<s>[INST] How to prove: $\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left\vert\frac{\sin{(n+1)x}}{\sin{x}}\right\vert dx-\frac{2\ln{n}}{\pi}\right)$ show that 

$$\mathop {\lim }\limits_{n \to \infty } \left( {\int\limits_0^{\frac{\pi }{2}} {\left\vert\frac{{\sin \left( {2n + 1} \right)x}}{{\sin x}}\right\vert\,dx - \frac{{2\ln n}}{\pi }} } \right) = \frac{{6\ln 2}}{\pi } + \frac{{2\gamma }}{\pi } + \frac{2}{\pi }\sum\limits_{k = 1}^\infty  {\frac{1}{{2k + 1}}\ln \left( {1 + \frac{1}{k}} \right)}\cdots (1) $$

I can prove $(1)$ it exsit it.and  also it is well kown that
$$I_{n}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin{(2n+1)x}}{\sin{x}}dx=\dfrac{\pi}{2}$$

proof:$$I_{n}-I_{n-1}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin{(2n+1)x}-\sin{(2n-1)x}}{\sin{x}}dx=2\int_{0}^{\frac{\pi}{2}}\cos{(2nx)}dx=0$$
  so
  $$I_{n}=I_{n-1}=\cdots=I_{0}=\dfrac{\pi}{2}$$
  But I can't prove  $(1)$,Thank you 

 [/INST] Notice for any continuous function $f(x)$ on $[0,\frac{\pi}{2}]$, we have:
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| f(x) dx = \frac{2}{\pi}\int_0^{\frac{\pi}{2}} f(x) dx$$
Apply this to $\frac{1}{\sin x} - \frac{1}{x}$, we get
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| \Big(\frac{1}{\sin x} - \frac{1}{x} \Big) dx
= \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \Big(\frac{1}{\sin x} - \frac{1}{x} \Big) dx\\
= \frac{2}{\pi} \left[\log\left(\frac{\tan(\frac{x}{2})}{x}\right)\right]_0^{\frac{\pi}{2}}
= \frac{2}{\pi} \left[\log\frac{2}{\pi} - \log{\frac12}\right] = \frac{2}{\pi} \log\frac{4}{\pi}
\tag{*1}$$
So it suffices to figure out the asymptotic behavior of following integral:
$$\int_0^{\frac{\pi}{2}}  \frac{|\sin((2n+1)x)|}{x} dx
= \int_0^{\pi(n+\frac12)} \frac{|\sin x|}{x} dx = \int_0^{\pi n} \frac{|\sin x|}{x} dx + O(\frac{1}{n})
$$
We can rewrite the rightmost integral as
$$\int_0^{\pi} \sin x \Big( \sum_{k=0}^{n-1} \frac{1}{x+k\pi} \Big) dx
= \int_0^1 \sin(\pi x) \Big( \sum_{k=0}^{n-1} \frac{1}{x+k} \Big) dx\\
= \int_0^1 \sin(\pi x) \Big( \psi(x+n) - \psi(x) \Big) dx
\tag{*2}
$$
where $\displaystyle \psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ is the 
digamma function.
Using following asymptotic expansion of $\psi(x)$ for large $x$:
$$\psi(x) = \log x - \frac{1}{2x} + \sum_{k=1}^{\infty}\frac{\zeta(1-2k)}{x^{2k}}$$
It is easy to verify
$$\int_0^1 \sin(\pi x)\psi(x+n) dx = \frac{2}{\pi} \log n + O(\frac{1}{n})\tag{*3}$$.
Substitute $(*3)$ into $(*2)$ and combine it with $(*1)$, we get
$$\lim_{n\to\infty} \left(\int_0^{\frac{\pi}{2}} \left|\frac{\sin((2n+1)x)}{\sin x}\right| dx - \frac{2}{\pi} \log n\right) = \frac{2}{\pi} \log\frac{4}{\pi} - \int_0^1 \sin(\pi x)\psi(x) dx \tag{*4}$$
To compute the rightmost integral of $(*4)$, we first integrate it by part:
$$\int_0^1 \sin(\pi x)\psi(x) dx = \int_0^1 \sin(\pi x)\,d\log\Gamma(x) =
-\pi\int_0^1 \cos(\pi x)\log\Gamma(x) dx
$$
We then apply following result$\color{blue}{^{[1]}}$

Kummer (1847) Fourier series for $\log\Gamma(x)$ for $x \in (0,1)$
  $$\log\Gamma(x) = \frac12\log\frac{\pi}{\sin(\pi x)} + (\gamma + \log(2\pi))(\frac12 - x) + \frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\log k}{k}\sin(2\pi k x)$$

Notice


*

*$\displaystyle \int_0^1 \cos(\pi x)\log \frac{\pi}{\sin(\pi x)} dx = 0\quad$ because of symmtry.

*$\displaystyle \int_0^1 \cos(\pi x)\Big(\frac12 - x\Big) dx = \frac{2}{\pi^2}$

*$\displaystyle \int_0^1 \cos(\pi x)\sin(2\pi k x) dx = \frac{4k}{(4k^2-1)\pi} $
We can evaluate RHS of $(*4)$ as
$$\begin{align}
\text{RHS}_{(*4)} = & \frac{2}{\pi}\log\frac{4}{\pi} + \pi \left[ 
\Big(\gamma + \log(2\pi)\Big)\frac{2}{\pi^2}
+ \frac{4}{\pi^2}\sum_{k=2}^{\infty}\frac{\log k}{4k^2-1}
\right]\\
= & \frac{2}{\pi}\left[\log 8 + \gamma + \sum_{k=2}^{\infty}\log k \left(\frac{1}{2k-1}-\frac{1}{2k+1}\right) \right]\\
= & \frac{6\log 2}{\pi} + \frac{2\gamma}{\pi} + \frac{2}{\pi}\sum_{k=1}\frac{\log(1+\frac{1}{k})}{2k+1}
\end{align}$$
Notes
$\color{blue}{[1]}$ For more infos about Kummer's Fourier series, please see
following paper by Donal F. Connon.
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