A question on limit of $L^1$ norm of a conjugate Dirichlet kernel Let $$L_n = \int_{-1}^1 \left|\frac{cos(x/2)-cos((n+\frac{1}{2})x)}{2sin(x/2)}\right|dx$$
What are $$\lim_{k\to\infty} \frac{L_{2k+1}}{\log(2k+1)}$$ and $$\lim_{k\to\infty} \frac{L_{2k}}{\log(2k)}$$
$k,n \in \mathbb{Z}$
 A: First, let $I_n$ be given by the integral 
$$\begin{align}
I_n&=\int_{-\pi}^{\pi}\left|\frac{\cos(x/2)-\cos((n+1/2)x}{\sin(x/2)}\right|\,dx\tag1
\end{align}$$
We exploit the evenness of the integrand in $(1)$ along with the trigonometric identity $1-\cos(\phi)=2\sin^2(\phi/2)$ to write
$$\begin{align}
I_n&=4\int_0^\pi\frac{\left|\sin^2(x/4)-\sin^2((2n+1)x/4)\right|}{\sin(x/2)}\,dx \tag 2\end{align}$$
Applying the triangle inequality to $(2)$ reveals the bounds
$$I_n\le 4\log(2) +4\int_0^\pi \frac{\sin^2((2n+1)x/4)}{\sin(x/2)}\,dx\tag 3$$
$$I_n\ge \left|4\log(2) -4\int_0^\pi \frac{\sin^2((2n+1)x/4)}{\sin(x/2)}\,dx\right|\tag 4$$
Now, we analyze further the integral $\displaystyle \int_0^\pi \frac{\sin^2((2n+1)x/4)}{\sin(x/2)}\,dx$.  
Enforcing the substitution $(2n+1)x/4\to x$ yields
$$\begin{align}
\int_0^\pi \frac{\sin^2((2n+1)x/4)}{\sin(x/2)}\,dx&= \int_0^{(2n+1)\pi/4} \frac{\sin^2(x)}{\frac{2n+1}{4}\sin\left(\frac{2x}{2n+1}\right)}\,dx
\end{align}$$ 
For asymptotically large $n$, the integrand behaves like $x/2$ and we have
$$\begin{align}
\int_0^\pi \frac{\sin^2((2n+1)x/4)}{\sin(x/2)}\,dx&\sim \int_0^{(2n+1)\pi/4} \frac{\sin^2(x)}{x/2}\,dx \\\\
&=\log(n+1/2)+O(1) \tag 5
\end{align}$$ 
Putting together $(3)-(5)$, we see that 
$$\lim_{n\to \infty}\frac{I_n}{\log(n)}=4$$
