2F1に関するKummerの定理, Gaussの第二定理, Baileyの定理

超幾何級数の $- 1$ における特殊値公式として, 以下のようなものがある.

Kummerの定理

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ 1 + a - b \end{array}; - 1] & = \frac{Γ (1 + \frac{a}{2}) Γ (1 + a - b)}{Γ (1 + a) Γ (1 + \frac{a}{2} - b)} \end{aligned}$

超幾何級数のEuler積分表示より,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ 1 + a - b \end{array}; - 1] & = \frac{Γ (1 + a - b)}{Γ (a) Γ (1 - b)} \int_{0}^{1} x^{a - 1} (1 - x)^{- b} (1 + x)^{- b} d x \\ = \frac{Γ (1 + a - b)}{Γ (a) Γ (1 - b)} \int_{0}^{1} x^{a - 1} (1 - x^{2})^{- b} d x \\ = \frac{Γ (1 + a - b)}{Γ (a) Γ (1 - b)} \frac{Γ (\frac{a}{2}) Γ (1 - b)}{Γ (1 + \frac{a}{2} - b)} \\ = \frac{Γ (1 + \frac{a}{2}) Γ (1 + a - b)}{Γ (1 + a) Γ (1 + \frac{a}{2} - b)} \end{aligned}$
と示される.

次に $\frac{1}{2}$ における特殊値を得るために以下の公式を示しておく.

Pfaffの変換公式

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ c \end{array}; x] & = (1 - x)^{- a}_{2} F_{1} [\begin{array}{c} a, c - b \\ c \end{array}; \frac{x}{x - 1}] \end{aligned}$

超幾何級数のEuler積分表示において, $t \mapsto 1 - t$ と変数変換すると
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ c \end{array}; x] & = \frac{Γ (c)}{Γ (b) Γ (c - b)} \int_{0}^{1} t^{b - 1} (1 - t)^{c - b - 1} (1 - x t)^{- a} d t \\ = (1 - x)^{- a} \frac{Γ (c)}{Γ (b) Γ (c - b)} \int_{0}^{1} t^{c - b - 1} (1 - t)^{b - 1} {(1 - \frac{x}{x - 1} t)}^{- a} d t \\ = (1 - x)^{- a}_{2} F_{1} [\begin{array}{c} a, c - b \\ c \end{array}; \frac{x}{x - 1}] \end{aligned}$
となって示される.

以下, 2つの $\frac{1}{2}$ における超幾何級数の特殊値公式を示す.

Gaussの第二定理

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ \frac{a + b + 1}{2} \end{array}; \frac{1}{2}] & = \frac{Γ (\frac{1}{2}) Γ (\frac{a + b + 1}{2})}{Γ (\frac{a + 1}{2}) Γ (\frac{b + 1}{2})} \end{aligned}$

Pfaffの変換公式より,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ \frac{a + b + 1}{2} \end{array}; \frac{1}{2}] & = 2^{a}_{2} F_{1} [\begin{array}{c} a, \frac{1 + a - b}{2} \\ \frac{a + b + 1}{2} \end{array}; - 1] \end{aligned}$
ここで, Kummerの定理とLegendreの倍角公式を用いると,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, \frac{1 + a - b}{2} \\ \frac{a + b + 1}{2} \end{array}; - 1] & = \frac{Γ (1 + \frac{a}{2}) Γ (\frac{a + b + 1}{2})}{Γ (1 + a) Γ (\frac{b + 1}{2})} \\ = \frac{Γ (\frac{1}{2}) Γ (\frac{a + b + 1}{2})}{2^{a} Γ (\frac{a + 1}{2}) Γ (\frac{b + 1}{2})} \end{aligned}$
となって示される.

Baileyの定理

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, 1 - a \\ c \end{array}; \frac{1}{2}] & = \frac{Γ (\frac{c}{2}) Γ (\frac{c + 1}{2})}{Γ (\frac{a + c}{2}) Γ (\frac{1 + c - a}{2})} \end{aligned}$

Pfaffの変換公式より,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, 1 - a \\ c \end{array}; \frac{1}{2}] & = 2^{a}_{2} F_{1} [\begin{array}{c} a + c - 1, a \\ c \end{array}; - 1] \end{aligned}$
ここで, Kummerの定理とLegendreの倍角公式を用いて,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a + c - 1, a \\ c \end{array}; - 1] & = \frac{Γ (\frac{a + c + 1}{2}) Γ (c)}{Γ (a + c) Γ (\frac{1 + c - a}{2})} \\ = 2^{- a} \frac{Γ (\frac{c}{2}) Γ (\frac{c + 1}{2})}{Γ (\frac{a + c}{2}) Γ (\frac{1 + c - a}{2})} \end{aligned}$
となるから定理を得る.