Desigualdad que $\frac{\sin x}{x}$

Question

¿Alguien puede explicar a mí, por qué la siguiente desigualdad es verdadera?

$$\sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi-\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi - \frac{\pi}{4} \right)\right|}{(k+1) \pi - \frac{\pi}{4}} \ \frac{\pi}{2} $$

La pregunta es motivado por el siguiente cálculo $$ \sum_{k=0}^{\infty} \int_{k \pi}^{(k+1) \pi} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \sum_{k=0}^{\infty} \int_{k \pi + \frac{\pi}{4}}^{(k+1)\pi\frac{\pi}{4}} \left| \frac{\sin \xi}{\xi} \right| \, \text{d} \xi \geq \\ \geq \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi \frac{\pi}{4} \right)\right|}{(k+1) \pi \frac{\pi}{4}} \ \frac{\pi}{2} = \sum_{k=0}^{\infty} \frac{\left| \sin\left( (k+1) \pi \frac{\pi}{4} \right)\right|}{2(k+1) - \frac{1}{2}} = \\ = \sum_{k=0}^{\infty} \frac{\sqrt{2}}{2} \frac{1}{2(k+1) - \frac{1}{2}} = \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k+3} \geq \sum_{k=0}^{\infty} \frac{\sqrt{2}}{4k} = \\ = \frac{\sqrt{2}}{4}\sum_{k=0}^{\infty} \frac{1}{k} = \infty $$ lo que muestra que $\frac{\sin \xi}{\xi} \notin \mathcal{L}^1(\mathbb{R})$.

Answer 1

Esto utiliza el hecho de que x}x\right|$$\inta^b |f(x)|\ge (b-a)\inf{x\in(a,b)} |f(x)|$ $and that $$ is minimised on $\left|\frac{\sin (k\pi + \frac\pi4, (k +1) \pi-\frac\pi4)$ when $x=(k+1) \pi-\frac\pi4$

Answer 2

Sugerencia

$x\in[k\pi+\pi/4,(k+1)\pi-\pi/4]$,

$$|\sin x|\ge|\sin\big((k+1)\pi-\pi/4\big)|$$

y

$$x\le(k+1)\pi-\pi/4$$