A simple proof of Slater’s transformations for bilateral series ${}_{r}\psi _{r},\, {}_{2r}\psi _{2r}$ and ${}_{2r-1}\psi _{2r-1}$

Abstract

We give a simple proof of Slater’s transformations for bilateral series ${}_r\psi _r$, ${}_{2r}\psi _{2r}$, and ${}_{2r-1}\psi _{2r-1}$, using the residue theorem only, without technical manipulation of the series.

1 Introduction

The bilateral basic hypergeometric series ${}_n\psi _n$ is defined by

$$\begin{aligned} {}_n\psi _n\left( \begin{array}{c}{a_1,\, \ldots ,\, a_n}\\ {b_1,\, \ldots ,\, b_n}\end{array};\,{q,\; x}\, \right) :=\sum _{-\infty<k<\infty } \frac{(a_1,\,\ldots ,\,a_n;\,q)_{k}}{(b_1,\,\ldots ,\,b_n;\,q)_{k}}\, x^k \end{aligned}$$

for

$$\begin{aligned} \left| \,b_1\cdots b_na_1^{-1}\cdots a_n^{-1}\,\right|<|x|<1, \quad |q|<1, \quad a_i,\;b_i,\;x,\; q\in {\mathbb {C}}, \ \text {and}\ a_i,\; qb_i^{-1}\notin q^{{\mathbb {Z}}_{>0}}, \end{aligned}$$

where

$$\begin{aligned}&(a_1,\,a_2,\,\ldots ,\,a_{n}\,;q)_k =(a_1\,;q)_k(a_2\,;q)_k\,\cdots \, (a_{n}\,;q)_k\,,\\&(a\,;q)_k=(a\,;q)_{\infty }/(aq^k\,;q)_{\infty }\,, \end{aligned}$$

and

$$\begin{aligned} (a\,;q)_{\infty }=\prod _{i\ge 0}(1-aq^i)\,. \end{aligned}$$

In [6, 7], Slater derived transformations for bilateral basic hypergeometric series. She applied the results by Sears [5] in [6], and she considered the basic analogue of the Barnes-type integrals in [7]. Exposition on the work by Slater is provided in Chapter 5 of the book by Gasper and Rahman [1]. On the other hand, Ito and Sanada gave a proof of the transformation for ${}_r\psi _r$ series (Theorem 1 below) and a proof of the transformation for very-well-poised-balanced ${}_{2r}\psi _{2r}$ series (Theorem 2 below) from the viewpoint of the connection problem associated with a Jackson integral [2].

The purpose of the present paper is to give a simple proof of these results by means of the residue theorem only, without technical manipulation of the series, and also to study the integrals which represent several functions in q-analysis. We refer the reader to our previous work [4] for the proof of Ramanujan’s ${}_1\psi _1$-sum and Bailey’s ${}_6\psi _6$-sum from the same point of view (See also [3]).

We also use the symbols

$$\begin{aligned}&(a_1,\,a_2,\,\ldots ,\,a_{n}\,;q)_{\infty } =(a_1\,;q)_{\infty }(a_2\,;q)_{\infty }\,\cdots \, (a_{n}\,;q)_{\infty }\,,\\&\theta (a)=(a,\, qa^{-1}\,;\,q)_{\infty } \end{aligned}$$

and

$$\begin{aligned}&\theta (a_1,\,\ldots ,\,a_n)=\theta (a_1)\cdots \theta (a_n)\,. \end{aligned}$$

In this paper, the base q is fixed to be a real number satisfying $0<q<1$ for simplicity.

2 ${}_{r}\psi _{r}$ series

Slater’s transformation for ${}_{r}\psi _{r}$ series is given by the following, which is (5.4.3) of [1] ( (4) of [6], (7.2.5) of [7], and (5.1) of [2] ):

