Rudin exercise 8.21 Following is an exercise in Rudin's Principles of Mathematical Analysis (exercise 8.21).
Let $$L_n = \frac1{2\pi}\int^\pi_{-\pi}\left|\frac{\sin(n+\frac12)x}{\sin\frac12x}\right|\,\text dx\space \space  \space(n=1,2,3,...)$$
Prove that there exists a constant $C>0$ such that $$L_n>C\log n$$
or, more precisely, that the sequence $$\left\{L_n-\frac4{\pi^2}\log n\right\}$$is bounded.

I cannot do this problem because the integral is too complicated. Also, why is the last result 'more precise'?
 A: By symmetry, we can integrate on $[0,\pi]$ and multiply by $2.$ Then let $x=2y.$ We see 
$$L_n = \frac{2}{\pi}\int_0^{\pi/2}\left|\frac{\sin(2n+1)y}{\sin y}\right|\, dy.$$
Now $\sin y \le y$ on this interval. So replace $\sin y$ downstairs by $y$ and the integral gets smaller. Then make the change of variables $y = z/(2n+1).$ We obtain
$$L_n \ge \frac{2}{\pi}\int_0^{n\pi}\left|\frac{\sin z}{z}\right|\, dy
\ge \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k\pi}\int_0^\pi\sin z\, dz = \frac{4}{\pi^2} \sum_{k=1}^{n}\frac{1}{k}\ge \frac{4}{\pi^2}\cdot \ln n.$$ Not sure about the "more precisely" part yet.
A: http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Dirichlet_kernel
$$
1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots+2\cos(nx) = \frac{ \sin\left[\left(n+\frac{1}{2}\right)x\right\rbrack }{ \sin\left(\frac{x}{2}\right) }
$$
A: This is a Dirichlet kernel..
$$D_n(x) = \frac{\sin(n+\frac12)x}{\sin\frac12x} = \sum\limits_{k=-n}^n e^{ikn}$$
You can complete the answer using this answer for norm of Dirichlet Kernel
