積分botの積分

積分,楕円積分,積分bot

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今回はこちらの式を証明します。
$K (x)$ を第一種完全楕円積分とします。

$\int_{0}^{1} \frac{K (x)}{\sqrt{1 - x^{2}}} d x = \frac{Γ {(\frac{1}{4})}^{4}}{16 π}$

Wallisの公式

$\int_{0}^{\frac{π}{2}} \sin^{2 n} θ d θ = \frac{π}{2} \frac{(2 n - 1)!!}{(2 n)!!}$

$I_{n} = \int_{0}^{\frac{π}{2}} \sin^{2 n} θ d θ$ とおく。すると、
$\begin{aligned} I_{n} & = \int_{0}^{\frac{π}{2}} \sin^{2 n} θ d θ \\ = \int_{0}^{\frac{π}{2}} \sin θ \sin^{2 n - 1} θ d θ \\ = [- \cos θ \sin^{2 n - 1} θ]_{0}^{\frac{π}{2}} + (2 n - 1) \int_{0}^{\frac{π}{2}} \cos^{2} θ \sin^{2 n - 2} θ d θ \\ = (2 n - 1) \int_{0}^{\frac{π}{2}} (1 - \sin^{2} θ) \sin^{2 (n - 1)} θ d θ \\ = (2 n - 1) \int_{0}^{\frac{π}{2}} \sin^{2 (n - 1)} θ d θ - (2 n - 1) \int_{0}^{\frac{π}{2}} \sin^{2 n} θ d θ \\ = (2 n - 1) I_{n - 1} - (2 n - 1) I_{n} \end{aligned}$
となるから $I_{n} = \frac{2 n - 1}{2 n} I_{n - 1}$ という漸化式が導かれる。
また、初項である $I_{0}$ は
$\begin{aligned} I_{0} & = \int_{0}^{\frac{π}{2}} \sin^{2 \cdot 0} θ d θ \\ = \int_{0}^{\frac{π}{2}} d θ \\ = \frac{π}{2} \end{aligned}$
だから
$\begin{aligned} I_{n} & = \frac{π}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \dots \frac{2 n - 1}{2 n} \\ = \frac{π}{2} \frac{(2 n - 1)!!}{(2 n)!!} \end{aligned}$

K (x)

の級数展開

$K (x) = \frac{π}{2} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{2} x^{2 n}$

$\begin{aligned} K (x) & = F (\frac{π}{2}, x) \\ = \int_{0}^{\frac{π}{2}} (1 - x^{2} \sin^{2} θ)^{- \frac{1}{2}} d θ \\ = \int_{0}^{\frac{π}{2}} (1 + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{2^{n}} \frac{(x^{2} \sin^{2} θ)^{n}}{n!}) d θ \\ = \frac{π}{2} + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} x^{2 n} \int_{0}^{\frac{π}{2}} \sin^{2 n} θ d θ \\ = \frac{π}{2} + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} x^{2 n} \frac{π}{2} \frac{(2 n - 1)!!}{(2 n)!!} \\ = \frac{π}{2} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{2} x^{2 n} \end{aligned}$

これで証明の準備が整いました。

まず、 $K (x)$ を級数展開して変形します。
$\begin{aligned} \int_{0}^{1} \frac{K (x)}{\sqrt{1 - x^{2}}} d x & = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} \frac{π}{2} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{2} x^{2 n} d x \\ = \frac{π}{2} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{2} \int_{0}^{1} \frac{x^{2 n}}{\sqrt{1 - x^{2}}} d x \\ = \frac{π}{2} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{2} \int_{0}^{\frac{π}{2}} \sin^{2 n} x d x \\ = \frac{π^{2}}{4} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{3} \end{aligned}$
なんだか $\sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{3}$ がとても厄介そうですね。
とりあえずできる範囲で式変形してみます。
$\begin{aligned} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{3} & = \sum_{n = 0}^{\infty} \frac{{(\frac{1}{2})}_{n}^{3}}{n!^{3}} \\ =_{3} F_{2} [\begin{array}{c} \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \\ 1, 1 \end{array}; 1] \end{aligned}$
ここでDixonの恒等式で $a = b = c = \frac{1}{2}$ とすれば
$\begin{aligned} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{3} & =_{3} F_{2} [\begin{array}{c} \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \\ 1, 1 \end{array}; 1] \\ = \frac{Γ (1 + \frac{1}{4}) Γ (1 + \frac{1}{4} - \frac{1}{2} - \frac{1}{2}) Γ (1 + \frac{1}{2} - \frac{1}{2}) Γ (1 + \frac{1}{2} - \frac{1}{2})}{Γ (1 + \frac{1}{2}) Γ (1 + \frac{1}{2} - \frac{1}{2} - \frac{1}{2}) Γ (1 + \frac{1}{4} - \frac{1}{2}) Γ (1 + \frac{1}{4} - \frac{1}{2})} \\ = \frac{\frac{1}{4} Γ (\frac{1}{4}) Γ (\frac{1}{4}) Γ (1) Γ (1)}{\frac{1}{2} Γ (\frac{1}{2}) Γ (\frac{1}{2}) Γ (\frac{3}{4}) Γ (\frac{3}{4})} \\ = \frac{1}{2 π} \frac{Γ {(\frac{1}{4})}^{2} Γ {(\frac{3}{4})}^{2}}{Γ {(\frac{3}{4})}^{4}} \\ = \frac{1}{2 π} \frac{2 π^{2}}{Γ {(\frac{3}{4})}^{4}} \\ = \frac{π}{Γ {(\frac{3}{4})}^{4}} \end{aligned}$
ここで相反公式より $Γ (\frac{3}{4}) = \frac{1}{Γ (\frac{1}{4})} \frac{π}{\sin (\frac{π}{4})} = \frac{\sqrt{2} π}{Γ (\frac{1}{4})}$ が成り立つので
$\begin{aligned} \int_{0}^{1} \frac{K (x)}{\sqrt{1 - x^{2}}} d x & = \frac{π^{2}}{4} \sum_{n = 0}^{\infty} {(\frac{(2 n - 1)!!}{(2 n)!!})}^{3} \\ = \frac{π^{2}}{4} \sum_{n = 0}^{\infty} \frac{{(\frac{1}{2})}_{n}^{3}}{n!^{3}} \\ = \frac{π^{2}}{4}_{3} F_{2} [\begin{array}{c} \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \\ 1, 1 \end{array}; 1] \\ = \frac{π^{3}}{Γ {(\frac{3}{4})}^{4}} \\ = \frac{Γ {(\frac{1}{4})}^{4}}{16 π} \end{aligned}$
レムニスケート周率 $ϖ$ を用いればこの積分は $\frac{ϖ^{2}}{2}$ と表現できます。