An improper Riemann integral that no Lebesgue integral reproduces: the sine integral
Statement
Let for and . Then the improper Riemann integral converges,
while
because on the integrand is at least , whose integral is , and the harmonic series diverges. Since the Lebesgue integral is absolute, not being integrable means : the Lebesgue integral of over does not exist, although the improper Riemann integral does.
Remarks
Not proved in this library. The divergence of 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 . The value 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 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 over with value ; see Henstock-Kurzweil versus Lebesgue: is Lebesgue integrable iff and 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.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 1 result over 1 level. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis, Ch. 8, Example 36 (standard reference, not scraped)
- Dirichlet integral (Wikipedia) (standard reference, not scraped)