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I have this function

$$f(x)=\begin{cases}x^a \cos\big(\frac{1}{x}\big)&x>0\\0&x=0\end{cases}$$

defined on $[0,1]$. I found that $f$ is bounded, differentiable and continuous and for the last part I need to prove that $f$ is Riemann integrable on $[0,1]$. I got a bit stuck (continuity and boundedness imply Riemann integrability, but this doesn't work the other way around, so I can't require $f$ to be continuous or bounded necessary). Can anyone help me?

eranreches
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user9068702
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    I dont understand... you said that $f$ is continuous and bounded, thus it is Riemann integrable. In other words... you are stuck on what exactly? – Masacroso Dec 18 '17 at 13:51
  • f is not continuous and bounded. At the previous parts of the problem, i was asked for what a is f continuous and for what a it is bounded and I did it. Now, as a separate question, I am asked for what a is f Riemann integrable (without having it continuous or bounded necessary). But I can't require f to be continuous and bounded now, as Riemann integrability doesn't imply continuity and boundedness. – user9068702 Dec 18 '17 at 14:01
  • if $f$ is unbounded then its not Riemann integrable. You need to check then the cases when $f$ is bounded and discontinuous (there is, at least, one $a$ where this happen). Also is important to add the information of your previous comment to the question, otherwise it is not clear what you are asking. – Masacroso Dec 18 '17 at 14:08
  • for this last task you can prove, by example, that an integral of Riemann in $[a,b]$ is the same that an integral of Riemann in $(a,b)$ – Masacroso Dec 18 '17 at 14:11
  • @Masacroso thank you! So i got that for all a>0, it is bounded and continuous and so integrable. For a<0, f it is not bounded as x goes to zero. I am stuck when a = 0, what should I do? – user9068702 Dec 18 '17 at 14:51
  • you can prove what happen for $a=0$ in many ways. By example what I said in my last comment. – Masacroso Dec 18 '17 at 16:20

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