Inequality involving $\frac{\sin x}{x}$ Can anybody explain me, why the following inequality is true?
$$\sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq  \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \  \frac{\pi}{2} $$
The question is motivated from the following calculation
$$
\sum_{k=0}^{\infty} \int_{k \pi}^{(k+1) \pi} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \\
\geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \  \frac{\pi}{2} =  \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{2(k+1)  - \frac{1}{2}} = \\
= \sum_{k=0}^{\infty}  \frac{\sqrt{2}}{2} \frac{1}{2(k+1)  - \frac{1}{2}} = \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k+3} \geq \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k} = \\
= \frac{\sqrt{2}}{4}\sum_{k=0}^{\infty} \frac{1}{k} = \infty
$$
which shows that $\frac{\sin \xi}{\xi} \notin \mathcal{L}^1(\mathbb{R})$.

Hint
For $x\in[k\pi+\pi/4,(k+1)\pi-\pi/4]$,
$$|\sin x|\ge|\sin\big((k+1)\pi-\pi/4\big)|$$
and
$$x\le(k+1)\pi-\pi/4$$
