Gegenbauer多項式の積の公式2

結構前の記事において, 以下を示した.

$l + m + n = 2 N$ が偶数のとき,
$\begin{aligned} \int_{- 1}^{1} C_{l}^{(a)} (x) C_{m}^{(a)} (x) C_{n}^{(a)} (x) (1 - x^{2})^{a - \frac{1}{2}} d x & = \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n}}{(N - l)! (N - m)! (N - n)! (a)_{N + 1}} \frac{2^{1 - 2 a} π Γ (N + 2 a)}{Γ (a)^{2}} \end{aligned}$

今回はこれに関して直接展開して項別積分するとどうなるのかについて考えてみたいと思う.
$\begin{aligned} C_{n}^{(a)} (x) & = \frac{(2 a)_{n}}{n!}_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; \frac{1 - x}{2}] \end{aligned}$
であるから, $x \to 1 - 2 x$ と変換すると,
$\begin{aligned} 2^{2 a} \frac{(2 a)_{l} (2 a)_{m} (2 a)_{n}}{l! m! n!} \int_{0}^{1}_{2} F_{1} [\begin{array}{c} - l, 2 a + l \\ a + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} - m, 2 a + m \\ a + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; x] (x (1 - x))^{a - \frac{1}{2}} d x \\ = \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n}}{(N - l)! (N - m)! (N - n)! (a)_{N + 1}} \frac{2^{1 - 2 a} π Γ (N + 2 a)}{Γ (a)^{2}} \end{aligned}$
となるから,
$\begin{aligned} \int_{0}^{1}_{2} F_{1} [\begin{array}{c} - l, 2 a + l \\ a + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} - m, 2 a + m \\ a + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; x] (x (1 - x))^{a - \frac{1}{2}} d x \\ = \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n} l! m! n!}{(N - l)! (N - m)! (N - n)! (a)_{N + 1} (2 a)_{l} (2 a)_{m} (2 a)_{n}} \frac{2^{1 - 4 a} π Γ (N + 2 a)}{Γ (a)^{2}} \end{aligned}$
である. 超幾何級数を展開して項別積分すると,
$\begin{aligned} \sum_{0 \leq i} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \frac{(- n, 2 a + n)_{k}}{k! {(a + \frac{1}{2})}_{k}} \frac{{(a + \frac{1}{2})}_{i + j + k}}{(2 a + 1)_{i + j + k}} \\ = \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n} l! m! n!}{(N - l)! (N - m)! (N - n)! (a)_{N + 1} (2 a)_{l} (2 a)_{m} (2 a)_{n}} \frac{2^{1 - 4 a} π Γ (N + 2 a)}{Γ (a)^{2}} \frac{Γ (2 a + 1)}{Γ {(a + \frac{1}{2})}^{2}} \\ = \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n} l! m! n! (2 a)_{N}}{(N - l)! (N - m)! (N - n)! (a + 1)_{N} (2 a)_{l} (2 a)_{m} (2 a)_{n}} \end{aligned}$
となる. つまり, 以下の三重超幾何級数の和公式が成り立つことになる.

$l + m + n = 2 N$ が偶数のとき,
$\begin{aligned} \sum_{0 \leq i} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \frac{(- n, 2 a + n)_{k}}{k! {(a + \frac{1}{2})}_{k}} \frac{{(a + \frac{1}{2})}_{i + j + k}}{(2 a + 1)_{i + j + k}} \\ = \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n} l! m! n! (2 a)_{N}}{(N - l)! (N - m)! (N - n)! (a + 1)_{N} (2 a)_{l} (2 a)_{m} (2 a)_{n}} \end{aligned}$
が成り立つ.

しかし, 上の和公式を直接示すのもそれほど簡単ではなさそうである. 積分の別の展開の仕方を考えてみる.
$\begin{aligned} \int_{0}^{1}_{2} F_{1} [\begin{array}{c} - l, 2 a + l \\ a + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} - m, 2 a + m \\ a + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; x] (x (1 - x))^{a - \frac{1}{2}} d x \\ = \sum_{0 \leq i, j} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \int_{0}^{1} x^{i + j}_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; x] (x (1 - x))^{a - \frac{1}{2}} d x \end{aligned}$
ここで,
$\begin{aligned} \int_{0}^{1} x^{M}_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; x] (x (1 - x))^{a - \frac{1}{2}} d x & = \frac{Γ {(a + \frac{1}{2})}^{2}}{Γ (2 a + 1)} \sum_{0 \leq k} \frac{(- n, 2 a + n)_{k}}{k! {(a + \frac{1}{2})}_{k}} \frac{{(a + \frac{1}{2} + M)}_{k}}{(2 a + 1 + M)_{k}} \end{aligned}$
であり, Saalschützの和公式より,
$\begin{aligned} \sum_{0 \leq k} \frac{(- n, 2 a + n)_{k}}{k! {(a + \frac{1}{2})}_{k}} \frac{{(a + \frac{1}{2} + M)}_{k}}{(2 a + 1 + M)_{k}} & = \frac{(a + \frac{1}{2}, 1 + M - n)_{n}}{(2 a + 1 + M, \frac{1}{2} - a - n)_{n}} \\ = \frac{(- M)_{n}}{(2 a + 1 + M)_{n}} \end{aligned}$
であるから, これを用いて,
$\begin{aligned} \sum_{0 \leq i, j} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \int_{0}^{1} x^{i + j}_{2} F_{1} [\begin{array}{c} - n, 2 a + n \\ a + \frac{1}{2} \end{array}; x] (x (1 - x))^{a - \frac{1}{2}} d x \\ = \frac{Γ {(a + \frac{1}{2})}^{2}}{Γ (2 a + 1)} \sum_{0 \leq i, j} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \frac{(- i - j)_{n}}{(2 a + 1 + i + j)_{n}} \\ = (- 1)^{n} \frac{Γ {(a + \frac{1}{2})}^{2}}{Γ (2 a + 1 + n)} \sum_{0 \leq i, j} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \frac{(2 a + 1)_{i + j} (i + j)!}{(2 a + 1 + n)_{i + j} (i + j - n)!} \end{aligned}$
つまり, 以下を得る.

$l + m + n = 2 N$ が偶数のとき,
$\begin{aligned} \sum_{0 \leq i, j} \frac{(- l, 2 a + l)_{i}}{i! {(a + \frac{1}{2})}_{i}} \frac{(- m, 2 a + m)_{j}}{j! {(a + \frac{1}{2})}_{j}} \frac{(2 a + 1)_{i + j} (i + j)!}{(2 a + 1 + n)_{i + j} (i + j - n)!} \\ = (- 1)^{n} \frac{(a)_{N - l} (a)_{N - m} (a)_{N - n} l! m! n! (2 a)_{N} (2 a + 1)_{n}}{(N - l)! (N - m)! (N - n)! (a + 1)_{N} (2 a)_{l} (2 a)_{m} (2 a)_{n}} \end{aligned}$
が成り立つ.