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??
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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.
Giovanni De Gaetano
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How could I use your answer to show that $f(x)=\frac{\sin\left(\frac{1}{t}\right)}{t}$ is not Lebesgue Integrable on (0,1)? https://math.stackexchange.com/questions/2235812/show-improperly-riemman-integrable-function-is-not-lebesgue-integrable – Alfredo Lozano Apr 16 '17 at 20:39
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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 $$\int_{\mathbb{R}} \left| \frac{\sin x}{x} \right| \, dx \geq \sum_{k=0}^{\infty} \frac{2}{5+6k}=\infty.$$
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https://en.wikipedia.org/wiki/Lp_space#Lp_spaces
– Giovanni De Gaetano Feb 10 '15 at 09:56