Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D / 4 / b / ii / Solution

No. On each interval $[k\pi ,(k+1)\pi ]$, for $k\ge 1$,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.