Let $D_n(x)$ be the Dirichlet kernel defined by $$D_n(x):=\frac{\sin\frac{(2n+1)x}{2}}{2\pi\sin\frac{x}{2}}$$where $D_n(0)$ can be set to $\frac{2n+1}{2\pi}$ if we desire it to be continuous. Another expression for $D_n$ is $$D_n(x)=\frac{1}{\pi}\Bigg(\frac{1}{2}+\sum_{k=1}^n \cos kx \Bigg).$$I read that $\lim_{n\to\infty}\int_{-\pi}^{\pi}|D_n(x)|dx=+\infty$, but I cannot prove it rigourously. How can it be done? Thank you so much!!!
[Math] Divergence $\int_{-\pi}^{\pi} |D_n(x)|dx$ for Dirichlet kernel as $n\to\infty$
fourier analysisintegrationlimits
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Claim If $L_n=\displaystyle\frac{1}{\pi}\int_0^{\pi}|D_n(t)|dt$, we have $L_n=\dfrac{4}{\pi^2}\log n+O(1)$.
Proof We begin by dividing the interval $[0,\pi]$ into the subintervals where $\sin\left(n+\dfrac 1 2\right)$ keeps its sign. Since $\sin t/2$ is always non-negative in said domain, we may write $$\frac{1}{\pi }\int\limits_0^\pi {\left| {{D_n}\left( t \right)} \right|} = \frac{1}{\pi }\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{{\sin t/2}}dt} + \frac{1}{\pi }\sum\limits_{k = 1}^{n - 1} {\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{{\sin t/2}}dt} } + \frac{1}{\pi }\int\limits_{\frac{{2n\pi }}{{2n + 1}}}^\pi {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{{\sin t/2}}dt} $$ The last integral goes to $0$ as $n$ goes to infinity. We may focus our attention to the middle sum. The function $g(t)=\dfrac{1}{{\sin t/2}} - \dfrac{1}{t/2}$ is continuous on $[0,\pi]$, thus the integrals $${{c_n} = \frac 1 \pi\int\limits_0^{\frac{{2n\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|g\left( t \right)dt} }$$ exist and are bounded above by a constant for any $n$. We can then consider $$d_n=\frac{2}{\pi }\sum\limits_{k = 1}^{n - 1} {\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\frac{{\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|}}{t}} } dt$$ But $t^{-1}$ is positive and decreasing, thus we have the bounds $$\frac{2}{\pi }\sum\limits_{k = 1}^{n - 1} {\frac{{2n + 1}}{{2\left( {k + 1} \right)\pi }}\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|} } dt \leqslant {d_n} \leqslant \frac{2}{\pi }\sum\limits_{k = 1}^{n - 1} {\frac{{2n + 1}}{{2k\pi }}\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|} } dt$$ But it is immediate that $${\int\limits_{\frac{{2k\pi }}{{2n + 1}}}^{\frac{{2\left( {k + 1} \right)\pi }}{{2n + 1}}} {\left| {\sin \left( {n + \frac{1}{2}} \right)t} \right|} }dt=\frac{4}{2n+1}$$ whence $$\frac{4}{{{\pi ^2}}}\sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \leqslant {d_n} \leqslant \frac{4}{{{\pi ^2}}}\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} $$
It is known that $\displaystyle\sum\limits_{k = 1}^{n - 1} {\frac{1}{k}} = \log n + O\left( 1 \right)$ (in fact, $=\log +n+\gamma+o(1)$), thus we can claim that ${d_n} = \dfrac{4}{{{\pi ^2}}}\log n + O\left( 1 \right)$, that is, $d_n-\dfrac{4}{\pi^2}\log n$ is bounded by a constant as $n\to\infty$. Finally, we look into the integral $$\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\sin \left( {n + \frac{1}{2}} \right)t}}{t}} dt$$ that we excluded from our sum. The integrand is positive, continuous and decreasing on the interval in question, and reaches a maximum at $t=0$ with value $n+\dfrac 1 2$, whence we may give the crude estimation: $$\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\sin \left( {n + \frac{1}{2}} \right)t}}{t}} dt \leqslant \left( {n + \frac{1}{2}} \right)\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {dt} = \pi $$
\begin{align*} &L_{n}=\dfrac{1}{\pi}\int_{0}^{\pi}\left|\dfrac{\sin{(n+\frac{1}{2})x}}{\sin{\frac{x}{2}}}\right|dx>\dfrac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left|\dfrac{\sin{(n+\frac{1}{2})2t}}{\sin{t}}\right|dt>\dfrac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\dfrac{|\sin{(n+\frac{1}{2})2t}|}{t}dt\\ &=\dfrac{2}{\pi}\int_{0}^{(2n+1)\pi/2}\dfrac{|\sin{u}|}{u}du>\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\dfrac{|\sin{u}|}{u}du\\ &=\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{0}^{\pi}\dfrac{\sin{u}}{u+k\pi}du>\dfrac{2}{\pi}\sum_{k=0}^{n-1}\dfrac{1}{(k+1)\pi}\int_{0}^{\pi}\sin{u}du\\ &=\dfrac{4}{\pi^2}\sum_{k=0}^{n-1}\dfrac{1}{k+1} >\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma \end{align*} where $\gamma$ is Euler's constant. other hand
$$\left|\dfrac{\sin(n+1/2)t}{\sin{t/2}}-\dfrac{\sin{nt}}{\tan{t/2}}\right|\le 1$$
so \begin{align*} &L_{n}\le 1+\dfrac{2}{\pi}\int_{0}^{\pi/2}\left|\dfrac{\sin{(2nt)}}{\tan{t}}\right|dt\\ &<1+\dfrac{2}{\pi}\int_{0}^{\pi/2}\dfrac{|\sin{2nt}|}{t}dt\\ &=1+\dfrac{2}{\pi}\sum_{k=1}^{n}\int_{0}^{\pi}\dfrac{\sin{t}}{t+(k-1)\pi}dt\\ &<1+\dfrac{2}{\pi}\int_{0}^{\pi}\dfrac{\sin{t}}{t}dt+\dfrac{4}{\pi^2}\sum_{k=2}^{n}\dfrac{1}{k-1}\\ &<C+\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma \end{align*} so your result is obvious.
