Let $D_N$ denote the Dirichlet kernel
$$D_N(\theta)=\sum_{n=-N}^Ne^{ik\theta}=\frac{\sin((N+\frac 1 2) \theta)}{\sin (\frac {\theta} 2)}$$ And define $$L_N=\frac 1 {2\pi}\int_{-\pi}^{\pi}|D_N(\theta)|d\theta$$ Prove that $L_N\geq c\log N$
Could you help me understand these 4 steps that I remarked below, they literally do not ring any bell to me..
Since $\frac x {\sin x}\geq 1 $ for $x\in [-\frac {\pi}2,\frac {\pi}2]$, it follows that
$$|D_N(\theta)|\geq 2\frac {|{\sin((N+\frac 1 2) \theta})|}{\theta}$$
Then, $$\int_{-\pi}^{\pi}|D_N(\theta)|d\theta \\ \geq 4\int_0^{\pi}\frac {|{\sin((N+\frac 1 2) \theta})|}{\theta} d\theta $$ $$=4\int_0^{(N+\frac 1 2)\pi}\frac {|{\sin \theta}|}{\theta} d\theta \\
\geq 4\int_0^{N\pi}\frac {|{\sin \theta}|}{\theta} d\theta$$
$$=4\sum_{k=0}^{N-1}\int_{k\pi}^{(k+1)\pi}\frac {|{\sin \theta}|}{\theta} d\theta $$
$$\geq 4\sum_{k=0}^{N-1} \frac 1 {(k+1)\pi}\int_{k\pi}^{(k+1)\pi}{|{\sin \theta}|} d\theta \tag 1$$
$$=\frac 8 {\pi}\sum_{k=0}^{N-1} \frac 1 {(k+1)} \tag 2$$
$$=\frac 8 {\pi}\log(N+1) \tag 3\\ \geq \frac 8 {\pi}\log N \\ \therefore L_N\geq c\log N$$
(1) $\frac{|\sin\theta|}{\theta} \geq \frac{|\sin\theta|}{(k+1)\pi}$ for $\theta\in [k\pi,\ (k+1)\pi]$.
(2) Calculate the integral of sine, which is either always positive or negative depending on $k$. (which is 2).
(3) I am not sure about equality, but you can use $\sum_1^{N}\frac{1}{k} \geq \int_1^{N+1} 1/k = \log(N+1)$