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Proof.
Given p , q > 0 p,q>0 B ( p , q ) B(p,q)
(A.18) B ( p , q ) ≡ ∫ 0 1 t p − 1 ( 1 − t ) q − 1 𝑑 t . B(p,q)\equiv\int_{0}^{1}t^{p-1}(1-t)^{q-1}dt.
Then the beta function satisfies B ( p , q ) = Γ ( p ) Γ ( q ) Γ ( p + q ) B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}
(A.19) ( β ) k ( α ) k \displaystyle\frac{(\fb)_{k}}{(\fa)_{k}} = \displaystyle= Γ ( β + k ) Γ ( β ) ⋅ Γ ( α ) Γ ( α + k ) \displaystyle\frac{\Gamma(\fb+k)}{\Gamma(\fb)}\cdot\frac{\Gamma(\fa)}{\Gamma(\fa+k)}
= \displaystyle= Γ ( α ) Γ ( β ) ⋅ B ( β + k , α − β ) Γ ( α − β ) \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)}\cdot\frac{B(\fb+k,\fa-\fb)}{\Gamma(\fa-\fb)}
= \displaystyle= Γ ( α ) Γ ( β ) Γ ( α − β ) ∫ 0 1 t β + k − 1 ( 1 − t ) α − β − 1 𝑑 t . \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}t^{\fb+k-1}(1-t)^{\fa-\fb-1}dt.
Now we return to the definition of Φ ♯ \Ku
(A.20) Φ ♯ ( β , α , y ) \displaystyle\Ku(\beta,\alpha,y) = \displaystyle= ∑ k = 0 ∞ ( β ) k ( α ) k ⋅ y k k ! \displaystyle\sum\limits_{k=0}^{\infty}\frac{(\beta)_{k}}{(\alpha)_{k}}\cdot\frac{y^{k}}{k!}
= \displaystyle= Γ ( α ) Γ ( β ) Γ ( α − β ) ∫ 0 1 t β − 1 ( 1 − t ) α − β − 1 ∑ k = 0 ∞ ( y t ) k − 1 k ! 𝑑 t \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}t^{\beta-1}(1-t)^{\fa-\fb-1}\sum\limits_{k=0}^{\infty}\frac{(yt)^{k-1}}{k!}dt
= \displaystyle= Γ ( α ) Γ ( β ) Γ ( α − β ) ∫ 0 1 e y t t β − 1 ( 1 − t ) α − β − 1 𝑑 t . \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\beta-1}(1-t)^{\fa-\fb-1}dt.
The proof is done.