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Following this proof Estimate the integral of the absolute value of the Dirichlet kernel

Prove for every $n\in \mathbb{N}$ : $\|D_n(x)\|_{L^1} \geq C_1+C_2\log(n)$ I don't understand $(1),(2),(3),(4)$.
Proof: \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\\ &>^{(1)}\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\dfrac{|\sin{u}|}{u}du\\ &=^{(2)}\dfrac{2}{\pi}\sum_{k=0}^{n-1}\int_{0}^{\pi}\dfrac{\sin{u}}{u+k\pi}du\\ &>^{(3)}\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}\\ &>^{(4)}\dfrac{4}{\pi^2}\ln{n}+\dfrac{4}{\pi^2}\gamma \end{align*}

lebong66
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1 Answers1

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(1) Instead of integrating over the entire interval $[0,2(n+1)\pi/2]$ we can take the integrals over $[0,\pi], [\pi,2\pi],\ldots,[(n-1)\pi,n\pi],[n\pi,2(n+1)\pi/2]$ and take their sum. The bounds of the last interval are not so nice so we drop it from the sum and that's where the strict inequality comes from.

(2) Here we substitute $u=v+k\pi$ in each of the integrals but we are a bit sloppy and still write $u$ for the new integration variable. This substitution has the effect that all integrals now go over the same range $[0,\pi]$ and that we can drop the absolute value for the sine since it is positive on this range.

(3) On $[0,\pi),$ we have that $$\frac{1}{u+k\pi}>\frac{1}{(k+1)\pi}$$ and we apply this bound to each of the integrals. This gives a strict inequality because both sides of the estimate are continuous.

(4) The difference between the harmonic series and the logarithm converges to the Euler-Mascheroni constant $\gamma.$

Stewan
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