \frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan n \pi}dt$$

Now we can calculate both sides and we get

$$ \frac{4\pi ^2 - \pi^2}{2\arctan 2\pi n} \leq I \leq \frac{4\pi ^2 - \pi^2}{2\arctan \pi n} $$ So $$ \frac{3\pi ^2 }{2\arctan 2\pi n} \leq I \leq \frac{3\pi ^2}{2\arctan \pi n} $$ At this point if you do not have to compute $I^n$ since $\lim _{x \rightarrow \infty} \arctan (x) = \pi/2$, you will get $ \lim _{n \rightarrow \infty} I= 3\pi$.

Which implies that $\frac{I}{3\pi} \rightarrow 1$.

So the limit $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$ is of the form $1^{\infty}$. We will apply L'H rule to calculate that. We change variable and instead of $n$ we work with $t$.

By doing the standard method to apply L'H rule it suffices to calcualte $$ \lim _{t\rightarrow \infty}t \ln ( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx ) $$ We calcualte \begin{align*} \lim _{t\rightarrow \infty}t \ln ( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx ) &= \lim _{t\rightarrow \infty}\frac{ \ln ( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx ) }{\frac{1}{t}} \\ &= \lim _{t\rightarrow \infty}\frac{ \frac{1} {( \frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(tx)} \ dx )} \frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{x}{\arctan(tx)} \right )' \ dx \ }{-\frac{1}{t^2}} \\ &= \lim _{t\rightarrow \infty}\frac{\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{x}{\arctan(tx)^2} \right ) \frac{x}{1+t^2x^2} \ dx \ }{\frac{1}{t^2}} \\ &=\lim _{t\rightarrow \infty}\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{x}{\arctan(tx)^2} \right ) \frac{t^2x}{1+t^2x^2} \ dx \\ &= \lim _{t\rightarrow \infty}\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{1}{\arctan(tx)^2} \right ) \frac{t^2x^2}{1+t^2x^2} \ dx \\ &= \lim _{t\rightarrow \infty}\frac{1}{3 \pi}\int_\pi^{2\pi}\left (\frac{1}{\left ( \frac{\pi}{2} \right )^2} \right )\cdot 1 \ dx \, \,\rm{by\,dominated \,convergence} \\ &=\frac{4}{3 \pi^2} \end{align*}

So we conclude the final answer is $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n=e^{\frac{4}{3 \pi^2}}$$

Asymptotics of the integral:

When $x$ is large, \begin{align} \frac{1}{\arctan{x}} =&\frac{2}{\pi}\frac{1}{1-\frac{2}{\pi}\arctan\left(\frac{1}{x}\right)}\tag1\\ =&\frac{2}{\pi}\frac{1}{1-\frac{2}{\pi}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)}\tag2\\ =&\frac{2}{\pi}\left(1+\frac{2}{\pi}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)+\frac{4}{\pi^2}\left(\frac{1}{x}-\frac{1}{3x^3}+\cdots\right)^2+\cdots\right)\tag3\\ =&\frac{2}{\pi}+\frac{4}{\pi^2 x}+\frac{8}{\pi^3 x^2}+\mathcal{O}\left(x^{-3}\right)\tag4 \end{align} Then \begin{align} \frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(nx)}{\rm d}x =&\frac{1}{3\pi}\int^{2\pi}_\pi\left(\frac{2x}{\pi}+\frac{4}{\pi^2 n}+\frac{8}{\pi^3 n^2x}+\cdots\right)\ {\rm d}x\\ =&\ 1+\frac{4}{3\pi^2n}+\frac{8\ln{2}}{3\pi^4n^2}+\mathcal{O}\left(n^{-3}\right) \end{align}

Computing the limit:

Your limit is thus \begin{align} \lim_{n\to\infty}\left(\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(nx)}{\rm d}x\right)^n =&\ \exp\left\{\lim_{n\to\infty}n\ln\left(\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(nx)}{\rm d}x\right)\right\}\tag5\\ =&\ \exp\left\{\lim_{n\to\infty}n\left(\frac{4}{3\pi^2n}+\mathcal{O}\left(n^{-2}\right)\right)\right\}\tag6\\ =&\ \large{\color{red}{\exp\left(\frac{4}{3\pi^2}\right)}}\normalsize\approx1.144645419236050\tag7\\ \end{align}

Numerical Verification:

Using Mathematica I am getting $$\left(\frac{1}{3\pi}\int^{2\pi}_\pi\frac{x}{\arctan(999999999x)}{\rm d}x\right)^{999999999}\approx1.144645419247325$$ which is consistent with the derived result.

Explanation:

$(1)$: Used the fact that $\displaystyle\arctan{x}=\frac{\pi}{2}-\arctan\left(\frac{1}{x}\right)$.
$(2)$: Used the series for $\arctan{x}$.
$(3)$: Used a geometric series.
$(4)$: Expanded the terms.
$(5)$: Used the fact that $\displaystyle\lim_{n\to\infty}a_n^{b_n}=\exp\lim_{n\to\infty}b_n\ln(a_n)$.
$(6)$: Used the series for $\ln(1+x)$.
$(7)$: As $n\to\infty$, only the constant term remains.