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Proof.
By definition,
(A.38)
I α − 1 ( 2 − y t ) = ∑ k = 0 ∞ ( − y t ) k + α − 1 2 k ! ⋅ Γ ( k + α ) . I_{\fa-1}(2\sqrt{-yt})=\sum\limits_{k=0}^{\infty}\frac{(-yt)^{k+\frac{\fa-1}{2}}}{k!\cdot\Gamma(k+\fa)}.
Integrating the above expansion, it follows that
(A.39)
Γ ( α ) Γ ( α − β ) ⋅ ∫ 0 ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − y t ) 𝑑 t \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
= \displaystyle=
Γ ( α ) Γ ( α − β ) ⋅ ( − y ) α − 1 2 ⋅ ∑ k = 0 ∞ ( − y ) k k ! ⋅ Γ ( k + α ) ⋅ ∫ 0 ∞ e − t ⋅ t α − β + k − 1 𝑑 t \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot(-y)^{\frac{\alpha-1}{2}}\cdot\sum\limits_{k=0}^{\infty}\frac{(-y)^{k}}{k!\cdot\Gamma(k+\fa)}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\fa-\fb+k-1}dt
= \displaystyle=
Γ ( α ) Γ ( α − β ) ⋅ ( − y ) α − 1 2 ⋅ ∑ k = 0 ∞ ( − y ) k ⋅ Γ ( α − β + k ) k ! ⋅ Γ ( k + α ) . \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot(-y)^{\frac{\alpha-1}{2}}\cdot\sum\limits_{k=0}^{\infty}\frac{(-y)^{k}\cdot\Gamma(\alpha-\beta+k)}{k!\cdot\Gamma(k+\fa)}.
By the recursive formula of the Gamma function, Γ ( α − β + k ) Γ ( k + α ) = ( α − β ) k ⋅ Γ ( α − β ) ( α ) k ⋅ Γ ( α ) \frac{\Gamma(\alpha-\beta+k)}{\Gamma(k+\alpha)}=\frac{(\alpha-\beta)_{k}\cdot\Gamma(\alpha-\beta)}{(\alpha)_{k}\cdot\Gamma(\alpha)} , so it follows that
(A.40)
Γ ( α ) Γ ( α − β ) ⋅ ∫ 0 ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − y t ) 𝑑 t \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
= \displaystyle=
( − y ) α − 1 2 ⋅ ∑ k = 0 ∞ ( α − β ) k ( − y ) k ( α ) k ⋅ k ! \displaystyle(-y)^{\frac{\alpha-1}{2}}\cdot\sum\limits_{k=0}^{\infty}\frac{(\alpha-\beta)_{k}(-y)^{k}}{(\alpha)_{k}\cdot k!}
= \displaystyle=
( − y ) α − 1 2 ⋅ Φ ♯ ( α − β , α , − y ) . \displaystyle(-y)^{\frac{\alpha-1}{2}}\cdot\Ku(\fa-\fb,\fa,-y).
Therefore,
(A.41)
Γ ( α ) Γ ( α − β ) ⋅ e y ( − y ) 1 − α 2 ⋅ ∫ 0 ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − y t ) 𝑑 t \displaystyle\frac{\Gamma(\fa)}{\Gamma(\fa-\fb)}\cdot e^{y}(-y)^{\frac{1-\fa}{2}}\cdot\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
= \displaystyle=
e y ⋅ Φ ♯ ( α − β , α , − y ) \displaystyle e^{y}\cdot\Ku(\fa-\fb,\fa,-y)
= \displaystyle=
Φ ♯ ( β , α , y ) . \displaystyle\Ku(\fb,\fa,y).
The last equality follows from Kummer’s transformation law.