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

For fixed $n>0$, Tonelli givesbecause the quotient is bounded near zero. Fubini therefore permits reversing the order for this finite $x$ interval. Direct integration in $x$ yieldsThus the desired integral equals $\pi /2$ minus the integral of the exponential remainder. Its absolute value is at mostConsequently $\boxed{\underset{n\to \infty }{\lim }{\int }_0^n\sin x/xdx=\pi /2}$. The truncation is essential: Fubini is not being applied to an absolutely integrable function over the whole infinite rectangle.

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.