Lebesgue-Integrability of $x \mapsto \frac{\sin x}{x}$ When we have the function $\frac{\sin x}{x}$ and we want to check it ass for the integrability do we have to do the following to see if it is in $L^1(\mathbb{R})$? $$\int_{\mathbb{R}}| \frac{\sin x}{x}|\leq \int_{\mathbb{R}} dx=\mu (\mathbb{R})$$ is it correct so far?? How do we continue??
 A: Hint
1) Prove the following inequality
$$ \int_\mathbb{R} \left| \frac{\sin(x)}{x} \right| dx \geq 2 \sum_{k=1}^\infty \frac{2}{k \pi}. $$
To prove it observe that 
$$\int_\mathbb{R} \left| \frac{\sin(x)}{x} \right| dx = 2 \int_0^\infty \frac{|\sin(x)|}{x} dx = 2 \sum_{k=0}^\infty \int_0^\pi \frac{|\sin(x+k\pi)|}{x+k\pi} dx.$$
2) Examine the behavior of the series.
A: Hint:


*

*Deduce from $$|\sin(x)| \geq \frac{1}{2} \qquad \text{for all} \, \, x \in \left[ \frac{1}{6}\pi, \frac{5}{6}\pi \right]$$ that $$|\sin(x)| \geq \frac{1}{2} \qquad \text{for all} \, \, x \in \left[ \frac{1}{6}\pi+ k \pi, \frac{5}{6}\pi +k \pi\right].$$

*Show that $$\int_{\mathbb{R}} \left| \frac{\sin x}{x} \right| \, dx \geq \int_{\{x \geq 0; |\sin(x)| \geq \frac{1}{2}\}} \frac{|\sin(x)|}{x} \, dx \geq \sum_{k=0}^{\infty} \frac{1}{2} \int_{\frac{1}{6}\pi+ k \pi}^{\frac{5}{6}\pi + k \pi} \frac{1}{x} \, dx.$$

*Conclude from $$ \int_{\frac{1}{6}\pi+ k \pi}^{\frac{5}{6}\pi + k \pi} \frac{1}{x} \, dx \geq \frac{1}{\frac{5}{6} \pi + k \pi} \int_{\frac{1}{6}\pi+ k \pi}^{\frac{5}{6}\pi + k \pi} \, dx$$ and step 2 that $$\int_{\mathbb{R}} \left| \frac{\sin x}{x} \right| \, dx \geq \sum_{k=0}^{\infty} \frac{2}{5+6k}=\infty.$$

