Non-terminating Jacksonの和公式

前回の記事 で示したBaileyの${}_8\phi_7$の3項変換公式は
\begin{align} W(a;b,c,d,e,f;x)=\Q87{a,\sqrt aq,-\sqrt aq,b,c,d,e,f}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f}{x} \end{align}
として,
\begin{align} &W\left(a;b,c,d,e,f;\frac{a^2q^2}{bcdef}\right)\\ &=\frac{(aq,aq/de,aq/df,aq/ef,eq/c,fq/c,b/a,bef/a;q)_{\infty}}{(aq/d,aq/e,aq/f,aq/def,q/c,efq/c,be/a,bf/a;q)_{\infty}}W\left(ef/c;aq/bc,aq/cd,ef/a,e,f;\frac{bd}a\right)\\ &\qquad +\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,d,e,f,aq/bc,bdef/a^2q,a^2q^2/bdef;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bd/a,be/a,bf/a,def/aq,aq^2/def,q/c,b^2q/a;q)_{\infty}}\\ &\qquad\qquad\cdot\,W\left(b^2/a;b,bc/a,bd/a,be/a,bf/a;\frac{a^2q^2}{bcdef}\right) \end{align}
というものだった. ここで, $a^2q=bcdef$とすると, $aq/bc\cdot aq/cd=efq/c$より, Rogersの和公式
\begin{align} W\left(a;b,c,d;\frac{aq}{bcd}\right)&=\Q65{a,\sqrt aq,-\sqrt aq,b,c,d}{\sqrt a,-\sqrt a,aq/b,aq/c,aq/d}{\frac{aq}{bcd}}\\ &=\frac{(aq,aq/bc,aq/bd,aq/cd;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/bcd;q)_{\infty}} \end{align}
を用いて,
\begin{align} &W\left(a;b,c,d,e,f;q\right)\\ &=\frac{(aq,aq/de,aq/df,aq/ef,eq/c,fq/c,b/a,bef/a;q)_{\infty}}{(aq/d,aq/e,aq/f,aq/def,q/c,efq/c,be/a,bf/a;q)_{\infty}}W\left(ef/c;ef/a,e,f;\frac{bd}a\right)\\ &\qquad +\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,d,e,f,aq/bc,1/c,cq;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bd/a,be/a,bf/a,a/bc,bcq/a,q/c,b^2q/a;q)_{\infty}}\\ &\qquad\qquad\cdot\,W\left(b^2/a;b,bc/a,bd/a,be/a,bf/a;q\right)\\ &=\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,b/a;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a;q)_{\infty}}\\ &\qquad +\frac ba\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a,b^2q/a;q)_{\infty}}\\ &\qquad\qquad\cdot\,W\left(b^2/a;b,bc/a,bd/a,be/a,bf/a;q\right)\\ \end{align}
よって, 以下を得る.

この左辺は, $b$を用いずに書くと,
\begin{align} \frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,aq/cdef;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,aq/cde,aq/cdf,aq/cef,aq/def;q)_{\infty}} \end{align}
と対称的な形で表すこともできる. 左辺は,
\begin{align} &W\left(a;b,c,d,e,f;q\right)-\frac ba\frac{(aq,bq/a,bq/c,bq/d,bq/e,bq/f,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a,b^2q/a;q)_{\infty}}\\ &\qquad\cdot\,W\left(b^2/a;b,bc/a,bd/a,be/a,bf/a;q\right)\\ &=\frac{(a,\sqrt aq,-\sqrt aq,b,c,d,e,f;q)_{\infty}}{(\sqrt a,-\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f,q;q)_{\infty}}\sum_{0\leq n}\frac{(\sqrt aq^n,-\sqrt aq^n,aq^{n+1}/b,aq^{n+1}/c,aq^{n+1}/d,aq^{n+1}/e,aq^{n+1}/f,q^{n+1};q)_{\infty}}{(aq^n,\sqrt aq^{n+1},-\sqrt aq^{n+1},bq^n,cq^n,dq^n,eq^n,fq^n;q)_{\infty}}q^n\\ &\qquad -\frac ba\frac{(aq,b^2/a,bq/\sqrt{a},-bq/\sqrt a,b,c,d,e,f;q)_{\infty}}{(b/\sqrt a,-b/\sqrt a,aq/b,aq/c,aq/d,aq/e,aq/f,b^2q/a,q;q)_{\infty}}\sum_{0\leq n}\frac{(bq^n/\sqrt a,-bq^n/\sqrt a,bq^{n+1}/a,bq^{n+1}/c,bq^{n+1}/d,bq^{n+1}/e,bq^{n+1}/f,q^{n+1};q)_{\infty}}{(b^2q^n/a,bq^{n+1}/\sqrt a,-bq^{n+1}/\sqrt a,bq^n,bcq^n/a,bdq^n/a,beq^n/a,bfq^n/a;q)_{\infty}}q^n\\ &=\frac{(aq,b,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q)_{\infty}}\sum_{0\leq n}\frac{(\sqrt aq^n,-\sqrt aq^n,aq^{n+1}/b,aq^{n+1}/c,aq^{n+1}/d,aq^{n+1}/e,aq^{n+1}/f,q^{n+1};q)_{\infty}}{(aq^n,\sqrt aq^{n+1},-\sqrt aq^{n+1},bq^n,cq^n,dq^n,eq^n,fq^n;q)_{\infty}}q^n\\ &\qquad -\frac ba\frac{(aq,b,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q)_{\infty}}\sum_{0\leq n}\frac{(bq^n/\sqrt a,-bq^n/\sqrt a,bq^{n+1}/a,bq^{n+1}/c,bq^{n+1}/d,bq^{n+1}/e,bq^{n+1}/f,q^{n+1};q)_{\infty}}{(b^2q^n/a,bq^{n+1}/\sqrt a,-bq^{n+1}/\sqrt a,bq^n,bcq^n/a,bdq^n/a,beq^n/a,bfq^n/a;q)_{\infty}}q^n\\ &=-\frac 1a\frac{(aq,b,c,d,e,f;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q)_{\infty}}\int_a^b\frac{(t/\sqrt a,-t/\sqrt a,tq/a,tq/b,tq/c,tq/d,tq/e,tq/f;q)_{\infty}}{(tq/\sqrt a,-tq/\sqrt a,t,bt/a,ct/a,dt/a,et/a,ft/a;q)_{\infty}}\,d_qt \end{align}
と表すことができる. よって,
\begin{align} &\int_a^b\frac{(t/\sqrt a,-t/\sqrt a,tq/a,tq/b,tq/c,tq/d,tq/e,tq/f;q)_{\infty}}{(tq/\sqrt a,-tq/\sqrt a,t,bt/a,ct/a,dt/a,et/a,ft/a;q)_{\infty}}\,d_qt\\ &=-a\frac{(aq/b,aq/c,aq/d,aq/e,aq/f,q;q)_{\infty}}{(aq,b,c,d,e,f;q)_{\infty}}\frac{(aq,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef,b/a;q)_{\infty}}{(aq/c,aq/d,aq/e,aq/f,bc/a,bd/a,be/a,bf/a;q)_{\infty}}\\ &=(b-a)\frac{(q,aq/b,bq/a,aq/cd,aq/ce,aq/cf,aq/de,aq/df,aq/ef;q)_{\infty}}{(b,c,d,e,f,bc/a,bd/a,be/a,bf/a;q)_{\infty}} \end{align}
つまり, 以下を得る.