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if $\displaystyle \sin \alpha = p \bigg\lfloor \int^{1}_{0}\{\ln x\}dx\bigg\rfloor \;, \alpha \in (0,2\pi)$ .Then $p$ is ?

given $\lfloor x \rfloor $ is floor function of $x$ and $\{x\} = x-\lfloor x \rfloor$

using $\{\ln x \} = \ln (x) - \lfloor \ln x \rfloor $

and $0

could somme help me with this, thanks

  • 0
    Wow where did you get that from?? – 2017-01-06
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    It is not a bounded variation function ! I think $$|\int^{1}_{0}\{\ln x\}\bigg|dx \to \infty$$ – 2017-01-06
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    It seems that for $m>-1$, we have $$\int_{0}^{1} \{\ln x\} x^m dx =\frac {e^{m+1}}{(m+1)(e^{m+1}-1)}-\frac {1}{(1+m)^2} $$ – 2017-01-06

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