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].

Author: mathtuition88

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