sin(1/x)/x is improper Riemann Integrable but not Lebesgue Integrable on (0,1]

Consider $f(x)=\frac{\sin(\frac 1x)}{x}$ .
$\begin{aligned} \int_0^1 \frac{\sin(\frac 1x)}{x}\,dx&=\int_\infty^1\frac{\sin u}{\frac 1u}(-\frac{1}{u^2})\,du\\ &=\int_1^\infty\frac{\sin u}{u}\,du. \end{aligned}$
$\begin{aligned} \int_1^R\frac{\sin u}{u}\,du&=[\frac{-\cos u}{u}]_1^R+\int_1^R\frac{-\cos u}{u^2}\,du\\ &=\frac{-\cos R}{R}+\frac{\cos 1}{1}-\int_1^R\frac{\cos u}{u^2}\,du. \end{aligned}$
Note that $\lim_{R\to\infty}\frac{\cos R}{R}=0$ , and $\displaystyle |\int_1^\infty\frac{\cos u}{u^2}\,du|\leq\int_1^\infty\frac{1}{u^2}\,du=1<\infty.$

Thus $\int_1^\infty\frac{\sin u}{u}\,du<\infty$ , and $f$ is improper Riemann integrable.

However note that
$\begin{aligned} \int_1^\infty |\frac{\sin u}{u}|\,du&\geq\int_\pi^{(N+1)\pi}|\frac{\sin u}{u}|\,du\\ &=\sum_{k=1}^N\int_{k\pi}^{(k+1)\pi}|\frac{\sin u}{u}|\,du\\ &=\sum_{k=1}^N\int_0^\pi|\frac{\sin(t+k\pi)}{t+k\pi}|\,dt\\ &=\sum_{k=1}^N\int_0^\pi\frac{|\sin t|}{t+k\pi}\,dt\\ &\geq\sum_{k=1}^N\frac{1}{(k+1)\pi}\int_0^\pi\sin t\,dt\\ &=\frac{2}{\pi}\sum_{k=1}^N\frac{1}{(k+1)} \end{aligned}$
which diverges as $N\to\infty$ (harmonic series).

Thus $f$ is not Lebesgue integrable on $(0,1]$ .

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