Estimate the integral of the absolute value of the Dirichlet kernel $D_n=\sum_{k=-n}^n e^{k i x}=\sin ((N+1/2)x)/\sin(x/2)$
Prove that $$L_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}|D_n|dx=\frac{4}{\pi^2}\log n+O(1)$$
A problem in "Fourier Analysis" by E. Stein and R. Shakarchi explains how to show $L_n\ge C \log n$. Afterwards, it says more careful estimate shows this equality.
I already have $$\int_{k\pi}^{(k+1)\pi}|\frac{\sin t}{t}|dt=\frac{2}{\pi}\frac{1}{k}+O(\frac{1}{k^2})$$
$$\sum_{k=1}^{n}\frac{1}{k}=\log n+\gamma+O(\frac{1}{n})$$
so that
$$\int_{\pi}^{(k+1)\pi}|\frac{\sin t}{t}|dt=\frac{2}{\pi}\log n+O(1)$$
If we can have $$L_n=\frac{2}{\pi}\int_{\pi}^{(n+1)\pi}|\frac{\sin t}{t}|dt+O(1)$$, then we are done. According are two steps


*

*$$2 \pi L_n=\int_{-\pi}^{\pi}|\frac{\sin ((N+1/2)x)}{t/2}|dt+O(1)$$

*$$\int_{-\pi}^{\pi}|\frac{\sin ((N+1/2)x)}{t/2}|dt=4\int_{\pi}^{N\pi}|\frac{\sin t}{t}|dt+O(1)$$
Now I think I need some help on these two steps.

 A: \begin{align*}
&L_{n}=\dfrac{1}{\pi}\int_{0}^{\pi}\left|\dfrac{\sin{(n+\frac{1}{2})x}}{\sin{\frac{x}{2}}}\right|dx>\dfrac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left|\dfrac{\sin{(n+\frac{1}{2})2t}}{\sin{t}}\right|dt>\dfrac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\dfrac{|\sin{(n+\frac{1}{2})2t}|}{t}dt\\
&=\dfrac{2}{\pi}\int_{0}^{(2n+1)\pi/2}\dfrac{|\sin{u}|}{u}du>\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\dfrac{|\sin{u}|}{u}du\\
&=\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{0}^{\pi}\dfrac{\sin{u}}{u+k\pi}du>\dfrac{2}{\pi}\sum_{k=0}^{n-1}\dfrac{1}{(k+1)\pi}\int_{0}^{\pi}\sin{u}du\\
&=\dfrac{4}{\pi^2}\sum_{k=0}^{n-1}\dfrac{1}{k+1}
>\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma
\end{align*}
where $\gamma$ is Euler's constant.
other hand
$$\left|\dfrac{\sin(n+1/2)t}{\sin{t/2}}-\dfrac{\sin{nt}}{\tan{t/2}}\right|\le 1$$
so
\begin{align*}
&L_{n}\le 1+\dfrac{2}{\pi}\int_{0}^{\pi/2}\left|\dfrac{\sin{(2nt)}}{\tan{t}}\right|dt\\
&<1+\dfrac{2}{\pi}\int_{0}^{\pi/2}\dfrac{|\sin{2nt}|}{t}dt\\
&=1+\dfrac{2}{\pi}\sum_{k=1}^{n}\int_{0}^{\pi}\dfrac{\sin{t}}{t+(k-1)\pi}dt\\
&<1+\dfrac{2}{\pi}\int_{0}^{\pi}\dfrac{\sin{t}}{t}dt+\dfrac{4}{\pi^2}\sum_{k=2}^{n}\dfrac{1}{k-1}\\
&<C+\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma
\end{align*}
so  your  result is obvious.
