How to prove that arc-length of x cos(1/x) is divergent?

Question

A function

$$f(x)=x\cos\left(\frac1x\right), \,\, x \in (0,1) $$

And I want to prove that length of the graph of $f$ over the interval $(\alpha,1)$ is divergent as $\alpha\to 0$.

I try to use a comparison test, but I have no idea.

Answer 1

Making the change of variable $t = \frac 1 x$, the length of the graph on $(\alpha, 1)$ is given by

$$\int \limits _\alpha ^1 \sqrt {1 + \left( \cos \frac 1 x + \frac 1 x \sin \frac 1 x \right)^2} \ \Bbb d x = \int \limits _1 ^{\frac 1 \alpha} \sqrt {1 + (\cos t + t \sin t)^2} \frac 1 {t^2} \ \Bbb d t .$$

For $\alpha \ne 0$ the integrand is a bounded continuous function on $[1 , \frac 1 \alpha]$, so its integral is finite.

On the other hand,

$$\int \limits _0 ^1 \sqrt {1 + \left( \cos \frac 1 x + \frac 1 x \sin \frac 1 x \right)^2} \ \Bbb d x = \int \limits _1 ^\infty \sqrt {1 + (\cos t + t \sin t)^2} \frac 1 {t^2} \ \Bbb d t \ge \int \limits _1 ^\infty \frac {|\cos t + t \sin t|} {t^2} \ \Bbb d t \ge \\ \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac {|\cos t + t \sin t|} {t^2} \ \Bbb d t = \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac {\cos t + t \sin t} {t^2} \ \Bbb d t \ge \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac {t \sin t} {t^2} \ \Bbb d t = \\ \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac {\sin t} t \ \Bbb d t \ge \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac {\sin^2 t} t \ \Bbb d t = \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac {1 - \cos 2t} {2t} \ \Bbb d t \ge \sum _{k = 1} ^\infty \int \limits _{2k \pi + \frac \pi 4} ^{2k\pi + \frac \pi 2} \frac 1 {2t} \ \Bbb d t = \\ \frac 1 2 \sum _{k = 1} ^\infty \ln \frac {2k\pi + \frac \pi 2} {2k\pi + \frac \pi 4} = \frac 1 2 \sum _{k = 1} ^\infty \ln \frac {k + \frac 1 4} {k + \frac 1 8} = \frac 1 2 \sum _{k = 1} ^\infty \ln \left( 1 + \frac {\frac 1 8} {k + \frac 1 8} \right) \sim \frac 1 2 \sum _{k = 1} ^\infty \frac {\frac 1 8} {k + \frac 1 8} \sim \sum _{k = 1} ^\infty \frac 1 k = \infty ,$$

where the tilde between two series means that those series have the same behaviour, i.e. either simultaneously convergent, or simultaneously divergent.