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Absolute convergence of the Liouville Upsilon defining integral

Statement

Let 0<γ<2, Q=2/γ+γ/2, and z∈C satisfy 0<Re⁡z<Q. Set a=Q/2−z. Then

Jγ(z):=∫0∞(a2e−t−sinh⁡2(at/2)sinh⁡(γt/4)sinh⁡(t/γ))dtt

converges absolutely. The convergence is uniform as z ranges over any compact subset of the open strip 0<Re⁡z<Q.

Facts & Assumptions

Given: 0<γ<2 and z in the open strip 0<Re⁡z<Q.

[F1]
[A1]

a=Q/2−z and Dγ(t)=sinh⁡(γt/4)sinh⁡(t/γ) for t>0.

Proof

1.1givenF1A1algebra

Fix a compact subset K of the strip. Uniformly for z∈K, Taylor expansions at t=0 give Dγ(t)=t2/4+O(t4), sinh⁡2(at/2)=a2t2/4+OK(t4), and a2e−t=a2−a2t+OK(t2). Hence sinh⁡2(at/2)/Dγ(t)=a2+OK(t2) and the integrand of Jγ is −a2+OK(t); it is bounded uniformly near 0.

2.1F1A1step 1.1algebra∎

From [F1], γ/4+1/γ=Q/2. On K, compactness and the strict strip inequalities give η:=min⁡z∈K(Q/2−∣Re⁡a∣)>0. For t≥1, Dγ(t)≥cγeQt/2 for a constant cγ>0, while ∣sinh⁡(at/2)∣2≤e∣Re⁡a∣t. Thus the second term inside parentheses is bounded by CKe−ηt; the first is bounded by CKe−t. After division by t, these bounds are integrable on [1,∞) uniformly for z∈K. Combining with step 1.1 proves the asserted absolute and locally uniform convergence.

Remarks

At either boundary Re⁡z=0 or Re⁡z=Q, ∣Re⁡a∣=Q/2, the quotient ∣sinh⁡(at/2)∣2/Dγ(t) tends to 1, and a2e−t tends to zero. The integrand is therefore asymptotic in magnitude to 1/t, so the dt/t tail is not absolutely integrable. The open-strip hypothesis is needed for this integral definition.

Depends on

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