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I guess that the following estimates are classical. Let $D_N$ be the $1D$ Dirichlet kernel, $$ D_N(t)=\frac{\sin((N+\frac12)t)}{\sin (t/2)}. $$ We have for $1<p<\infty$, \begin{align} \Vert D_N\Vert_{L^p(0,2π)}&\sim_{N\rightarrow +\infty} N^{1-\frac 1p} \frac{2}{π^{1/p}} \left( \int_{0}^{+\infty} {\frac{\vert\sin t\vert^p}{t^p} }dt\right)^{1/p}, \end{align} and $ \Vert D_N\Vert_{L^1(0,2π)}\sim_{N\rightarrow +\infty}\frac{4 }{π^2}\ln N. $ Of course, here $a_N\sim b_N$ means $$ a_N=b_N(1+\epsilon _N), \quad \lim \epsilon_N=0. $$ Although I found in the literature some sort of equivalence, say the $L^1$ norm is between $c_1\ln N$ and $c_2\ln N$, the above form is not so easy to locate.

My question: I would be interested in a full expansion, using some type of Euler-Maclaurin formula, say for the $L^p$ norm of the Dirichlet kernel, with $$ \Vert D_N\Vert_{L^p(0,2π)}= c_0 N^{1-\frac1p}+c_1 N^{-\frac1p}+O(N^{-1-\frac1p}). $$ Is it written anywhere?

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  • $\begingroup$ If I may ask, why are you specifically interested in the rate of convergence? $\endgroup$
    – Rajesh D
    Commented Jun 25, 2019 at 16:25
  • $\begingroup$ Possibly related : math.stackexchange.com/q/27517/2987 $\endgroup$
    – Rajesh D
    Commented Jun 25, 2019 at 16:27
  • $\begingroup$ @RajeshDachiraju In the answer mentioned in your comment, the method which is presented does not even provide an equivalent in my sense. However to get that is not very difficult: you may write for $\theta\in [0,π/2]$, $sin\theta=\theta/(1+\theta^2 \sigma(\theta))$ and thus $(sin\theta)^{-1}=\theta^{-1}(1+\theta^2 \sigma(\theta))$. The term $1$ in the parenthesis gives you the main term, you only have to prove that the contribution of the other term is asymptotically smaller, which is not difficult. $\endgroup$
    – Bazin
    Commented Jun 25, 2019 at 19:39
  • $\begingroup$ @RajeshDachiraju In fact it seems to be a very natural example to study the classical Euler-Maclaurin formula in a situation where the amplitude has pointwise singularities because of the $\vert \sin t\vert$. It is interesting to see that in spite of that inconvenience, an asymptotic expansion can still be reached. Moreover, I insist on the fact that a direct Euler-Maclaurin style proof is simpler and more precise than the proof provided in that answer. $\endgroup$
    – Bazin
    Commented Jun 25, 2019 at 19:43
  • $\begingroup$ Understood, thats why I said "related" and not "duplicate". What I don't understand is your motivation. Is your motivation is to see if Euler-Maclaurin is useful here? or do you really have the application of this result in any larger problem? Normally, one is interested in the asymptotic result and don't want to look at other smaller terms. $\endgroup$
    – Rajesh D
    Commented Jun 26, 2019 at 1:47

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