A short way to determine the divergence of the integral: $\int\limits_0^1 \left| \frac{1}{x}\cos\left(\frac{1}{x}\right) \right| dx$

Question

The integral $$\int_0^1 \left| \frac{1}{x}\cos\left(\frac{1}{x}\right) \right| dx$$I can use the subsets in the form $$\left[ \frac{1}{(2k+7/3)\pi} , \frac{1}{(2k+5/3)\pi}\right]$$ where the cosine function is bigger than 1/2 in these subsets we will get the summation $$\sum_{0}^{\infty} \ln \left( 1+ \frac{2}{6k+5}\right) $$ I am trying to show that $f(x)=x^2\sin(1/x^2)$ on the interval $[0,1]$ is not of bounded variation by showing that $f’$ is not integrable. ( Royden book 4th edition page 114)

Answer 1

Hmm, how about instead of bounding the cosine in the integral and integrating the $\frac{1}{x}$, you in fact bound both of them.

$$\int_{((2k+7/3) \pi)^{-1}}^{((2k+5/3)\pi)^{-1}} \left|\frac{1}{x}\cos(\frac{1}{x})\right| dx \geq \left(\frac{1}{(2k+5/3)\pi}-\frac{1}{(2k+7/3) \pi}\right)(2k+\frac{5}{3})\pi* \cos(\frac{7\pi}{3} ) = \frac{2\pi C(2k+\frac{5}{3})\pi}{3(2k+5/3)\pi*(2k+7/3)\pi} = \frac{C'}{(2k+7/3)} $$

Then you can sum these to show divergence.