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Proof.
The proof is straightforward.
For example, we only prove
(A.28) Φ ♯ ( β , α , y ) ∼ Γ ( α ) Γ ( α − β ) ⋅ ( − y ) − β \Ku(\fb,\fa,y)\sim\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{-\beta}
as y → − ∞ y\to-\infty u = − y t u=-yt
Φ ♯ ( β , α , y ) \displaystyle\Ku(\fb,\fa,y) = 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^{\fb-1}(1-t)^{\fa-\fb-1}dt
(A.29) = Γ ( α ) Γ ( β ) Γ ( α − β ) ⋅ ( − y ) − β ⋅ ∫ 0 − y e − u u β − 1 ( 1 + u y ) α − β − 1 d u . \displaystyle=\frac{\Gamma(\fa)}{\Gamma(\fb)\Gamma(\fa-\fb)}\cdot(-y)^{-\fb}\cdot\int_{0}^{-y}e^{-u}u^{\fb-1}\Big(1+\frac{u}{y}\Big)^{\fa-\fb-1}du.
Since α − β − 1 > 0 \fa-\fb-1>0 − 1 ≤ u y ≤ 0 -1\leq\frac{u}{y}\leq 0 ( 1 + u y ) α − β − 1 ≤ 1 (1+\frac{u}{y})^{\fa-\fb-1}\leq 1
(A.30) lim y → − ∞ ∫ 0 − y e − u u β − 1 ( 1 + u y ) α − β − 1 𝑑 u = ∫ 0 ∞ e − u u β − 1 𝑑 u = Γ ( β ) . \lim\limits_{y\to-\infty}\int_{0}^{-y}e^{-u}u^{\fb-1}\Big(1+\frac{u}{y}\Big)^{\fa-\fb-1}du=\int_{0}^{\infty}e^{-u}u^{\fb-1}du=\Gamma(\beta).
Therefore, as y → − ∞ y\to-\infty
(A.31) Φ ♯ ( β , α , y ) ∼ Γ ( α ) Γ ( α − β ) ⋅ ( − y ) − β . \Ku(\fb,\fa,y)\sim\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot(-y)^{-\fb}.