A question on limit of $L^1$ norm of a conjugate Dirichlet kernel

Question

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 $$\lim_{k\to\infty} \frac{L_{2k}}{\log(2k)}$$

$k,n \in \mathbb{Z}$

Answer 1

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

$$\lim_{n\to \infty}\frac{I_n}{\log(n)}=4$$