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Askey-Wilson積分の証明

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Askey-Wilson積分
$\begin{aligned} \frac{1}{2 π i} \int_{C} \frac{(t^{2}, 1 / t^{2}; q)_{\infty}}{(a t, a / t, b t, b / t, c t, c / t, d t, d / t; q)_{\infty}} \frac{d t}{t} & = \frac{2 (a b c d; q)_{\infty}}{(q, a b, a c, a d, b c, b d, c d; q)_{\infty}} \end{aligned}$
を示す. ここで, $C$ は単位円 $| t | = 1$ とする.

$| a |, | b |, | c |, | d | < 1$ とするとき,
$\begin{aligned} \frac{1}{2 π i} \int_{C} \frac{(t^{2}, 1 / t^{2}; q)_{\infty}}{(a t, a / t, b t, b / t, c t, c / t, d t, d / t; q)_{\infty}} \frac{d t}{t} & = \frac{2 (a b c d; q)_{\infty}}{(q, a b, a c, a d, b c, b d, c d; q)_{\infty}} \end{aligned}$
が成り立つ.

$\begin{array}{r} h (x; a) := \prod_{n = 0}^{\infty} (1 - 2 a x q^{n} + a^{2} q^{2 n}) = (a e^{i θ}, a e^{- i θ}; q)_{\infty}, x = \cos θ \end{array}$
として, $h (x; a_{1}, \dots, a_{r}) := h (x; a_{1}) \dots h (x; a_{r})$ とすると,
$\begin{aligned} h (x; 1, - 1, \sqrt{q}, - \sqrt{q}) \\ = (e^{i θ}, e^{- i θ}, - e^{i θ}, - e^{- i θ}, e^{i θ} \sqrt{q}, e^{- i θ} \sqrt{q}, - e^{i θ} \sqrt{q}, - e^{- θ} \sqrt{q}; q)_{\infty} \\ = (e^{2 i θ}, e^{- 2 i θ}, e^{2 i θ} q, e^{- 2 i θ} q; q^{2})_{\infty} \\ = (e^{2 i θ}, e^{- 2 i θ}; q)_{\infty} \end{aligned}$
となる. よって,
$\begin{aligned} \frac{1}{2 π i} \int_{C} \frac{(t^{2}, 1 / t^{2}; q)_{\infty}}{(a t, a / t, b t, b / t, c t, c / t, d t, d / t; q)_{\infty}} \frac{d t}{t} & = \frac{1}{2 π} \int_{- π}^{π} \frac{h (x; 1, - 1, \sqrt{q}, - \sqrt{q})}{h (x; a, b, c, d)} d θ \\ = \frac{1}{π} \int_{0}^{π} \frac{h (x; 1, - 1, \sqrt{q}, - \sqrt{q})}{h (x; a, b, c, d)} d θ \\ = \frac{1}{π} \int_{- 1}^{1} \frac{h (x; 1, - 1, \sqrt{q}, - \sqrt{q})}{h (x; a, b, c, d)} \frac{d x}{\sqrt{1 - x^{2}}} \end{aligned}$
を用いて, Askey-Wilson積分は
$\begin{aligned} \int_{- 1}^{1} \frac{h (x; 1, - 1, \sqrt{q}, - \sqrt{q})}{h (x; a, b, c, d)} \frac{d x}{\sqrt{1 - x^{2}}} & = \frac{2 π (a b c d; q)_{\infty}}{(q, a b, a c, a d, b c, b d, c d; q)_{\infty}} \end{aligned}$
という実積分に変形できる. 以下はGasper-Rahmanの本, Basic Hypergeometric Seriesに書かれている証明である.

