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Proof.
To prove this estimate, we need the following integral representation formula for Φ ♯ ( β , α , y ) \Ku(\fb,\fa,y)
(5.100) Φ ♯ ( β , α , y ) = Γ ( α ) Γ ( α − β ) ⋅ e y ( − y ) 1 − α 2 ⋅ ∫ 0 ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − yt ) dt . \Ku(\fb,\fa,y)=\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.
The proof is included in Lemma A.2 A
The key point in the proof of (5.98 I α − 1 I_{\alpha-1} 5.5 α = 1 − 1 n \alpha=1-\frac{1}{n} α − 1 = − 1 n ≥ − 1 2 \alpha-1=-\frac{1}{n}\geq-\frac{1}{2} I α − 1 I_{\alpha-1} 5.48 5.5
(5.101) I α − 1 ( 2 − y t ) = I − 1 n ( 2 − y t ) \displaystyle I_{\alpha-1}(2\sqrt{-yt})=I_{-\frac{1}{n}}(2\sqrt{-yt}) ≤ \displaystyle\leq C n ⋅ max { ( 2 − y t ) − 1 n , ( 2 − y t ) − 1 2 ⋅ e 2 − y t } \displaystyle C_{n}\cdot\max\Big\{(2\sqrt{-yt})^{-\frac{1}{n}},(2\sqrt{-yt})^{-\frac{1}{2}}\cdot e^{2\sqrt{-yt}}\Big\}
≤ \displaystyle\leq C n ⋅ ( − y t ) − 1 4 ⋅ e 2 − y t . \displaystyle C_{n}\cdot(-yt)^{-\frac{1}{4}}\cdot e^{2\sqrt{-yt}}.
Substituting the above in (5.100
(5.102) ∫ 0 ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − y t ) 𝑑 t \displaystyle\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt ≤ \displaystyle\leq C n ⋅ ∫ 0 ∞ e − t + 2 − y t ⋅ t 2 α − 3 4 − β 𝑑 t \displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-t+2\sqrt{-yt}}\cdot t^{\frac{2\alpha-3}{4}-\beta}dt
= \displaystyle= C n ⋅ ∫ 0 ∞ e − t + 2 − y t + ( 2 α − 3 4 − β ) log t 𝑑 t \displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-t+2\sqrt{-yt}+(\frac{2\alpha-3}{4}-\beta)\log t}dt
= \displaystyle= C n ⋅ ∫ 0 ∞ e − u 2 + 2 − y u + ( α − 2 β − 1 2 ) log u 𝑑 u . \displaystyle C_{n}\cdot\int_{0}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.
Therefore,
(5.103) Φ ♯ ( β , α , y ) \displaystyle\Ku(\fb,\fa,y) ≤ \displaystyle\leq C n ⋅ e y ( − y ) 1 − 2 α 4 Γ ( α − β ) ⋅ ∫ 0 ∞ e − u 2 + 2 − y u + ( α − 2 β − 1 2 ) log u 𝑑 u . \displaystyle C_{n}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{0}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.
Next, Φ ♯ \Ku t ≥ 1 − y t\geq\frac{1}{-y} y ≤ − 1 y\leq-1
(5.104) I α − 1 ( 2 − y t ) ≥ C n − 1 ⋅ e 2 − y t ( − y t ) 1 4 , I_{\alpha-1}(2\sqrt{-yt})\geq C_{n}^{-1}\cdot\frac{e^{2\sqrt{-yt}}}{(-yt)^{\frac{1}{4}}},
and hence
∫ 0 ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − y t ) 𝑑 t \displaystyle\int_{0}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt ≥ ∫ 1 − y ∞ e − t ⋅ t α − 1 2 − β ⋅ I α − 1 ( 2 − y t ) 𝑑 t \displaystyle\geq\int_{\frac{1}{-y}}^{\infty}e^{-t}\cdot t^{\frac{\fa-1}{2}-\fb}\cdot I_{\fa-1}(2\sqrt{-yt})dt
≥ C n − 1 ⋅ ∫ 1 − y ∞ e − t + 2 − y t + ( 2 α − 3 4 − β ) log t 𝑑 t \displaystyle\geq C_{n}^{-1}\cdot\int_{\frac{1}{-y}}^{\infty}e^{-t+2\sqrt{-yt}+(\frac{2\alpha-3}{4}-\beta)\log t}dt
(5.105) = C n − 1 ⋅ ∫ 1 − y ∞ e − u 2 + 2 − y u + ( α − 2 β − 1 2 ) log u 𝑑 u . \displaystyle=C_{n}^{-1}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.
Therefore,
(5.106) Φ ♯ ( β , α , y ) ≥ C n − 1 ⋅ e y ( − y ) 1 − 2 α 4 Γ ( α − β ) ⋅ ∫ 1 − y ∞ e − u 2 + 2 − y u + ( α − 2 β − 1 2 ) log u du . \Ku(\fb,\fa,y)\geq C_{n}^{-1}\cdot\frac{e^{y}(-y)^{\frac{1-2\alpha}{4}}}{\Gamma(\alpha-\beta)}\cdot\int_{\frac{1}{\sqrt{-y}}}^{\infty}e^{-u^{2}+2\sqrt{-y}u+(\alpha-2\beta-\frac{1}{2})\log u}du.
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