ScalingStacks

Proof. [0575]

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Proof.

By definition,

(A.38) Iα−1​(2​−y​t)=∑k=0∞(−y​t)k+α−12k!⋅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⋅tα−12−β⋅Iα−1​(2​−y​t)​𝑑t\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−12⋅∑k=0∞(−y)kk!⋅Γ⁡(k+α)⋅∫0∞e−t⋅tα−β+k−1​𝑑t\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−12⋅∑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⋅tα−12−β⋅Iα−1​(2​−y​t)​𝑑t\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−12⋅∑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−12⋅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)⋅ey​(−y)1−α2⋅∫0∞e−t⋅tα−12−β⋅Iα−1​(2​−y​t)​𝑑t\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= ey⋅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.

∎

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