= Solution
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 $\varepsilon>0$, choose $0<\delta<\min\{1,\varepsilon/4\}$. The cell $[0,\delta]$ contributes at most $2\delta<\varepsilon/2$ 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 $\varepsilon/2$, by the introductory proof. Joining that partition to the cell $[0,\delta]$ gives
$$
\boxed{U(g,P)-L(g,P)<2\delta+\varepsilon/2<\varepsilon.}
$$
Thus \b[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.
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