The sine integral is improperly Riemann integrable and not Lebesgue integrable

The function $f$ is continuous on $[0,1]$, so its proper integral there exists. By [L1], the improper integral ${\int }_1^{\mathrm{\infty }}\frac{\sin x}{x}dx$ converges. For $R>1$, [L3] gives ${\int }_0^{R}f(x)dx={\int }_0^{1}f(x)dx+{\int }_1^R\frac{\sin x}{x}dx\mathrm{.}$ Passing to the limit as $R\to \mathrm{\infty }$ and using [L4], the improper integral ${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx$ converges. In the terminology of [L6], the sine integral is convergent.

Apply [L2] with the nonincreasing function $g(x)=1\mathrm{/}x$, the bounded-gap sequence $x_j=j\pi$, and the locally integrable oscillatory factor $u(x)=\sin x$. Since ${\int }_{j\pi }^{(j+1)\pi }\mathrm{\mid }\sin x\mathrm{\mid }dx=2(j\ge 1),$ [L2] yields divergence of ${\int }_1^{\mathrm{\infty }}\frac{\mathrm{\mid }\sin x\mathrm{\mid }}{x}dx\mathrm{.}$ Because $\mathrm{\mid }f\mathrm{\mid }$ is continuous on $[0,1]$, its proper integral there exists, so [L3] and [L4] show that ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }f(x)\mathrm{\mid }dx$ also diverges. Thus the improper integral of $\sin x\mathrm{/}x$ is not absolutely convergent.

By [L6], steps 1.1 and 1.2 show that the sine integral is only conditionally convergent. If $f$ were Lebesgue integrable on $[0,\mathrm{\infty })$, then [L5] would force ${\int }_{[0,\mathrm{\infty })}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1<+\mathrm{\infty },$ so $\mathrm{\mid }f\mathrm{\mid }$ would be Lebesgue integrable on the half-line. For each natural number $n\ge 1$, put $h_n:=\mathrm{\mid }f\mathrm{\mid }{\chi }_{[0,n]}\mathrm{.}$ Then $0\le h_n\le h_{n+1}$ and $h_n(x)\uparrow \mathrm{\mid }f(x)\mathrm{\mid }$ for every $x\ge 0$. Since $\mathrm{\mid }f\mathrm{\mid }$ is continuous on $[0,n]$, [L7] makes it Riemann integrable there, and [L8] gives ${\int }_{[0,\mathrm{\infty })}{h}_{n}d{\lambda }_1={\int }_{[0,n]}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1={\int }_0^n\mathrm{\mid }f(x)\mathrm{\mid }dx\mathrm{.}$ Because $h_n\uparrow \mathrm{\mid }f\mathrm{\mid }$, [L9] yields ${\lim }_{n\to \mathrm{\infty }}{\int }_0^n\mathrm{\mid }f(x)\mathrm{\mid }dx={\lim }_{n\to \mathrm{\infty }}{\int }_{[0,\mathrm{\infty })}{h}_{n}d{\lambda }_1={\int }_{[0,\mathrm{\infty })}\mathrm{\mid }f\mathrm{\mid }d{\lambda }_1<+\mathrm{\infty }\mathrm{.}$ By [L4], that finite limit is exactly the improper integral ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }f(x)\mathrm{\mid }dx,$ so the absolute improper integral converges, contradicting step 1.2. Therefore $f$ is not Lebesgue integrable on the half-line, and the statement is false.