Non-terminating Jacksonの和公式

前回の記事 で示したBaileyの${}_8{\varphi }_7$の3項変換公式は
$$\begin{array}{r}W(a;b,c,d,e,f;x)={}_8{\varphi }_7[\begin{array}{c}a,\sqrt{a}q,-\sqrt{a}q,b,c,d,e,f\\ \sqrt{a},-\sqrt{a},aq/b,aq/c,aq/d,aq/e,aq/f\end{array};x]\end{array}$$
として,
$$\begin{array}{rl}& W(a;b,c,d,e,f;\frac{a^2{q}^2}{bcdef})\\ & =\frac{(aq,aq/de,aq/df,aq/ef,eq/c,fq/c,b/a,bef/a;q{)}_{\mathrm{\infty }}}{(aq/d,aq/e,aq/f,aq/def,q/c,efq/c,be/a,bf/a;q{)}_{\mathrm{\infty }}}W(ef/c;aq/bc,aq/cd,ef/a,e,f;\frac{bd}{a})\\ & +\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,d,e,f,aq/bc,bdef/a^{2}q,a^2{q}^2/bdef;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bd/a,be/a,bf/a,def/aq,a{q}^2/def,q/c,b^{2}q/a;q{)}_{\mathrm{\infty }}}\\ & \cdot W(b^2/a;b,bc/a,bd/a,be/a,bf/a;\frac{a^2{q}^2}{bcdef})\end{array}$$
というものだった. ここで, $a^{2}q=bcdef$とすると, $aq/bc\cdot aq/cd=efq/c$より, Rogersの和公式
$$\begin{array}{rl}W(a;b,c,d;\frac{aq}{bcd})& ={}_6{\varphi }_5[\begin{array}{c}a,\sqrt{a}q,-\sqrt{a}q,b,c,d\\ \sqrt{a},-\sqrt{a},aq/b,aq/c,aq/d\end{array};\frac{aq}{bcd}]\\ & =\frac{(aq,aq/bc,aq/bd,aq/cd;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/bcd;q{)}_{\mathrm{\infty }}}\end{array}$$
を用いて,
$$\begin{array}{rl}& W(a;b,c,d,e,f;q)\\ & =\frac{(aq,aq/de,aq/df,aq/ef,eq/c,fq/c,b/a,bef/a;q{)}_{\mathrm{\infty }}}{(aq/d,aq/e,aq/f,aq/def,q/c,efq/c,be/a,bf/a;q{)}_{\mathrm{\infty }}}W(ef/c;ef/a,e,f;\frac{bd}{a})\\ & +\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,d,e,f,aq/bc,1/c,cq;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bd/a,be/a,bf/a,a/bc,bcq/a,q/c,b^{2}q/a;q{)}_{\mathrm{\infty }}}\\ & \cdot W(b^2/a;b,bc/a,bd/a,be/a,bf/a;q)\\ & =\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,b/a;q{)}_{\mathrm{\infty }}}{(aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a;q{)}_{\mathrm{\infty }}}\\ & +\frac{b}{a}\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,c,d,e,f;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a,b^{2}q/a;q{)}_{\mathrm{\infty }}}\\ & \cdot W(b^2/a;b,bc/a,bd/a,be/a,bf/a;q)\end{array}$$
よって, 以下を得る.

