Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .
Here is what i have tried :
I found out that for all $x>0$ , $x-\frac{x^3}{6} < \sin x
Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you .
Answer
Your solution is not good, because the left side, $M-\frac{M^3}{18}$, does not go to $\infty$, but rather to $-\infty$.
An alternative solution would include
- the fact that on any interval $[(k-1)\pi, k\pi]$, you have $$\frac{|\sin x|}{x} \geq \frac{|\sin x|}{k \pi}$$
- The knowledge that $$\int_{(k-1)\pi}^{k\pi} |\sin x| dx$$ is independent of $k$, i.e. that it is always equal to the same nonzero constant $C$
- The knowledge that the series $$\sum_{k=1}^{\infty}\frac1k$$ diverges (even if multiplied by a nonzero constant).
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