An improper Riemann integral that no Lebesgue integral reproduces: the sine integral

Let $f(x)={\displaystyle \frac{\sin x}{x}}$ for $x>0$ and $f(0)=1$. Then the improper Riemann integral converges,

${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx:={\lim }_{T\to \mathrm{\infty }}{\int }_0^T\frac{\sin x}{x}dx=\frac{\pi }{2},$

while

${\int }_0^{\mathrm{\infty }}\mid \frac{\sin x}{x}\mid dx=+\mathrm{\infty },$

because on $[k\pi ,(k+1)\pi ]$ the integrand is at least $\frac{\mathrm{\mid }\sin x\mathrm{\mid }}{(k+1)\pi }$, whose integral is $\frac{2}{(k+1)\pi }$, and the harmonic series diverges. Since the Lebesgue integral is absolute, $\mathrm{\mid }f\mathrm{\mid }$ not being integrable means $f\mathrm{\notin }{L}^1(0,\mathrm{\infty })$: the Lebesgue integral of $f$ over $(0,\mathrm{\infty })$ does not exist, although the improper Riemann integral does.

Not proved in this library. The divergence of ${\int }_0^{\mathrm{\infty }}\mathrm{\mid }f\mathrm{\mid }$ and the convergence of the improper integral are both in scope; what is recorded here is the conclusion about the Lebesgue integral (Lebesgue measure and the Lebesgue integral ‡), which cannot be stated without it.

What would prove it. The comparison with the harmonic series shown above, the Dirichlet test for the convergence of the improper integral, and the definition of the Lebesgue integral as an absolute integral, that is $f\in L^1⟺\mathrm{\mid }f\mathrm{\mid }\in L^1$. The value $\pi \mathrm{/}2$ needs a separate argument, by Feynman's trick, by contour integration, or by the Dirichlet kernel.

Which page it serves. The later improper-integrals page proves general Dirichlet and tail-mass criteria from which convergence of this integral and divergence of its absolute integral follow once the required trigonometric estimates are supplied. That concrete application and the value $\pi \mathrm{/}2$ are not proved there, so this item remains a cited boundary statement. The observation that no Lebesgue integral reproduces it belongs here, and it is the standard warning against the belief that the Lebesgue integral extends every integral in use.

The right repair, which is not deferred. The Henstock-Kurzweil integral is a non-absolute integral and integrates $f$ over $(0,\mathrm{\infty })$ with value $\pi \mathrm{/}2$; see Henstock-Kurzweil versus Lebesgue: $f$ is Lebesgue integrable iff $f$ and $\mathrm{\mid }f\mathrm{\mid }$ are both HK integrable ‡. The HK integral is in scope in this library, so the pair of statements can eventually be made here in full, with only the Lebesgue half quoted.