Best Answer
We know the sine rather well, that allows us to transform the integral
$$\int_{-\pi}^\pi \lvert D_n(x)\rvert\,dx$$
into something whose behaviour we can easier recognise. By symmetry, we need only consider the interval $[0,\pi]$. I don't like to type too many fractions either, so make the substitution $y = \frac{x}{2}$. Then we have
$$C_n := \int_{-\pi}^\pi \lvert D_n(x)\rvert\,dx = \frac{2}{\pi}\int_0^{\pi/2} \frac{\lvert\sin ((2n+1)y)\rvert}{\sin y}\,dy.$$
We can better analyse the behaviour if we get rid of the sine in the denominator. Replacing $\sin y$ with $y$ doesn't change much near $0$, where the most interesting things happen, so let's write
$$\frac{\pi}{2} C_n = \underbrace{\int_0^{\pi/2} \frac{\lvert \sin ((2n+1)y)\rvert}{y}\,dy}_{A_n} + \underbrace{\int_0^{\pi/2} \lvert \sin ((2n+1)y)\rvert \biggl(\frac{1}{\sin y} - \frac{1}{y}\biggr)\,dy}_{B_n}.$$
Since $\frac{1}{\sin y} - \frac{1}{y}$ is analytic in $\{ y : \lvert y\rvert < \pi\}$ and strictly positive on the open interval $(0,\pi)$, $B_n$ is positive and bounded above by $$B := \int_0^{\pi/2} \frac{1}{\sin y} - \frac{1}{y}\,dy.$$
If we care, it is not hard to show that $B_n \to \frac{2}{\pi}B$.
Now we can use that we know the zeros of the sine, and split the integral $A_n$ into parts between the zeros of $\sin ((2n+1)y)$. Let $z_k = \frac{k\pi}{2n+1}$, then
$$A_n = \int_0^{z_1}\frac{\sin ((2n+1)y)}{y}\,dy + \sum_{k=1}^{n-1} \int_{z_{k}}^{z_{k+1}} \frac{\lvert\sin ((2n+1)y)\rvert}{y}\,dy + \int_{z_n}^{\frac{\pi}{2}} \frac{\lvert\sin ((2n+1)y)\rvert}{y}\,dy.$$
We can neglect the last integral, since the denominator is bounded above by $1$ over the whole interval of integration there, and the length of the interval is $\frac{\pi}{2(2n+1)}$, so the contribution of that to $A_n$ is $O(n^{-1})$. The first integral is a constant independent of $n$, substituting $u = (2n+1)y$, we find
$$\int_0^{\frac{\pi}{2n+1}} \frac{\sin ((2n+1)y)}{y}\,dy = \int_0^\pi \frac{\sin u}{u}\,du.$$
We can now estimate
$$\frac{2}{(2n+1)z_{k+1}} < \int_{z_k}^{z_{k+1}} \frac{\lvert \sin ((2n+1)y)\rvert}{y}\,dy < \frac{2}{(2n+1)z_k}$$
using the strict monotonicity of $\frac{1}{y}$ and
$$\int_{z_k}^{z_{k+1}} \lvert \sin ((2n+1)y)\rvert\,dy = \frac{2}{2n+1}.$$
Summation yields
$$\frac{2}{\pi}(H_n - 1) < \sum_{k=1}^{n-1} \int_{z_{k}}^{z_{k+1}} \frac{\lvert\sin ((2n+1)y)\rvert}{y}\,dy < \frac{2}{\pi} H_{n-1}$$
where $H_m = \sum\limits_{k = 1}^m \frac{1}{k}$ is the $m$-th harmonic number. As $H_m = \log m + \gamma + O(m^{-1})$ with the Euler-Mascheroni constant $\gamma$, we altogether obtain $A_n = \dfrac{2}{\pi}\log n + O(1)$, and therefore
$$C_n = \frac{4}{\pi^2}\log n + O(1).$$