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Explicit sectorial remainder bound for the right-half-plane log-Gamma branch

Statement

Fix $0 < δ \leq π / 2$ and $z \in C$ with $Re z > 0$ . For the branch $ℓ_{+}$ from Log-Gamma branch on the right half-plane, Binet integral identity for the right-half-plane log-Gamma branch gives $ℓ_{+} (z) = (z - \frac{1}{2}) Log z - z + \frac{1}{2} \log (2 π) + R (z), R (z) = \int_{0}^{\infty} \frac{Φ (u)}{(u + z)^{2}} d u,$ where $Φ (u) = \int_{0}^{u} (⌊ v ⌋ - v + 1 / 2) d v$ . Then $∣ R (z) ∣ \leq \frac{1}{8 \sin^{2} (δ / 2) ∣ z ∣} .$ For $δ = π / 2$ , the bound is $1 / (4 ∣ z ∣)$ . The relative error after exponentiating the three main terms is at most $\exp (∣ R (z) ∣) - 1$ .

Facts & Assumptions

Given: $0 < δ \leq π / 2$ and $Re z > 0$ .

[F1]

Binet integral identity for the right-half-plane log-Gamma branch: the exact branch identity and its remainder representation hold, with $Φ (n + t) = t (1 - t) / 2$ on each unit interval.

[F2]

Real and imaginary parts, complex conjugation, and modulus: for $z = x + i y$ , $∣ z ∣^{2} = x^{2} + y^{2}$ .

[F3]

Signs, monotonicity intervals, and ranges of sine and cosine and Trigonometric identities, Machin's formula and reduced Taylor tails: cosine decreases on $[0, π]$ and $\cos (π / 2) = 0$ , hence $\cos δ \geq 0$ for $0 < δ \leq π / 2$ .

[F4]

Trigonometric identities, Machin's formula and reduced Taylor tails: the cosine addition identity and $\sin^{2} x + \cos^{2} x = 1$ give $\sin^{2} (δ / 2) = (1 - \cos δ) / 2$ and $\cos^{2} (δ / 2) = (1 + \cos δ) / 2$ .

Proof

1.1F1algebra

The formula in [F1] defines $Φ (n + t) = t (1 - t) / 2$ for integer $n \geq 0$ and $0 \leq t \leq 1$ . Hence $0 \leq Φ (u) \leq 1 / 8$ for every $u \geq 0$ .

1.2givenF2F3F4algebra

Write $z = x + i y$ , with $x = Re z > 0$ , and put $r = ∣ z ∣$ . By [F2], $∣ u + z ∣^{2} = (u + x)^{2} + y^{2} = u^{2} + r^{2} + 2 u x \geq u^{2} + r^{2} \geq u^{2} + r^{2} - 2 u r \cos δ,$ since $u \geq 0$ and $\cos δ \geq 0$ by [F3]. The difference between the last expression and $\sin^{2} (δ / 2) (u + r)^{2}$ is $\cos^{2} (δ / 2) (u - r)^{2} \geq 0$ by [F4]. Thus $∣ u + z ∣^{- 2} \leq \sin^{- 2} (δ / 2) (u + r)^{- 2}$ .

2.1F1step 1.1step 1.2algebra∎

Applying steps 1.1 and 1.2 to the remainder in [F1] gives $∣ R (z) ∣ \leq \frac{1}{8 \sin^{2} (δ / 2)} \int_{0}^{\infty} \frac{d u}{(u + r)^{2}} = \frac{1}{8 \sin^{2} (δ / 2) r} .$ For any complex $w$ , $∣ e^{w} - 1 ∣ \leq e^{∣ w ∣} - 1$ by the exponential series; apply this with $w = R (z)$ .

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Sources

DLMF, §5.9(i), Binet's integral representations (standard reference, not scraped)
DLMF, §5.11(i), Stirling's expansion for log-Gamma (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2–5 (standard reference, not scraped)