<s>[INST] 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?
 [/INST] 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 $$\lim_{n\rightarrow\infty}\left(\frac{1}{3\pi}\int_\pi^{2\pi}\frac{x}{\arctan(nx)} \ dx\right)^n=e^{\frac{4}{3 \pi^2}}$$
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