Documentation

TauCeti.Analysis.Contour.WorkedExamples.HalfDisc.SineIntegral

The Dirichlet integral ∫₀^∞ sin x / x dx = π / 2 #

WorkedExamples/HalfDisc/Dirichlet.lean evaluates the Hungerbühler--Wasem motivating example in its complex form: the Cauchy principal values of e^{iaz}/z along the real segment [-R, R] converge to π i as R → ∞, the pole at the origin sitting on the contour and contributing only half its residue. This file cashes that in for the classical real statement it exists to prove — the Dirichlet integral

lim_{R → ∞} ∫_0^R sin (a x) / x dx = π / 2 for every frequency a > 0,

and in particular lim_{R → ∞} ∫_0^R sin x / x dx = π / 2. The limit is genuinely improper: x ↦ sin x / x is not Lebesgue integrable on [0, ∞), so there is no Bochner integral over Ioi 0 to state this about, and the result is a statement about the truncated integrals.

The bridge is a second, independent evaluation of the same principal value. Along the real axis the Dirichlet integrand splits as

e^{i a t} / t = cos (a t) / t + i · sin (a t) / t,

and the symmetric ε-excision at the origin treats the two parts completely differently:

So the principal value equals i · ∫_{-R}^{R} sin (a t) / t dt, and comparing with the half-disc evaluation of Dirichlet.lean through uniqueness of the principal value (HasCauchyPV.unique) turns the complex identity into a real one. Jordan's lemma disposes of the arc as R → ∞, and evenness of t ↦ sin (a t) / t halves the symmetric integral.

The frequency a is arbitrary in the principal-value identity — neither the cancellation nor the bound above needs anything of it — and only the passage to the limit R → ∞ requires a > 0, since that is where Jordan's lemma enters.

Main results #

References #

The Dirichlet integrand along the real axis #

The real and imaginary parts of the Dirichlet integrand along the real axis. For real t, e^{iat}/t = cos (a t)/t + i · sin (a t)/t. Both sides are 0 at t = 0 under the division-by-zero convention, so no hypothesis on t is needed.

Integrability of the excised real parts #

The two halves of the excised integral #

The principal value along the real segment, evaluated twice #

theorem TauCeti.Contour.hasCauchyPVAt_realSegment_dirichlet (a R : ℝ) :
HasCauchyPVAt (fun (t : ℝ) => ↑t) (-R) R (dirichletIntegrand a) 0 (↑(∫ (t : ℝ) in -R..R, Real.sin (a * t) / t) * Complex.I)

The principal value along the real segment, evaluated directly. For every frequency a and every real R, the Cauchy principal value of e^{iat}/t along [-R, R], excising the origin symmetrically, is i times the ordinary integral of sin (a t) / t.

Both defining clauses come from the split of the integrand into cos (a t)/t + i · sin (a t)/t: the excised real part integrates to 0 by oddness for every ε, while the imaginary part is bounded and its excised integrals converge by dominated convergence. Reversing a negative-radius interval negates both sides. No positivity of a is needed.

The two evaluations agree. Comparing the direct evaluation above with the half-disc evaluation of Dirichlet.lean through uniqueness of the principal value turns the complex identity into a real one: for each radius, i · ∫_{-R}^{R} sin (a t)/t dt is π i minus the arc contribution.

The improper integral #

The symmetric Dirichlet integral. For a positive frequency the integrals over [-R, R] converge to π: Jordan's lemma kills the arc contribution of integral_sin_mul_div_neg_self_eq.

theorem TauCeti.Contour.tendsto_integral_sin_mul_div_atTop {a : ℝ} (ha : 0 < a) :
Filter.Tendsto (fun (R : ℝ) => ∫ (x : ℝ) in 0..R, Real.sin (a * x) / x) Filter.atTop (nhds (Real.pi / 2))

The Dirichlet integral, for an arbitrary positive frequency. ∫_0^R sin (a x) / x dx → π / 2 as R → ∞, for every a > 0 — the value is independent of the frequency. This is the Hungerbühler--Wasem motivating application: the pole of e^{iaz}/z sits on the contour of integration, so the classical residue theorem does not apply directly to this contour. The generalized theorem handles it without an indentation and weights the residue by the winding number ½ of a point on a smooth arc.

The limit is genuinely improper: x ↦ sin (a x) / x is not Lebesgue integrable on [0, ∞), so this is a statement about the truncated integrals and not about an integral over Ioi 0.

The Dirichlet integral: ∫_0^R sin x / x dx → π / 2 as R → ∞.

The Dirichlet integral in Mathlib's Real.sinc spelling. The two integrands differ only at the single point 0, so the integrals agree.