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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

Did
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Stu
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    You forgot a minus before. $1/2t^2$. – hamam_Abdallah Aug 25 '17 at 19:41
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    You have a sign error but otherwise it is OK. If I were nitpicking then I would say you need to show $\ln(\cos(1/t))+1/(2t^2)$ (note the +) is integrable still. – Ian Aug 25 '17 at 19:44
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    I would say that $|\ln(\cos \frac 1t)|< \frac 1{t^2}$ for all $t$ in the region of integration. And $\int \frac {1}{t^2} dt$ converges over that region. $\int \ln(\cos \frac 1t)\ dt$ converges by the comparison test. – Doug M Aug 25 '17 at 19:44
  • ok thanks every one – Stu Aug 25 '17 at 19:46

3 Answers3

1

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 .

hamam_Abdallah
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$$ 2(x-1) < \ln x \text{ if } 0<x<1. $$ $$ 1 - \frac{x^2} 2 \le \cos x $$

$$ \text{Therefore } 0 \ge \int_{3/\pi}^\infty \ln\left( \cos \frac 1 t \right) \, dt \ge \int_{3/\pi} 2 \left( \left(\cos\frac 1 t\right) -1 \right) dt \ge \int_{3/\pi} ^\infty \frac {-1} {t^2} \, dt > - \infty. $$

Michael Hardy
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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 $\frac{3\sqrt{3}}{\pi}$.

Jack D'Aurizio
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