Lower $ L^{1} $ bound for the Dirichlet kernel: $ \frac{1}{2\pi}\int _0^{2\pi} \left| \frac{\sin((N+1/2)x)}{\sin(x/2)}\right| dx \geq C \log N $

Question

I want to bound the $ L^{1} $ norm of the Dirichlet kernel from below. It is well known that $$ \frac{1}{2\pi}\int _0^{2\pi} \left| \frac{\sin((N+1/2)x)}{\sin(x/2)}\right| dx \geq C \log N $$ for some $C>0$ .

But some steps in the calculation are not clear:

1) $\displaystyle\frac{1}{2\pi}\int _0^{2\pi} \left|\frac{\sin((N+1/2)x)}{\sin(x/2)}\right| dx = \frac{1}{\pi}\int _0^{\pi} \left| \frac{\sin((N+1/2)x)}{\sin(x/2)}\right| dx$

2) $\displaystyle\frac{1}{\pi}\int _0^{\pi} \frac{\vert \sin((N+1/2)x)\vert}{\vert \sin(x/2)\vert} dx \geq \frac{1}{\pi}\int _0^{\pi} \frac{\vert \sin((N+1/2)x)\vert}{x} dx$

Are these steps right? Why are they true?

Answer 1

1. This is due to the symmetry of the sine function. By periodicity the integral can be done on $[-\pi,\pi]$, and the integrand is an even function.
2. This is because $\sin x\le x$, $0\le x\le\pi$.