About the order of the $L^1$ norm of the Dirichlet kernel. Reading this text from Wikipedia, I found the following statement about the Dirichlet kernel:
$$\| D_n \|_{L^1} \approx \log n, $$
where $\approx$ denotes "is of the order". I think that this mean that
$$\displaystyle{\lim_{n\to\infty} \frac{\|D_n\|_{L^1}}{\log n} = 1},$$
as usual. I have trying to prove this, but without success. Someone can to say me how to show this? Or maybe, indicate to me a text where I can find a proof of this?
Thanks!
Note 1: $D_n$ is the Dirichlet kernel, i.e.,
$$ D_n(t) := \sum_{k=-n}^{n} e^{i kt} = 1 + 2\sum_{k=1}^{n} \cos kt. $$
Note 2: $\|D_n\|_{L^1}$ is the $L^1$ norm of $D_n$, i.e.,
$$\displaystyle{\|D_n\|_{L^1} := \int_{-\pi}^{\pi} |D_n(t)|dt }.$$
 A: 
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: $$\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {\frac{{\sin \left( {n + \frac{1}{2}} \right)t}}{t}} dt \leqslant \left( {n + \frac{1}{2}} \right)\int\limits_0^{\frac{{2\pi }}{{2n + 1}}} {dt}  = \pi $$
