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The Liouville constant-mode integral and total-charge bound

Statement

Let $γ > 0$ , $μ > 0$ , an integer $n \geq 0$ , and $α_{1}, \dots, α_{n} \in R$ (with the empty sum $A = 0$ when $n = 0$ ). Let $R > 0$ . Put $Q = 2 / γ + γ / 2$ , $A = \sum_{i} α_{i}$ , and $s = (A - 2 Q) / γ$ . Then

$I : = \int_{R} \exp ⁣ ((A - 2 Q) c - μ R e^{γ c}) d c$

is finite exactly when $A > 2 Q$ ; in that case

$I = \frac{1}{γ} (μ R)^{- s} Γ (s), Γ (s) : = \int_{0}^{\infty} t^{s - 1} e^{- t} d t .$

Facts & Assumptions

Given: $γ, μ, R > 0$ and real charges $α_{i}$ ; $A$ and $s$ are the displayed sums.

[A1]

$Q = 2 / γ + γ / 2$ , so $A - 2 Q = γ s$ .

Proof

1.1givenA1algebra

With $t = μ R e^{γ c}$ , one has $d c = d t / (γ t)$ and $e^{(A - 2 Q) c} = (t / (μ R))^{s}$ . Therefore $I = γ^{- 1} (μ R)^{- s} \int_{0}^{\infty} t^{s - 1} e^{- t} d t$ whenever this improper integral converges.

2.1givenstep 1.1algebra∎

The tail at $t = \infty$ is integrable for every real $s$ because exponential decay dominates every power. At $t ↓ 0$ , the integrand is comparable to $t^{s - 1}$ , whose integral is finite exactly for $s > 0$ ; when $s = 0$ it has a logarithmically divergent integral, and when $s < 0$ it diverges more strongly. Hence $I < \infty$ exactly when $s > 0$ , equivalent to $A > 2 Q$ because $γ > 0$ .

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Colin Guillarmou, Antti Kupiainen, Rémi Rhodes, Vincent Vargas, Conformal Bootstrap in Liouville Theory, arXiv:2005.11530v1