Does $\int_{3/\pi}^{+\infty}\ln\left(\cos1/t\right)dt$ converge? Good evening

Does $\displaystyle \int_{\frac{3}{\pi}}^{+\infty}\ln\left(\cos\frac {1}{t} \right) \, dt$ converge?

My solution :
I use this integrale as a reference : $\displaystyle \int_1^{+\infty}\frac {1}{t^{\alpha}} \, dt$ converges if $\alpha>1$
$$\ln\left(\cos\frac {1}{t}\right)=\ln\left(1+[\cos\frac {1}{t}-1]\right)\;\sim_{+\infty}\;\cos\frac {1}{t}-1\;\sim_{+\infty}\frac{1}{2t^2}$$
$\displaystyle \int_{\frac{3}{\pi}}^{+\infty}\frac{1}{2t^2}dt$ converges thus $\displaystyle \int_{\frac{3}{\pi}}^{+\infty}\ln\left(\cos\frac {1}{t}\right)dt$ converges.
I haven't got the correction, so I would like to know if it is correct?
Thanks
 A: Correct if you add a minus before $\frac {1}{2t^2}$.
Other method:
With the substitution $u=1/t $ , the integrale has the same nature than
$$\int_0\frac {1}{u^2}\ln (\cos (u))du $$
but as you almost said,  when $u\to 0^+,$
$$\ln (\cos (u))\sim \cos (u)-1\sim -\frac {u^2}{2} .$$
thus, the integrand of this integrale has a finite limit. So it converges .
A: $$ 2(x-1) < \ln x \text{ if } 0<x<1. $$ $$ 1 - \frac{x^2} 2 \le  \cos x $$
$$
\text{Therefore } 0 \ge \int_{3/\pi}^\infty \ln\left( \cos \frac 1 t \right) \, dt \ge \int_{3/\pi} 2 \left( \left(\cos\frac 1 t\right) -1 \right) dt \ge \int_{3/\pi} ^\infty \frac {-1} {t^2} \, dt > - \infty.
$$
A: By enforcing the substitution $t\mapsto\frac{1}{\theta}$ we are left with
$$ \int_{0}^{\pi/3}\frac{\log\cos\theta}{\theta^2}d\theta\stackrel{\text{IBP}}{=}\frac{3\log 2}{\pi}-\int_{0}^{\pi/3}\frac{\tan\theta}{\theta}\,d\theta$$
where $\frac{\tan\theta}{\theta}$ is a continuous, positive, increasing and convex function on $\left(0,\tfrac{\pi}{3}\right)$, bounded between $1$ and $\frac{3\sqrt{3}}{\pi}$.
