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Is $\text{ sgn}(\sin(\frac{\pi}{x}))$ fit the basic setting of Riemann Integrability on $[0,1]$(bounded function, closed and bounded interval)? Since $sgn(x)$ is bounded for all $x$, so $\text{ sgn}(\sin(\frac{\pi}{x}))$ is bounded on $[0,1]$, but $sin(\frac{\pi}{x})$ is undefined for $x=0$, so is it improper integral? or it is ok for basic settings of Riemann Integrability.

Also, is $\text{ sgn}(\sin(\frac{\pi}{x}))$ integrable on $[0,1]$? I know we can use alternating series to find the integral of $\text{ sgn}(\sin(\frac{\pi}{x}))$ by assuming the integrability, but any other proof for integrability of $\text{ sgn}(\sin(\frac{\pi}{x}))$ on $[0,1]$ without using the trick of finding the alternating series.

It would be appreciated if someone can tell me how to calculate the upper integral without using alternating series.

Thank you very much.

shilov
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Belive
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  • It's not defined at zero, but it can be extended to a bounded function on [0,1] by defining it to be some value at zero. The set of zeros $sin(\pi/x)$ are the points $x_k = 1/k$ for some positive integer $k$. This is a countable set, so the function is Riemann integrable by the Lebesgue criterion. – Philip Hoskins Apr 29 '16 at 08:35
  • Taking any partition of $[0,1]$, you can refine it to include some those points. I would start there for the upper integral. – Philip Hoskins Apr 29 '16 at 08:37
  • @PhilipHoskins Like $\text{ sgn}(\sin(\frac{\pi}{x})) = \text{ sgn}(\sin(\frac{\pi}{x}))$ on$(0,1]$ and $\text{ sgn}(\sin(\frac{\pi}{x})) = 0$ for $x=0$ ? – Belive Apr 29 '16 at 09:50
  • That would work. The value you pick won't affect the integral. – Philip Hoskins Apr 29 '16 at 19:26

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