文献あり

Dijksma-Koornwinderの積分公式

前の記事において, Gegenbauerの加法定理を示した. それは積分
$\begin{aligned} \int_{- 1}^{1} C_{n}^{(a)} (s t + x \sqrt{(1 - s^{2}) (1 - t^{2})}) C_{m}^{(a - \frac{1}{2})} (x) (1 - x^{2})^{a - 1} d x \\ = \frac{\sqrt{π} Γ (a)}{Γ (a + \frac{1}{2})} \frac{(2 a - 1)_{m} (n - m)! (a)_{m}^{2}}{m! (2 a)_{n + m}} {(4 \sqrt{(1 - s^{2}) (1 - t^{2})})}^{m} C_{n - m}^{(a + m)} (s) C_{n - m}^{(a + m)} (t) \end{aligned}$
として表される. 特に $m = 0$ の場合
$\begin{aligned} \int_{- 1}^{1} C_{n}^{(a)} (s t + x \sqrt{(1 - s^{2}) (1 - t^{2})}) (1 - x^{2})^{a - 1} d x \\ = \frac{\sqrt{π} Γ (a)}{Γ (a + \frac{1}{2})} \frac{n!}{(2 a)_{n}} C_{n}^{(a)} (s) C_{n}^{(a)} (t) \end{aligned}$
となって, これは右辺の積の積分表示と考えられる. 前の記事においてもその積分をJacobi多項式に拡張したが, それは積の形にはならなかった. 今回は二重積分によってJacobi多項式の積を表す公式を示す. まずJacobi多項式は, 以下のように定義される.
$\begin{aligned} P_{n}^{(a, b)} (x) & = \frac{(a + 1)_{n}}{n!}_{2} F_{1} [\begin{array}{c} - n, a + b + n + 1 \\ a + 1 \end{array}; \frac{1 - x}{2}] \end{aligned}$
すると, 次が成り立つ.

Dijksma-Koornwinder(1971)

非負整数 $n$ に対し,
$\begin{aligned} P_{n}^{(a, b)} (1 - 2 s^{2}) P_{n}^{(a, b)} (1 - 2 t^{2}) \\ = \frac{Γ (n + a + 1) Γ (n + b + 1)}{π n! (a + b + 1)_{n} Γ (a + \frac{1}{2}) Γ (b + \frac{1}{2})} \\ \cdot \int_{- 1}^{1} \int_{- 1}^{1} C_{2 n}^{(a + b + 1)} (s t u + v \sqrt{(1 - s^{2}) (1 - t^{2})}) (1 - u^{2})^{a - \frac{1}{2}} (1 - v^{2})^{b - \frac{1}{2}} d u d v \end{aligned}$
が成り立つ.

$\begin{aligned} C_{2 n}^{(a)} (x) & = (- 1)^{n} \frac{(a)_{n}}{n!} \sum_{k = 0}^{n} \frac{(- n, a + n)_{k}}{k! {(\frac{1}{2})}_{k}} x^{2 k} \end{aligned}$
と表されるので, 右辺の積分の部分は
$\begin{aligned} \int_{- 1}^{1} \int_{- 1}^{1} C_{2 n}^{(a + b + 1)} (s t u + v \sqrt{(1 - s^{2}) (1 - t^{2})}) (1 - u^{2})^{a - \frac{1}{2}} (1 - v^{2})^{b - \frac{1}{2}} d u d v \\ = (- 1)^{n} \frac{(a + b + 1)_{n}}{n!} \sum_{k = 0}^{n} \frac{(- n, a + b + n + 1)_{k}}{k! (\frac{1}{2})_{k}} \\ \cdot \int_{- 1}^{1} \int_{- 1}^{1} {(s t u + v \sqrt{(1 - s^{2}) (1 - t^{2})})}^{2 k} (1 - u^{2})^{a - \frac{1}{2}} (1 - v^{2})^{b - \frac{1}{2}} d u d v \\ = (- 1)^{n} \frac{(a + b + 1)_{n}}{n!} \sum_{0 \leq i, j} \frac{(2 i + 2 j)! (- n, a + b + n + 1)_{i + j}}{(i + j)! (2 i)! (2 j)! (\frac{1}{2})_{i + j}} (s t)^{2 i} ((1 - s^{2}) (1 - t^{2}))^{j} \\ \cdot \int_{- 1}^{1} \int_{- 1}^{1} u^{2 i} v^{2 j} (1 - u^{2})^{a - \frac{1}{2}} (1 - v^{2})^{b - \frac{1}{2}} d u d v \\ = (- 1)^{n} \frac{(a + b + 1)_{n}}{n!} \sum_{0 \leq i, j} \frac{(2 i + 2 j)! (- n, a + b + n + 1)_{i + j}}{(i + j)! (2 i)! (2 j)! (\frac{1}{2})_{i + j}} (s t)^{2 i} ((1 - s^{2}) (1 - t^{2}))^{j} \\ \cdot \frac{Γ (\frac{1}{2} + i) Γ (a + \frac{1}{2})}{Γ (a + i + 1)} \frac{Γ (\frac{1}{2} + j) Γ (b + \frac{1}{2})}{Γ (b + j + 1)} \\ = (- 1)^{n} \frac{(a + b + 1)_{n}}{n!} \sum_{0 \leq i, j} \frac{(- n, a + b + n + 1)_{i + j}}{i! j! {(\frac{1}{2})}_{i} {(\frac{1}{2})}_{j}} (s t)^{2 i} ((1 - s^{2}) (1 - t^{2}))^{j} \\ \cdot \frac{Γ (\frac{1}{2} + i) Γ (a + \frac{1}{2})}{Γ (a + i + 1)} \frac{Γ (\frac{1}{2} + j) Γ (b + \frac{1}{2})}{Γ (b + j + 1)} \\ = (- 1)^{n} \frac{(a + b + 1)_{n}}{n!} \frac{π Γ (a + \frac{1}{2}) Γ (b + \frac{1}{2})}{Γ (a + 1) Γ (b + 1)} \sum_{0 \leq i, j} \frac{(- n, a + b + n + 1)_{i + j}}{i! j! (a + 1)_{i} (b + 1)_{j}} (s t)^{2 i} ((1 - s^{2}) (1 - t^{2}))^{j} \end{aligned}$
ここで, Watsonの積公式より
$\begin{aligned} \sum_{0 \leq i, j} \frac{(- n, a + b + n + 1)_{i + j}}{i! j! (a + 1)_{i} (b + 1)_{j}} (s t)^{2 i} ((1 - s^{2}) (1 - t^{2}))^{j} \\ = (- 1)^{n} \frac{(a + 1)_{n}}{(b + 1)_{n}}_{2} F_{1} [\begin{array}{c} - n, a + b + n + 1 \\ a + 1 \end{array}; s^{2}]_{2} F_{1} [\begin{array}{c} - n, a + b + n + 1 \\ b + 1 \end{array}; t^{2}] \\ = (- 1)^{n} \frac{n!^{2}}{(a + 1, b + 1)_{n}} P_{n}^{(a, b)} (1 - 2 s^{2}) P_{n}^{(a, b)} (1 - 2 t^{2}) \end{aligned}$
となるから, これを代入して定理を得る.