- Thread starter
- #1

#### Spam

##### New member

- Feb 20, 2012

- 2

Hello everybody !

I was just doing a small exercice on integration :

Study the convergence of this integral : $$\displaystyle \int_{\frac{2}{\pi}}^{+\infty} \ln(\cos \frac{1}{t})dt$$

So.

Ok, the study in \( +\infty \) does not cause me much trouble, it is the others which seems much complicated.

Indeed, let \( f(t) = \ln\big(\cos \frac{1}{t}\big) \). Then, I want to study \( f( \frac{2}{\pi} + t) \) when \( t \to 0^+ \).

But, when I compute some Taylor series, i found that \( f( \frac{2}{\pi} + t) = \ln t + \ln \big(\frac{\pi^2}{4} + o(1)\big) \sim_0 \ln t \).

So, since \(t \mapsto \ln t \) is not integrable on \( [\frac{2}{\pi}, 1]\), the integral does not converge. Yet, the correction seems to think the opposite. But the fact that \(\lim_0 \ln t = - \infty \) implies that they cannot say that.

So, there's something I didn't catch here. This may be something stupid, but I keep thinking about it without understand why.

Thank you for taking your time to read this and help me !

Bye.

I was just doing a small exercice on integration :

Study the convergence of this integral : $$\displaystyle \int_{\frac{2}{\pi}}^{+\infty} \ln(\cos \frac{1}{t})dt$$

So.

Ok, the study in \( +\infty \) does not cause me much trouble, it is the others which seems much complicated.

Indeed, let \( f(t) = \ln\big(\cos \frac{1}{t}\big) \). Then, I want to study \( f( \frac{2}{\pi} + t) \) when \( t \to 0^+ \).

But, when I compute some Taylor series, i found that \( f( \frac{2}{\pi} + t) = \ln t + \ln \big(\frac{\pi^2}{4} + o(1)\big) \sim_0 \ln t \).

So, since \(t \mapsto \ln t \) is not integrable on \( [\frac{2}{\pi}, 1]\), the integral does not converge. Yet, the correction seems to think the opposite. But the fact that \(\lim_0 \ln t = - \infty \) implies that they cannot say that.

So, there's something I didn't catch here. This may be something stupid, but I keep thinking about it without understand why.

Thank you for taking your time to read this and help me !

Bye.

*I've just realize this a first fail since I didn't place this topic into the proper section... I'm trying to delete it but I haven't manage to do se yet. Well... I'll keep trying*
Last edited: