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Logarithm integrable ?

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'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

Administrator
Staff member
Jan 26, 2012
4,043
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
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.

$\int_{\frac{2}{\pi}}^1 ln(t)dt= - 1- (\frac{2}{\pi}ln(\frac{2}{\pi})- \frac{2}{\pi})$
 
Jan 31, 2012
54
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
Feb 20, 2012
2
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 :)