Non-terminating $q$-Whippleの変換公式
の記事で示した積分表示
\begin{align}
&\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}{\frac{a^2q^2}{bcdef}}\\
&=\frac{1}{def-aq}\frac{(aq,aq/bc,d,e,f,aq/de,aq/df,aq/ef;q)_{\infty}}{(aq/b,aq/c,aq/d,aq/e,aq/f,aq^2/def,def/a,q;q)_{\infty}}\int_{aq}^{def}\frac{(atq/bdef,atq/cdef,tq/def,t/a;q)_{\infty}}{(t/de,t/df,t/ef,atq/bcdef;q)_{\infty}}\,d_qt
\end{align}
を使いやすいように整理しておく.
\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}
のように書くことにすると,
\begin{align}
&\int_{aq}^{def}\frac{(atq/bdef,atq/cdef,tq/def,t/a;q)_{\infty}}{(t/de,t/df,t/ef,atq/bcdef;q)_{\infty}}\,d_qt\\
&=(def-aq)\frac{(aq/b,aq/c,aq/d,aq/e,aq/f,aq^2/def,def/a,q;q)_{\infty}}{(aq,aq/bc,d,e,f,aq/de,aq/df,aq/ef;q)_{\infty}}W\left(a;b,c,d,e,f;\frac{a^2q^2}{bcdef}\right)
\end{align}
である. これは,
\begin{align}
&\int_{agq}^{defg}\frac{(atq/bdefg,atq/cdefg,tq/defg,t/ag;q)_{\infty}}{(t/deg,t/dfg,t/efg,atq/bcdefg;q)_{\infty}}\,d_qt\\
&=(defg-agq)\frac{(aq/b,aq/c,aq/d,aq/e,aq/f,aq^2/def,def/a,q;q)_{\infty}}{(aq,aq/bc,d,e,f,aq/de,aq/df,aq/ef;q)_{\infty}}W\left(a;b,c,d,e,f;\frac{a^2q^2}{bcdef}\right)
\end{align}
と同値である.
$a\mapsto a/gq, defg=h$とすると,
\begin{align}
&\int_{a}^{h}\frac{(at/bgh,at/cgh,tq/h,tq/a;q)_{\infty}}{(td/h,te/h,tf/h,at/bcgh;q)_{\infty}}\,d_qt\\
&=(h-a)\frac{(a/bg,a/cg,a/dg,a/eg,a/fg,aq/h,hq/a,q;q)_{\infty}}{(a/g,a/bcg,d,e,f,ad/h,ae/h,af/h;q)_{\infty}}W\left(a/gq;b,c,d,e,f;\frac{a^2}{bcgh}\right)
\end{align}
$b\mapsto a/bgh, c\mapsto a/cgh,d\mapsto dh,e\mapsto eh,f\mapsto fh$とすると, 条件は$defgh^2=1$となり,
\begin{align}
&\int_{a}^{h}\frac{(bt,ct,tq/h,tq/a;q)_{\infty}}{(dt,et,ft,bcght/a;q)_{\infty}}\,d_qt\\
&=(h-a)\frac{(bh,ch,a/dgh,a/egh,a/fgh,aq/h,hq/a,q;q)_{\infty}}{(a/g,bcgh^2/a,dh,eh,fh,ad,ae,af;q)_{\infty}}W\left(a/gq;a/bgh,a/cgh,dh,eh,fh;bcgh\right)
\end{align}
だから, $k=bcgh/a$とすれば, 条件は$adefhk=bc$となり, このとき,
\begin{align}
&\int_{a}^{h}\frac{(bt,ct,tq/h,tq/a;q)_{\infty}}{(dt,et,ft,kt;q)_{\infty}}\,d_qt\\
&=(h-a)\frac{(bh,ch,bc/dk,bc/ek,bc/fk,aq/h,hq/a,q;q)_{\infty}}{(bch/k,hk,dh,eh,fh,ad,ae,af;q)_{\infty}}W\left(bch/kq;b/k,c/k,dh,eh,fh;ak\right)
\end{align}
よって変数を付け替えて, 以下を得る.
$cd=abefgh$のとき,
\begin{align}
&\int_{a}^b\frac{(tq/a,tq/b,ct,dt;q)_{\infty}}{(et,ft,gt,ht;q)_{\infty}}\,d_qt\\
&=(b-a)\frac{(bc,bd,cd/eh,cd/fh,cd/gh,aq/b,bq/a,q;q)_{\infty}}{(bcd/h,be,bf,bg,bh,ae,af,ag;q)_{\infty}}W\left(bcd/hq;c/h,d/h,be,bf,bg;ah\right)
\end{align}
が成り立つ.