By Pugh G.R.
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Extra info for An analysis of the Lanczos Gamma approximation
The derivation consists of three steps, which in a nutshell can be broken down as ∞ Γ(z + 1/2) = tz−1/2 e−t dt 0 √ = (z + r + 1/2)z+1/2 e−(z+r+1/2) 2 e [v(1 − log v)]z−1/2 v r dv × (step I) 0 √ = (z + r + 1/2)z+1/2 e−(z+r+1/2) 2 √ π/2 2 v(θ)r sin θ 2z × dθ cos θ log v(θ) −π/2 (step II) √ = (z + r + 1/2)z+1/2 e−(z+r+1/2) 2 π/2 a0 (r) × + cos θ 2 −π/2 ∞ 2z ak (r) cos (2kθ) dθ . 1) then follows from the last integral upon integrating term by term. The derivation is first retraced more or less along the same lines as Lanczos uses in his paper .
However, the integral F (z) diverges with, for example, z = −3/4 and a choice of σ = 5/4. To see this, note that v(1 − log v) ≤ (e − v) on [0, e], so that [v/(e − v)]5/4 ≤ [v(1 − log v)]−5/4 v 5/4 , whence e e [v/(e − v)]5/4 dv ≤ 0 But the left hand side [v(1 − log v)]−5/4 v 5/4 dv . 0 e 0 [v/(e − v)]5/4 dv = ∞. 43 Chapter 3. The Lanczos Formula leading terms similar in form to Stirling’s formula, while factoring the integrand into separate powers of z and r. 4). His steps are perhaps clues to the motivation behind his method and so these are reproduced in Appendix A.
There the author makes the replacement z → z − 1/2 and states (using σ to signify what is here denoted r): Γ(z + 1/2) = (z + σ + 1/2)z+1/2 exp [−(z + σ + 1/2)]F (z), F (z) = e 0 [v(1 − log v)]z−1/2 v σ dv, Re(z + σ + 1/2) > 0 . However, the integral F (z) diverges with, for example, z = −3/4 and a choice of σ = 5/4. To see this, note that v(1 − log v) ≤ (e − v) on [0, e], so that [v/(e − v)]5/4 ≤ [v(1 − log v)]−5/4 v 5/4 , whence e e [v/(e − v)]5/4 dv ≤ 0 But the left hand side [v(1 − log v)]−5/4 v 5/4 dv .