まず, $q$ 積分を
$\begin{aligned} \int_{a}^{b} f (t) d_{q} t & := \sum_{0 \leq n} (b q^{n} f (b q^{n}) - a q^{n} f (a q^{n})) \end{aligned}$
とすると, 前回の記事で示したAl-Salam-Verma積分
$\begin{aligned} \int_{a}^{b} \frac{(t q / a, t q / b, c t; q)_{\infty}}{(d t, e t, f t; q)_{\infty}} d_{q} t & = (b - a) \frac{(a q / b, b q / a, c / d, c / e, c / f, q; q)_{\infty}}{(a d, a e, a f, b d, b e, b f; q)_{\infty}}, (c = a b d e f) \end{aligned}$
を用いて,
$\begin{aligned} \frac{h (x; 1)}{h (x; a, b)} & = \frac{(1 / a, 1 / b; q)_{\infty}}{(b - a) (q, a q / b, b q / a, a b; q)_{\infty}} \int_{a}^{b} \frac{(u q / a, u q / b, u; q)_{\infty}}{(u / a b; q)_{\infty}} \frac{d_{q} u}{h (x; u)} \\ \frac{h (x; - 1)}{h (x; c, d)} & = \frac{(- 1 / c, - 1 / d; q)_{\infty}}{(d - c) (q, c q / d, d q / c, c d; q)_{\infty}} \int_{c}^{d} \frac{(v q / c, v q / d, - v; q)_{\infty}}{(- v / c d; q)_{\infty}} \frac{d_{q} v}{h (x; v)} \\ \frac{h (x; - \sqrt{q})}{h (x; u, v)} & = \frac{\sqrt{q} (- \sqrt{q} / u, - \sqrt{q} / v; q)_{\infty}}{(v - u) (q, u q / v, v q / u, u v; q)_{\infty}} \int_{u / \sqrt{q}}^{v / \sqrt{q}} \frac{(t q^{\frac{3}{2}} / u, t q^{\frac{3}{2}} / v, - t q; q)_{\infty}}{(- t q / u v; q)_{\infty}} \frac{d_{q} t}{h (x; t \sqrt{q})} \end{aligned}$
が成り立つ. また,
$\begin{aligned} \frac{1}{2} \int_{- π}^{π} \frac{h (x; \sqrt{q})}{h (x; t q^{\frac{1}{2}})} d θ & = \frac{1}{2} \int_{- π}^{π} \frac{(e^{i θ} \sqrt{q}, e^{- i θ} \sqrt{q}; q)_{\infty}}{(t e^{i θ} \sqrt{q}, t e^{- i θ} \sqrt{q}; q)_{\infty}} d θ \\ = \frac{1}{2} \int_{- π}^{π} \sum_{0 \leq n, m} \frac{(1 / t; q)_{n} (1 / t; q)_{m}}{(q; q)_{n} (q; q)_{m}} (t \sqrt{q})^{n + m} e^{i (n - m)} d θ \\ = π \sum_{0 \leq n} \frac{(1 / t, 1 / t; q)_{n}}{(q; q)_{n}^{2}} (t^{2} q)^{n} \\ = π \frac{(t q, t q; q)_{\infty}}{(q, t^{2} q; q)_{\infty}} \\ = π \frac{(t q, t q; q)_{\infty}}{(q; q)_{\infty} (t \sqrt{q}, - t \sqrt{q}, t q, - t q; q)_{\infty}} \\ = \frac{π (t q; q)_{\infty}}{(q, t \sqrt{q}, - t \sqrt{q}, - t q; q)_{\infty}} \end{aligned}$
である. よって, Al-Salam-Verma積分を反復して適用することによって,
$\begin{aligned} \frac{1}{2} \int_{- π}^{π} \frac{h (x; 1, - 1, \sqrt{q}, - \sqrt{q})}{h (x; a, b, c, d)} d θ \\ = \frac{(1 / a, 1 / b, - 1 / c, - 1 / d; q)_{\infty}}{2 (b - a) (d - c) (q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \int_{- π}^{π} h (x; \sqrt{q}, - \sqrt{q}) d x \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b, u; q)_{\infty}}{(u / a b; q)_{\infty}} \frac{d_{q} u}{h (x; u)} \int_{c}^{d} \frac{(v q / c, v q / d, - v; q)_{\infty}}{(- v / c d; q)_{\infty}} \frac{d_{q} v}{h (x; v)} \\ = \frac{(1 / a, 1 / b, - 1 / c, - 1 / d; q)_{\infty}}{2 (b - a) (d - c) (q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \int_{- π}^{π} d x \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b, u; q)_{\infty}}{(u / a b; q)_{\infty}} d_{q} u \int_{c}^{d} \frac{(v q / c, v q / d, - v; q)_{\infty}}{(- v / c d; q)_{\infty}} d_{q} v \\ \cdot \frac{\sqrt{q} (- \sqrt{q} / u, - \sqrt{q} / v; q)_{\infty}}{(v - u) (q, u q / v, v q / u, u v; q)_{\infty}} \int_{u / \sqrt{q}}^{v / \sqrt{q}} \frac{(t q^{\frac{3}{2}} / u, t q^{\frac{3}{2}} / v, - t q; q)_{\infty}}{(- t q / u