Prove the inequality $\int_{0}^{(n+(1/2))π} \lvert \sinα \rvert /α > (2/π^2) \log n$

Question

This relation is used for calculating the lower bound of Lebesgue’s constant $$\int_{0}^{(n+\frac12)π} \frac{\lvert \sin α \rvert}{α} > \frac2{π^2} \log n. $$ But I don’t know how do I get from the left side of the inequality to the right side?

Answer 1

Note $$ \alpha\in[(k-\frac12)\pi,(k+\frac12)π]\subset[(k-\frac12)\pi,(k+1)π]\Rightarrow\frac1\alpha>\frac1{\pi(k+1)} $$ and $$ \frac{\sin \alpha}{\alpha}\ge\frac{2}{\pi}, \ x\in[0,\frac\pi2], \ \sum_{k=1}^n\frac1k>\log n. $$ In fact \begin{eqnarray} &&\int_{0}^{(n+\frac12)π} \frac{\lvert \sin α \rvert}{ α}dα\\ &=&\int_{0}^{\fracπ2} \frac{\lvert \sin α \rvert}{ α}dα+\sum_{k=1}^n\int_{(k-\frac12)\pi}^{(k+\frac12)π} \frac{\lvert \sin α \rvert}{ α}dα\\ &\ge& \int_{0}^{\fracπ2} \frac{\lvert \sin α \rvert}{ α}dα+\sum_{k=1}^n\frac{1}{\pi(k+1)}\int_{(k-\frac12)\pi}^{(k+\frac12)π}\lvert \sin α \rvert dα\\ &\ge& 1+\sum_{k=1}^n\frac{2}{\pi(k+1)}\\ &>&\sum_{k=0}^n\frac{2}{\pi(k+1)}\\ &>&\sum_{k=1}^n\frac{2}{\pi k}\\ &>& \frac2{π} \log n\\ &>& \frac2{π^2} \log n. \end{eqnarray}

Answer 2

We have $$\int\limits_0^{(n+{1\over 2})\pi} {|\sin x|\over x}\,dx=\sum_{k=1}^{2n+1}\int\limits_{(k-1)\pi/2}^{k\pi/2} {|\sin x|\over x}\,dx\ge \sum_{k=1}^{2n+1}{2\over k\pi}\int\limits_{(k-1)\pi/2}^{k\pi/2}|\sin x|\,dx\\ ={2\over \pi}\left (\int\limits_{0}^{\pi/2}\sin x\,dx\right )\,\sum_{k=1}^{2n+1}{1\over k}={2\over \pi}\sum_{k=1}^{2n+1}{1\over k}> {2\over \pi}\log(2n+2) $$