ScalingStacks

Proof. [0573]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context · Original author HTML

Proof.

First, we temporarily assume α>β>0\fa>\fb>0. By Lemma A.2,

(A.36) ey⋅Φ♯⁡(α−β,α,−y)\displaystyle e^{y}\cdot\Ku(\fa-\fb,\fa,-y) =\displaystyle= Γ⁡(α)Γ⁡(α−β)​Γ​(β)​∫01ey⁡(1−t)​tα−β−1​(1−t)β−1​𝑑t\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)\Gamma(\fb)}\int_{0}^{1}e^{y(1-t)}t^{\fa-\fb-1}(1-t)^{\fb-1}dt
=\displaystyle= Γ⁡(α)Γ⁡(α−β)​Γ​(β)​∫01ey​s​(1−s)α−β−1​sβ−1​𝑑s\displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)\Gamma(\fb)}\int_{0}^{1}e^{ys}(1-s)^{\fa-\fb-1}s^{\fb-1}ds
=\displaystyle= Φ♯⁡(β,α,y).\displaystyle\Ku(\fb,\fa,y).

Now we prove the general case. Since both ey⋅Φ♯⁡(α−β,α,−y)Γ⁡(α)\frac{e^{y}\cdot\Ku(\fa-\fb,\fa,-y)}{\Gamma(\fa)} and Φ♯⁡(β,α,y)Γ⁡(α)\frac{\Ku(\fb,\fa,y)}{\Gamma(\fa)} are entire functions in ℂ\mathbb{C}, so the standard analytic continuation theorem implies that Φ♯⁡(β,α,y)=ey⋅Φ♯⁡(α−β,α,−y)\Ku(\fb,\fa,y)=e^{y}\cdot\Ku(\fa-\fb,\fa,-y) holds for any arbitrary β∈ℝ\beta\in\mathbb{R} and α∈ℝ∖{0,−1,−2,−3,…}\alpha\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\}. ∎

Original mathematics by the credited authors. Source-backed reader collection; mathematical self-containment is not assessed.