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True or false?

If the function $f$ is defined as follows: $\sin^{ \varepsilon }(x)\ln(x)$ with $\varepsilon>0$. Is $f$ Riemann integrable in the interval $]0,1[$ ? I know piecewise continuity is a sufficient condition and I think $f$ is continuous on the open interval between 0 and 1. The only point where there would be a "problem" is $x=0$ because $\ln(0)$ is undefined, but this point is outside the interval. Could it be that easy? I think there must be something more that I am missing.

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up vote 2 down vote accepted

If $\varepsilon\gt0$, then since $\lim\limits_{x\to0}x\log(x)=0$, $$ \begin{align} \lim_{x\to0}\sin^\varepsilon(x)\log(x) &=\lim_{x\to0}\color{#C00000}{\frac{\sin^\varepsilon(x)}{x^\varepsilon}}x^\varepsilon\log(x)\\ &=\color{#C00000}{1^\varepsilon}\lim_{x\to0}\color{#00A000}{(x^\varepsilon\log(\varepsilon x)-x^\varepsilon\log(\varepsilon))}\\ &=1\color{#00A000}{(0-0)}\\ &=0 \end{align} $$ Therefore, $\sin^\varepsilon(x)\log(x)$ is bounded on $(0,1)$.

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So the answer is yes, because the function is bounded in 0 and in 1? Thank you ! – tim_a Jan 25 '13 at 11:41
@tim_a: That's right. You're welcome. – robjohn Jan 25 '13 at 11:42

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