Calculating a limit of integral Computing the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n$$
I made the substiution $t=nx$ then, we have: $$I=\frac{1}{n^2}\int_{n\pi}^{2n\pi}\frac{t}{\arctan t}dt$$ where $I$ is the inside integral. How would you continue?
 A: 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 $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n=e^{\frac{4}{3 \pi^2}}$$
A: 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.
A: you might try  $tan (y )= nx $ so  $dy (1 + tan^2 y) = n dx $
so 
$$I = \lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n
=\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_{actan(n\pi}^{arctan(2n\pi)}\frac{tan (y )}{n y} \ dy (1 + tan^2 y)\right)^n
$$
and you can have sup et inf or use $arctan(n\pi)=\frac{\pi}{2}-\frac{1}{n\pi}$ from $tan(\frac{\pi}{2}-\alpha)= \frac{sin(\frac{\pi}{2}-\alpha)}{cos(\frac{\pi}{2}-\alpha)} = \frac{cos(\alpha)}{sin(\alpha)} \ approx \frac{1}{\alpha}$