この左辺は, $b$を用いずに書くと,
$$\begin{array}{r}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,aq/cdef;q{)}_{\mathrm{\infty }}}{(aq/c,aq/d,aq/e,aq/f,aq/cde,aq/cdf,aq/cef,aq/def;q{)}_{\mathrm{\infty }}}\end{array}$$
と対称的な形で表すこともできる. 左辺は,
$$\begin{array}{rl}& W(a;b,c,d,e,f;q)-\frac{b}{a}\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,c,d,e,f;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a,b^{2}q/a;q{)}_{\mathrm{\infty }}}\\ & \cdot W(b^2/a;b,bc/a,bd/a,be/a,bf/a;q)\\ & =\frac{(a,\sqrt{a}q,-\sqrt{a}q,b,c,d,e,f;q{)}_{\mathrm{\infty }}}{(\sqrt{a},-\sqrt{a},aq/b,aq/c,aq/d,aq/e,aq/f,q;q{)}_{\mathrm{\infty }}}\sum _{0\le n}\frac{(\sqrt{a}{q}^n,-\sqrt{a}{q}^n,a{q}^{n+1}/b,a{q}^{n+1}/c,a{q}^{n+1}/d,a{q}^{n+1}/e,a{q}^{n+1}/f,q^{n+1};q{)}_{\mathrm{\infty }}}{(a{q}^n,\sqrt{a}{q}^{n+1},-\sqrt{a}{q}^{n+1},b{q}^n,c{q}^n,d{q}^n,e{q}^n,f{q}^n;q{)}_{\mathrm{\infty }}}{q}^n\\ & -\frac{b}{a}\frac{(aq,b^2/a,bq/\sqrt{a},-bq/\sqrt{a},b,c,d,e,f;q{)}_{\mathrm{\infty }}}{(b/\sqrt{a},-b/\sqrt{a},aq/b,aq/c,aq/d,aq/e,aq/f,b^{2}q/a,q;q{)}_{\mathrm{\infty }}}\sum _{0\le n}\frac{(b{q}^n/\sqrt{a},-b{q}^n/\sqrt{a},b{q}^{n+1}/a,b{q}^{n+1}/c,b{q}^{n+1}/d,b{q}^{n+1}/e,b{q}^{n+1}/f,q^{n+1};q{)}_{\mathrm{\infty }}}{(b^2{q}^n/a,b{q}^{n+1}/\sqrt{a},-b{q}^{n+1}/\sqrt{a},b{q}^n,bc{q}^n/a,bd{q}^n/a,be{q}^n/a,bf{q}^n/a;q{)}_{\mathrm{\infty }}}{q}^n\\ & =\frac{(aq,b,c,d,e,f;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q{)}_{\mathrm{\infty }}}\sum _{0\le n}\frac{(\sqrt{a}{q}^n,-\sqrt{a}{q}^n,a{q}^{n+1}/b,a{q}^{n+1}/c,a{q}^{n+1}/d,a{q}^{n+1}/e,a{q}^{n+1}/f,q^{n+1};q{)}_{\mathrm{\infty }}}{(a{q}^n,\sqrt{a}{q}^{n+1},-\sqrt{a}{q}^{n+1},b{q}^n,c{q}^n,d{q}^n,e{q}^n,f{q}^n;q{)}_{\mathrm{\infty }}}{q}^n\\ & -\frac{b}{a}\frac{(aq,b,c,d,e,f;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q{)}_{\mathrm{\infty }}}\sum _{0\le n}\frac{(b{q}^n/\sqrt{a},-b{q}^n/\sqrt{a},b{q}^{n+1}/a,b{q}^{n+1}/c,b{q}^{n+1}/d,b{q}^{n+1}/e,b{q}^{n+1}/f,q^{n+1};q{)}_{\mathrm{\infty }}}{(b^2{q}^n/a,b{q}^{n+1}/\sqrt{a},-b{q}^{n+1}/\sqrt{a},b{q}^n,bc{q}^n/a,bd{q}^n/a,be{q}^n/a,bf{q}^n/a;q{)}_{\mathrm{\infty }}}{q}^n\\ & =-\frac{1}{a}\frac{(aq,b,c,d,e,f;q{)}_{\mathrm{\infty }}}{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q{)}_{\mathrm{\infty }}}{\int }_a^b\frac{(t/\sqrt{a},-t/\sqrt{a},tq/a,tq/b,tq/c,tq/d,tq/e,tq/f;q{)}_{\mathrm{\infty }}}{(tq/\sqrt{a},-tq/\sqrt{a},t,bt/a,ct/a,dt/a,et/a,ft/a;q{)}_{\mathrm{\infty }}}{d}_{q}t\end{array}$$
と表すことができる. よって,
$$\begin{array}{rl}& {\int }_a^b\frac{(t/\sqrt{a},-t/\sqrt{a},tq/a,tq/b,tq/c,tq/d,tq/e,tq/f;q{)}_{\mathrm{\infty }}}{(tq/\sqrt{a},-tq/\sqrt{a},t,bt/a,ct/a,dt/a,et/a,ft/a;q{)}_{\mathrm{\infty }}}{d}_{q}t\\ & =-a\frac{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q{)}_{\mathrm{\infty }}}{(aq,b,c,d,e,f;q{)}_{\mathrm{\infty }}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,b/a;q{)}_{\mathrm{\infty }}}{(aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a;q{)}_{\mathrm{\infty }}}\\ & =(b-a)\frac{(q,aq/b,bq/a,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef;q{)}_{\mathrm{\infty }}}{(b,c,d,e,f,bc/a,bd/a,be/a,bf/a;q{)}_{\mathrm{\infty }}}\end{array}$$
つまり, 以下を得る.