Solution
= Solution
\b[No.] On each <interval> $[k\pi,(k+1)\pi]$, for $k\ge1$,
$$
\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{x}\,dx\ge\frac1{(k+1)\pi}\int_{k\pi}^{(k+1)\pi}|\sin x|\,dx=\frac2{(k+1)\pi}.
$$
The resulting harmonic lower bound diverges. Therefore $\sin x/x\notin L^1(0,\infty)$. Its convergent improper integral does not make it <Lebesgue integrable>; its positive and negative parts both have infinite integral.