v; q)_{\infty}} \frac{h (x; \sqrt{q}) d_{q} t}{h (x; t \sqrt{q})} \\ = \frac{(1 / a, 1 / b, - 1 / c, - 1 / d; q)_{\infty}}{(b - a) (d - c) (q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b, u; q)_{\infty}}{(u / a b; q)_{\infty}} d_{q} u \int_{c}^{d} \frac{(v q / c, v q / d, - v; q)_{\infty}}{(- v / c d; q)_{\infty}} d_{q} v \\ \cdot \frac{\sqrt{q} (- \sqrt{q} / u, - \sqrt{q} / v; q)_{\infty}}{(v - u) (q, u q / v, v q / u, u v; q)_{\infty}} \int_{u / \sqrt{q}}^{v / \sqrt{q}} \frac{(t q^{\frac{3}{2}} / u, t q^{\frac{3}{2}} / v, - t q; q)_{\infty}}{(- t q / u v; q)_{\infty}} d_{q} t \frac{π (t q; q)_{\infty}}{(q, t \sqrt{q}, - t \sqrt{q}, - t q; q)_{\infty}} \\ = \frac{π (1 / a, 1 / b, - 1 / c, - 1 / d; q)_{\infty}}{(b - a) (d - c) (q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b, u; q)_{\infty}}{(u / a b; q)_{\infty}} d_{q} u \int_{c}^{d} \frac{(v q / c, v q / d, - v; q)_{\infty}}{(- v / c d; q)_{\infty}} d_{q} v \\ \cdot \frac{\sqrt{q} (- \sqrt{q} / u, - \sqrt{q} / v; q)_{\infty}}{(v - u) (q, q, u q / v, v q / u, u v; q)_{\infty}} \int_{u / \sqrt{q}}^{v / \sqrt{q}} \frac{(t q^{\frac{3}{2}} / u, t q^{\frac{3}{2}} / v, t q; q)_{\infty}}{(- t q / u v, t \sqrt{q}, - t \sqrt{q}; q)_{\infty}} d_{q} t \\ = \frac{π (1 / a, 1 / b, - 1 / c, - 1 / d; q)_{\infty}}{(b - a) (d - c) (q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b, u; q)_{\infty}}{(u / a b; q)_{\infty}} d_{q} u \int_{c}^{d} \frac{(v q / c, v q / d, - v; q)_{\infty}}{(- v / c d; q)_{\infty}} d_{q} v \\ \cdot \frac{\sqrt{q} (- \sqrt{q} / u, - \sqrt{q} / v; q)_{\infty}}{(v - u) (q, q, u q / v, v q / u, u v; q)_{\infty}} \frac{(v - u) (u q / v, v q / u, - u v, \sqrt{q}, - \sqrt{q}, q; q)_{\infty}}{\sqrt{q} (- \sqrt{q} / v, u, - u, - \sqrt{q} / u, v, - v; q)_{\infty}} \\ = \frac{π (1 / a, 1 / b, - 1 / c, - 1 / d, \sqrt{q}, - \sqrt{q}; q)_{\infty}}{(b - a) (d - c) (q, q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b; q)_{\infty}}{(u / a b, - u; q)_{\infty}} d_{q} u \int_{c}^{d} \frac{(v q / c, v q / d, - u v; q)_{\infty}}{(- v / c d, v, u v; q)_{\infty}} d_{q} v \\ = \frac{π (1 / a, 1 / b, - 1 / c, - 1 / d, \sqrt{q}, - \sqrt{q}; q)_{\infty}}{(b - a) (d - c) (q, q, q, a q / b, b q / a, a b, c q / d, d q / c, c d; q)_{\infty}} \\ \cdot \int_{a}^{b} \frac{(u q / a, u q / b; q)_{\infty}}{(u / a b, - u; q)_{\infty}} d_{q} u \frac{(d - c) (c q / d, d q / c, c d u, - u, - 1, q; q)_{\infty}}{(- 1 / d, c, c u, - 1 / c, d, d u; q)_{\infty}} \\ = \frac{π (1 / a, 1 / b, \sqrt{q}, - \sqrt{q}, - 1; q)_{\infty}}{(b - a) (q, q, a q / b, b q / a, a b, c d, c, d; q)_{\infty}} \int_{a}^{b} \frac{(u q / a, u q / b, c d u; q)_{\infty}}{(u / a b, c u, d u; q)_{\infty}} d_{q} u \\ = \frac{π (1 / a, 1 / b, \sqrt{q}, - \sqrt{q}, - 1; q)_{\infty}}{(b - a) (q, q, a q / b, b q / a, a b, c d, c, d; q)_{\infty}} \frac{(b - a) (a q / b, b q / a, a b c d, c, d, q; q)_{\infty}}{(1 / b, a c, a d, 1 / a, b c, b d; q)_{\infty}} \\ = \frac{π (\sqrt{q}, - \sqrt{q}, - 1, a b c d; q)_{\infty}}{(q, a b, a c, a d, b c, b d, c d; q)_{\infty}} \\ = \frac{2 π (a b c d; q)_{\infty}}{(q, a b, a c, a d, b c, b d, c d; q)_{\infty}} \end{aligned}$
となって示される.