I'm having a hard time showing that $\displaystyle \lim_{c \to \0} \int_{c}^{1}|\frac{sin(\frac{1}{x})}{x}|dx $ fails to converge absolutely, any insight would be greatly appreciated.

Printable View

- May 14th 2011, 06:11 PMeastmanAbsolute Convergence of Integral
I'm having a hard time showing that $\displaystyle \lim_{c \to \0} \int_{c}^{1}|\frac{sin(\frac{1}{x})}{x}|dx $ fails to converge absolutely, any insight would be greatly appreciated.

- May 14th 2011, 06:57 PMProve It
Try writing $\displaystyle \displaystyle \sin{\left(\frac{1}{x}\right)}$ as a Taylor Series, then simplifying $\displaystyle \displaystyle \left|\frac{\sin{\left(\frac{1}{x}\right)}}{x} \right|$.

- May 14th 2011, 07:06 PMJose27
Another way: If you know that $\displaystyle \int_1^{\infty} \left| \frac{\sin (z)}{z} \right|dz$ is infinite, just apply a change of variables.

- May 15th 2011, 06:57 AMeastmanThanks
Thanks to both of you for your help. I did try one way before I read this, maybe you can tell me this works...

$\displaystyle \lim_{c\to\0}\int_{c}^{1}|\frac{sin(\frac{1}{x})}{ x}|dx\geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }|\frac{sin(\frac{1}{x})}{x}|dx \geqslant \sum_{k = 1}^\infty\int_{\frac{2}{(4k+1)\pi} }^{\frac{2}{(4k-1)\pi} }\frac{dx}{x}$

...and then just evaluate, I could be way off. (Wondering) - May 15th 2011, 11:21 AMJose27
- May 15th 2011, 12:40 PMeastman
Sorry, I'm slightly confused, could you elaborate?

- May 15th 2011, 02:23 PMJose27
Disregard the second part of my last message, could you explain how you got the last inequality?

- May 16th 2011, 01:01 PMeastman
You're right the last inequality does not hold, sin vanishes over those intervals.. I ended up doing it with the change of variables like you orginally said. Thanks for your help.