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.