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∫sin(1/x) over [0,1]by mathdunce
Tags: ∫sin1 or x 
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#1
Nov2910, 05:24 PM

P: 16

1. The problem statement, all variables and given/known data
2. Relevant equations Prove that f is integrable over [0,1] 3. The attempt at a solution I have seen some quite complicated solution on the web. For example, http://www.astarmathsandphysics.com/...ided_by_x.html My solution is a little different. Could you please check whether mine is correct? Thank you! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Nov2910, 05:42 PM

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There's something fishy about your proof. But I can't immediately put my hand on it...
Can you please state the squeeze theorem for me? My version of the squeeze theorem says something about limits, but I guess that's not what you're using here... Your proof would actually imply that the function [tex][0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{l} 0 ~ x\in \mathbb{Q}\\ 1 ~ x\in \mathbb{R}\setminus \mathbb{Q} \end{array}\right.[/tex] is Riemannintegrable. But this is not the case... 


#3
Nov2910, 06:27 PM

P: 16

Here is the rigorous statement: http://en.wikibooks.org/wiki/Real_An...nn_integration 


#4
Nov2910, 06:41 PM

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P: 18,218

∫sin(1/x) over [0,1]
Ah fun, I haven't seen that theorem before.
But what's fishy about your proof is that you used [tex]\epsilon[/tex] in two fundamentally different ways. First you split the problem up in [tex][0,\epsilon/4]\cup [\epsilon/4,1][/tex]. Let's not use epsilon here, but let's use a. So we split up the problem in [tex][0,a]\cup [a,1][/tex]. In order to apply the squeeze theorem here, you'll need to find for every epsilon, two functions p and q such that [tex]\int_0^a{(qp)}<\epsilon[/tex] But the functions p and q do not apply in this case. Hmm, I have a little trouble to explain what is wrong in this case. I hope you can see it... 


#5
Nov2910, 06:55 PM

P: 16




#6
Nov2910, 06:58 PM

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P: 18,218

Your function is dependant of epsilon. But the function in the squeeze theorem should be independent of epsilon. That is the main problem.



#7
Nov2910, 07:18 PM

P: 16

1. If you look at the proof of the squeeze theorem, I changed the [tex]\int_a^b[/tex] to [tex]\int_0^{\frac{1}{4\varepsilon}}[/tex]. Since p and q are integrable over any [a,b] (a, b [tex]\in\mathbb{R}[/tex]), p and q are also integrable over the interval I picked. [tex]\varepsilon[/tex] here only means that interval can be arbitrarily small, but as long as it is positive, the integrability still holds. 2. The proof at http://www.astarmathsandphysics.com/...ided_by_x.html uses essentially very similar idea. In fact, they also split the interval and use something like a squeeze theorem. 


#8
Nov2910, 07:23 PM

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P: 18,218

Yes, the link you provided is correct. I still disagree with your proof however. Maybe another member can shine some light here...



#9
Nov2910, 07:27 PM

P: 16




#10
Nov2910, 07:36 PM

P: 256

You are abusing notation, terribly.
So, in your proof, you want to show f is integrable in [0,e/4]. In order to apply this squeeze theorem [tex] \int_{0}^{\frac{\epsilon}{4}} q  p < \delta [/tex] For every delta > 0. BUT! [tex] \int_{0}^{\frac{\epsilon}{4}} q  p = \epsilon /2 [/tex] Let [tex] \delta = \epsilon /4. [/tex] The thing is, your epsilon is fixed, while I can vary the delta. 


#11
Nov2910, 09:33 PM

P: 16

l'Hôpital and Micromass, Thank you! 


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