证明

$$I= \lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)=\frac{6\ln{2}}{\pi}+\frac{2\gamma}{\pi}+\frac{2}{\pi}\cdot\sum_{k=1}^{\infty}\frac{1}{2k+1}\ln\left(1+\frac{1}{k} \right)$$

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证明:
我们应该有结论:对定义在$\left[0,\frac{\pi}{2}\right]$任意连续函数$f(x)$,有
$$\lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}}|\sin nx|f(x)dx=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}f(x)dx$$
特别的，对于奇数的$n$也成立.这时
$$\lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}}|\sin(2n+1)x|\left(\frac{1}{\sin{x}}-\frac{1}{x}\right)dx=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{\sin{x}}-\frac{1}{x} \right)dx$$
注意到
$$\frac{1}{\sin{x}}=\frac{1}{x}+\sum_{n=1}^{\infty}(-1)^{n}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)$$
就有
$$\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{\sin{x}}-\frac{1}{x} \right)dx=\sum_{n=1}^{\infty}(-1)^{n}\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)dx=\frac{2}{\pi}\ln\frac{4}{\pi}$$
而
$$\int_{0}^{\frac{\pi}{2}}\frac{|\sin(2n+1)x|}{x}dx=\int_{0}^{\frac{\pi}{2}(2n+1)}\frac{|\sin{x}|}{x}dx=\int_{0}^{n\pi}\frac{|\sin{x}|}{x}dx+\int_{n\pi}^{n\pi+\frac{\pi}{2}}\frac{|\sin{x}|}{x}dx$$
同时
$$\int_{n\pi}^{n\pi+\frac{\pi}{2}}\frac{|\sin{x}|}{x}dx\sim o\left(\frac{1}{n}\right) \qquad (n\to\infty)$$
所以
$$\int_{0}^{\frac{\pi}{2}}\frac{|\sin(2n+1)x|}{x}dx=\int_{0}^{n\pi}\frac{|\sin{x}|}{x}dx+o\left(\frac{1}{n}\right) \qquad (n\to\infty)$$
$$\begin{align*} \int_{0}^{n\pi}\frac{|\sin{x}|}{x}dx&=\int_{0}^{\pi}\left(\sum_{k=0}^{n-1}\frac{1}{x+k\pi}\right)\sin{x}dx\\ &=\int_{0}^{1}\left(\sum_{k=0}^{n-1}\frac{1}{x+k}\right)\sin{\pi x}dx\\ &=\int_{0}^{1}\sin{\pi x}\left(\psi(x+n)-\psi(x)\right)dx \end{align*}$$
这里
$$\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$$
我们知道它有渐进展开式
$$\psi(x)=\ln{x}-\frac{1}{2x}-\sum_{r=1}^{n}\frac{(-1)^{r-1}B_{r}}{2r}x^{-2r}+O(x^{-2n-2})$$
而
$$\zeta(1-2m)=\frac{(-1)^{m}B_{m}}{2m}\qquad (m\geq 1)$$
所以
$$\psi(x)=\ln{x}-\frac{1}{2x}+\sum_{k=1}^{\infty}\frac{\zeta(1-2k)}{x^{2k}}+O(x^{-2n-2})$$
另一方面，有
$$\int_0^1 \sin(\pi x)\psi(x+n) dx = \frac{2}{\pi} \ln n + O\left(\frac{1}{n}\right)\qquad (n\to\infty)$$
于是
$$\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)=\frac{2}{\pi}\ln\frac{4}{\pi}-\int_{0}^{1}\sin{\pi x}\psi(x)dx$$
下面来弄后面那个积分,分部积分有
$$\int_{0}^{1}\sin\pi{x}\cdot \psi(x)dx=\int_{0}^{1}\sin{\pi x}d\ln\Gamma(x)=-\pi\int_{0}^{1}\cos\pi x\ln\Gamma(x)dx$$
然后把$\ln\Gamma(x)$的Fourier展开，就是
$$\ln\Gamma(x)=\frac{1}{2}\ln\frac{\pi}{\sin{\pi x}}+[\gamma+\ln{2\pi}]\left(\frac{1}{2}-x\right)+\frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\ln{k}}{k}\sin(2\pi kx)$$
接着就是3个式子

$$\int_0^1 \cos\pi x\ln\frac{\pi}{\sin(\pi x)} dx = 0  \tag{1}$$
$$\int_{0}^{1}\cos\pi x\left(\frac{1}{2}-x\right)dx=\frac{2}{\pi^2} \tag{2}$$
  $$\int_{0}^{1}\cos\pi x\sin2\pi kx dx=\frac{4k}{(4k^2-1)\pi} \tag{3}$$

代进去计算得到
$$\begin{align*}   I&=\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)\\ &=\frac{2}{\pi}\ln\frac{4}{\pi}+\pi\left[\frac{2}{\pi^2}(\gamma+\ln2\pi)+\frac{4}{\pi^2}\sum_{k=2}^{\infty}\frac{\ln{k}}{4k^2-1} \right]\\ &=\frac{2}{\pi}\left[\ln{8}+\gamma+\sum_{k=2}^{\infty}\ln{k}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\right]\\ &=\frac{6\ln{2}}{\pi}+\frac{2\gamma}{\pi}+\frac{2}{\pi}\cdot\sum_{k=1}^{\infty}\frac{1}{2k+1}\ln\left(1+\frac{1}{k} \right) \end{align*}$$