Theorem 1

Suppose that

$$\begin{aligned} c_i,\; c_ic_j^{-1}\notin q^{{\mathbb {Z}}},\;\; 1\le i\ne j\le r, \end{aligned}$$

for $r\ge 1.$ Then we have

$$\begin{aligned}&\frac{\theta (Ax)\,\prod _{i=1}^{r}(b_i,\,qa_i^{-1}\,;\,q)_{\infty }}{\prod _{i=1}^{r}\theta (c_i)}\, {}_{r}\psi _{r}\left( \begin{array}{c}{a_1,\, a_2,\,\ldots ,\,a_r}\\ {b_1,\,b_2,\,\ldots ,\,b_r}\end{array};\,{q,\; x}\, \right) \nonumber \\&=\sum _{s=1}^{r}\frac{q}{c_s} \frac{\theta (Axc_sq^{-1})}{\theta (c_s)} \frac{\prod _{i=1}^{r} (qb_ic_s^{-1},\,c_sa_i^{-1}\,;\,q)_{\infty }}{\prod _{\begin{array}{c} 1\le i\le r\\ i \ne s \end{array}} \theta (c_sc_i^{-1})}\,\nonumber \\&\quad \times {}_{r}\psi _{r}\left( \begin{array}{c}{qa_1c_s^{-1},\, qa_2c_s^{-1},\,\ldots ,\,qa_rc_s^{-1}}\\ {qb_1c_s^{-1},\, qb_2c_s^{-1},\,\ldots ,\,qb_rc_s^{-1}}\end{array};\,{q,\; x}\, \right) , \end{aligned}$$

(2.1)

where $A=\prod _{i=1}^ra_ic_i^{-1}$ and $\left| \,b_1\cdots b_ra_1^{-1}\cdots a_r^{-1}\,\right|<|\,x\,|<1$.

Proof

Let F(t) be a function defined by

$$\begin{aligned} F(t)&=\frac{1}{t} \frac{\theta (Axt^{-1})}{\theta (t^{-1})} \, \prod _{i=1}^{r}\frac{ (b_it,\,qa_i^{-1}t^{-1}\,;\,q)_{\infty }}{\theta (c_i\,t)}\,. \end{aligned}$$

(2.2)

First, for $k\in {\mathbb {Z}}$, we have

$$\begin{aligned}&\textrm{Res}_{t=q^k}F(t)\,dt =\lim _{t\rightarrow q^k }(t-q^k)F(t)\nonumber \\&= \dfrac{\theta (Ax)\, \prod _{i=1}^{r}\theta (a_i)}{(q\,;q)_{\infty }^2\,\prod _{i=1}^r\theta (c_i)} \prod _{i=1}^{r}\frac{(b_iq^k\,;q)_{\infty }}{(a_iq^k\,;q)_{\infty }}\,x^k\,, \end{aligned}$$

(2.3)

and

$$\begin{aligned}&\textrm{Res}_{t=c_s^{-1}q^{1+k}}F(t)\,dt= \lim _{t\rightarrow c_s^{-1}q^{1+k}}(t-c_s^{-1}q^{1+k})F(t)\nonumber \\&=-\frac{q}{c_s} \dfrac{\theta (Axc_sq^{-1})\, \prod _{i=1}^{r}\theta (c_sa_i^{-1})}{(q\,;q)_{\infty }^2\,\theta (c_s)\, \prod _{\begin{array}{c} 1\le i\le r\\ i\ne s \end{array}} \theta (c_sc_i^{-1})}\, \prod _{i=1}^{r} \frac{(qb_ic_s^{-1}q^{k}\,;q)_{\infty }}{(qa_ic_s^{-1}q^{k}\,;q)_{\infty }}\,x^k\,. \end{aligned}$$

(2.4)

Secondly, for real positive numbers $R_1$ and $R_2$ satisfying

$$\begin{aligned} R_j,\,c_iR_j\notin q^{{\mathbb {Z}}}, \quad 1\le i\le r,\;\;j=1,\,2, \end{aligned}$$

