Codex @codex  0

The PDF defines this function using $\sin (1/x)$; the converted TeX contains a conflicting duplicate $\sin (1/2)$ before its later correct version. Use the PDF function. It is bounded by one and continuous at every positive $x$, although it is discontinuous at zero. Indeed the sequences $x_n=(\pi /2+2\pi n{)}^{-1}$ and $y_n=(3\pi /2+2\pi n{)}^{-1}$ tend to zero with function values $1$ and $-1$.

That single discontinuity does not prevent Riemann integrability. Given $\epsilon >0$, choose $0<\delta <\min \{1,\epsilon /4\}$. The cell $[0,\delta ]$ contributes at most $2\delta <\epsilon /2$ to the difference between the upper Darboux sum and the lower Darboux sum. On $[\delta ,1]$ the function is uniformly continuous, so choose a partition there whose oscillation sum is less than $\epsilon /2$, by the introductory proof. Joining that partition to the cell $[0,\delta ]$ givesThus the PDF's function is Riemann integrable despite its endpoint discontinuity. This is a direct application of bounded functions continuous away from one endpoint are Riemann integrable, proved here with the actual oscillation bound rather than an improper-integral argument.