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Why is $x^a \sin(x)$ Lebesgue-integrable on $[0,\infty[$? It's obviously measurable, so why $\int |f|<\infty$... You can divide the integral by the period of sine, but I don't know how to cope with $x\to0$...

EDIT: $-2<a<-1$

copper.hat
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Greg P.
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  • is it indeed? What is $a$ ? – Mirko Nov 28 '14 at 00:49
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    near 0, $\sin x$ behaves like $x$ – David Holden Nov 28 '14 at 00:59
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    improper integral of $\frac1{x^p}$ from 1 to $\infty$ is convergent when $p>1$, so this is taken care of by $a<-1$. Improper integral of $\frac1{x^p}$ from 0 to 1 is convergent if $p<1$. Near 0 you may replace $\sin x$ with $x$, like $x^a x=x^{a+1}$ so $-1<a+1<0$, that is $p<1$ and the integral is convergent. – Mirko Nov 28 '14 at 01:02
  • In order to derive Lebesgue-Integrability from Rieman-Integrability, how do you get $\lim_{\alpha\to 0} \int_{\alpha}^1 x^a*sin(x) = \int_a^1 x^(a+1)$, only given $\lim sin/x = 1$ – Greg P. Nov 28 '14 at 02:38

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