Does this integral exist?

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} I & = \int_\frac 23^\infty \frac {\log \left(\cos \frac 1x\right)}x dx & \small \color{#3D99F6} \text{Let }u = \frac 1x \implies du = - \frac {dx}{x^2} \\ & = \int_0^\frac 32 \frac {\log(\cos u)}u du \end{aligned}$

We note that the integrand $\dfrac {\log(\cos u)}u$ is defined over $u \in \left(0, \frac 32\right)$ , if $\displaystyle \lim_{u \to 0} \frac {\log(\cos u)}u$ exists.

$\begin{aligned} L & = \lim_{u \to 0} \frac {\log(\cos u)}u & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{u \to 0} \frac {-\tan u}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = 0 \end{aligned}$

Therefore, the integrand is defined over the interval and the integral $I$ converges .

The integrand function $f(x)$ is defined, continue and always negative over the interval $\left[ \frac{2}{3}, +\infty \right)$ , since $\cos \left( \frac{1}{x} \right)>0$ on $\left( \frac{2}{\pi}, +\infty \right) \supseteq \left[ \frac{2}{3}, +\infty \right)$ and $\cos \left( \frac{1}{x} \right)\le1 \Rightarrow \frac{1}{x}\log\cos \left( \frac{1}{x} \right) \le 0\ \forall x\in \left( \frac{2}{\pi}, +\infty \right) \supseteq \left[ \frac{2}{3}, +\infty \right)$ Thus, we can apply the asymptotic comparison criteria.

When $x\to +\infty$ , $f(x) \sim -\frac{1}{2x^3}$ , therefore the integral $\boxed{\mathrm{converges}}$ .