## ∫sin(1/x) over [0,1]

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
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 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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 $$[0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{l} 0 ~ x\in \mathbb{Q}\\ 1 ~ x\in \mathbb{R}\setminus \mathbb{Q} \end{array}\right.$$ is Riemann-integrable. But this is not the case...

 Quote by micromass 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 $$[0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{l} 0 ~ x\in \mathbb{Q}\\ 1 ~ x\in \mathbb{R}\setminus \mathbb{Q} \end{array}\right.$$ is Riemann-integrable. But this is not the case...
Hello Micromass. Here is the integral version of the squeeze theorem. Basically, that theorem says if you can show the function is between two integrable functions and the integral of the difference of those two functions is smaller than any positive number, then the function in between is also integrable.
Here is the rigorous statement:
http://en.wikibooks.org/wiki/Real_An...nn_integration

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## ∫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 $$\epsilon$$ in two fundamentally different ways.

First you split the problem up in $$[0,\epsilon/4]\cup [\epsilon/4,1]$$. Let's not use epsilon here, but let's use a. So we split up the problem in $$[0,a]\cup [a,1]$$.

In order to apply the squeeze theorem here, you'll need to find for every epsilon, two functions p and q such that

$$\int_0^a{(q-p)}<\epsilon$$

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...

 Quote by micromass Ah fun, I haven't seen that theorem before. But what's fishy about your proof is that you used $$\epsilon$$ in two fundamentally different ways. First you split the problem up in $$[0,\epsilon/4]\cup [\epsilon/4,1]$$. Let's not use epsilon here, but let's use a. So we split up the problem in $$[0,a]\cup [a,1]$$. In order to apply the squeeze theorem here, you'll need to find for every epsilon, two functions p and q such that $$\int_0^a{(q-p)}<\epsilon$$ 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...
Thank you! But I can not see what is wrong with my proof. Yes, $$\epsilon$$ is used to split the interval and to construct the condition in the squeeze theorem. However, these two uses are consistent, because for any $$0<\epsilon<1$$ we can always split the interval like this. Anyway, I am not absolutely sure that I am right. That is why i posted. So please give me a clearer explanation.
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Your function is dependant of epsilon. But the function in the squeeze theorem should be independent of epsilon. That is the main problem.

 Quote by micromass Your function is dependant of epsilon. But the function in the squeeze theorem should be independent of epsilon. That is the main problem.
Sorry, but I disagree.
1. If you look at the proof of the squeeze theorem, I changed the $$\int_a^b$$ to $$\int_0^{\frac{1}{4\varepsilon}}$$. Since p and q are integrable over any [a,b] (a, b $$\in\mathbb{R}$$), p and q are also integrable over the interval I picked. $$\varepsilon$$ 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.
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, the link you provided is correct. I still disagree with your proof however. Maybe another member can shine some light here...

 Quote by micromass Yes, the link you provided is correct. I still disagree with your proof however. Maybe another member can shine some light here...
Anyway, you might be right, but I could not see it. Thank you!
 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 $$\int_{0}^{\frac{\epsilon}{4}} q - p < \delta$$ For every delta > 0. BUT! $$\int_{0}^{\frac{\epsilon}{4}} q - p = \epsilon /2$$ Let $$\delta = \epsilon /4.$$ The thing is, your epsilon is fixed, while I can vary the delta.

 Quote by l'Hôpital 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 $$\int_{0}^{\frac{\epsilon}{4}} q - p < \delta$$ For every delta > 0. BUT! $$\int_{0}^{\frac{\epsilon}{4}} q - p = \epsilon /2$$ Let $$\delta = \epsilon /4.$$ The thing is, your epsilon is fixed, while I can vary the delta.
OK, now I see why I was wrong.
l'Hôpital and Micromass, Thank you!