# Logarithm integrable ?

#### Spam

##### New member
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'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:

#### Jameson

Staff member
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 If this happens next time just report the post so the Moderators are aware of it. As it stands if you get any infractions over this I'll reverse them since you are making an effort.

Jameson

• Mr Fantastic

#### HallsofIvy

##### Well-known member
MHB Math Helper
I don't understand why you say "since t-> ln(t) is not integrable on $[\frac{2}{\pi}, 1]$". The anti-derivative of ln(t) is t ln(t)- t and that is defined for all in that interval.[FONT=MathJax_Main-Web][[/FONT][FONT=MathJax_Main-Web]2[/FONT] [FONT=MathJax_Math-italic-Web]π[/FONT] [FONT=MathJax_Main-Web],[/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]][/FONT]

You say also "$\lim_{t\to 0} ln(t)= -\infty$". That's true but irrelevant. 0 is not in the given interval.

$\int_{\frac{2}{\pi}}^1 ln(t)dt= - 1- (\frac{2}{\pi}ln(\frac{2}{\pi})- \frac{2}{\pi})$

#### Also sprach Zarathustra

##### Member
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'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 After substitution $x=\frac{1}{t}$, you will get:

$$\int_{0}^{\frac{\pi}{2}} \frac{\ln\cos (x)}{x^2}dx$$

Which can be written as:

$$\int_{0}^{\frac{\pi}{4}} \frac{\ln\cos (x)}{x^2}dx+ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln\cos (x)}{x^2}dx$$

Could you proceed? (Hint: this integral is converges)

Last edited:

#### Spam

##### New member
I don't understand why you say "since t-> ln(t) is not integrable on $[\frac{2}{\pi}, 1]$". The anti-derivative of ln(t) is t ln(t)- t and that is defined for all in that interval.[FONT=MathJax_Main-Web][[/FONT][FONT=MathJax_Main-Web]2[/FONT] [FONT=MathJax_Math-italic-Web]π[/FONT] [FONT=MathJax_Main-Web],[/FONT][FONT=MathJax_Main-Web]1[/FONT][FONT=MathJax_Main-Web]][/FONT]
Oh yeah. This was a mistake. This isn't exactly what I meant. I meant : "since $$t\mapsto \ln\, t$$ is not integrable on $[0,1]$"

You say also "$\lim_{t\to 0} ln(t)= -\infty$". That's true but irrelevant. 0 is not in the given interval.
I said that because we are studying the behaviour of $$f(\frac{2}{\pi} + t)$$ when $$t\to 0^+$$. And we have found that : $$f(\frac{2}{\pi} + t) \sim \ln\, t$$

So I assumed that it is indeed in 0 that we have to see how $$\ln$$ behaves.

Obviously, there is still something I didn't understand. But thank you all for helping me 