we have

$$\begin{aligned}&F\left( R_1\,e^{\sqrt{-1}\,\theta }q^{-l}\right) \nonumber \\&=\frac{\left( AxR_1^{-1}\,e^{-\sqrt{-1}\,\theta }q^l,\, qA^{-1}x^{-1}R_1\,e^{\sqrt{-1}\,\theta }\,;q\right) _{\infty }}{R_1\,e^{\sqrt{-1}\,\theta }q^{-l}\, \left( R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l,\, qR_1\,e^{\sqrt{-1}\theta }\,;q\right) _{\infty }}\nonumber \\&\quad \times \prod _{i=1}^{r} \frac{ \left( b_iR_1\,e^{\sqrt{-1}\theta },\, qa_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l\,;q\right) _{\infty }}{\left( c_iR_1\,e^{\sqrt{-1}\theta },\, qc_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l\,;q\right) _{\infty }}\, \nonumber \\&\quad \times \frac{\left( AxR_1^{-1}\,e^{-\sqrt{-1}\,\theta }\,;q\right) _l}{\left( R_1^{-1}\,e^{-\sqrt{-1}\theta }\,;q\right) _l} \prod _{i=1}^r \frac{\left( qb_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }\,;q\right) _l}{\left( qc_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }\,;q\right) _l} \left( \frac{b_1\cdots b_r}{a_1\cdots a_r} \frac{1}{x}\right) ^l,\nonumber \\&F\left( R_2\,e^{\sqrt{-1}\,\theta }q^{l}\right) \nonumber \\&=\frac{\left( AxR_2^{-1}\,e^{-\sqrt{-1}\,\theta },\, qA^{-1}x^{-1}R_2\,e^{\sqrt{-1}\,\theta }q^l\,;q\right) _{\infty }}{R_2\,e^{\sqrt{-1}\,\theta }q^{l}\, (R_2^{-1}\,e^{-\sqrt{-1}\theta },\, qR_2\,e^{\sqrt{-1}\theta }q^l\,;q)_{\infty }}\nonumber \\&\quad \times \prod _{i=1}^{r} \frac{ \left( b_iR_2\,e^{\sqrt{-1}\theta }q^l,\, qa_i^{-1}R_2^{-1}\,e^{-\sqrt{-1}\theta }\,;q\right) _{\infty }}{\left( c_iR_2\,e^{\sqrt{-1}\theta }q^l,\, qc_i^{-1}R_2^{-1}\,e^{-\sqrt{-1}\theta }\,;q\right) _{\infty }}\, \nonumber \\&\quad \times \frac{\left( qA^{-1}x^{-1}R_2\,e^{\sqrt{-1}\,\theta }\,;q\right) _l}{\left( R_2^{-1}\,e^{\sqrt{-1}\theta },\,;q\right) _l} \prod _{i=1}^r \frac{\left( a_iR_2\,e^{\sqrt{-1}\theta }\,;q\right. )_l}{\left( c_iR_2\,e^{\sqrt{-1}\theta }\,;q\right) _l}\,x^l\,, \end{aligned}$$

and thus

$$\begin{aligned}&\left| \,\int _{C_1^{(l)}} F(t)\,dt\,\right| \nonumber \\&=\left| \,\int _0^{2\pi }F(R_1\,e^{\sqrt{-1}\theta }q^{-l}) \,R_1\,\sqrt{-1}\,e^{\sqrt{-1}\theta }q^{-l} \,\textrm{d}\theta \,\right| \le M_1 \left| \,\frac{b_1\cdots b_r}{a_1\cdots a_r} \frac{1}{x}\, \right| ^l, \end{aligned}$$

(2.5)

$$\begin{aligned}&\left| \,\int _{C_2^{(l)}}F(t)\,dt\,\right| =\left| \,\int _0^{2\pi }F(R_2\,e^{\sqrt{-1}\theta }q^l)\, R_2\,\sqrt{-1}\,e^{\sqrt{-1}\theta }q^l\,\textrm{d}\theta \,\right| \le M_2\, \left| \,x\, \right| ^l, \end{aligned}$$

(2.6)

where $l\in {\mathbb {Z}}_{\ge 0},$ $C_1^{(l)}= \{\,R_1\,e^{\sqrt{-1}\theta }q^{-l}\in {\mathbb {C}}\mid 0\le \theta \le 2\pi \}, C_2^{(l)}= \{\,R_2\,e^{\sqrt{-1}\theta }q^l\in {\mathbb {C}}\mid 0\le \theta \le 2\pi \},$ and $M_1,\,M_2$ are positive numbers independent of l.

The residue theorem combined with inequalities (2.5) and (2.6) leads to

$$\begin{aligned}&\sum _{-\infty<k<\infty } \textrm{Res}_{t=q^k}F(t)\,dt +\sum _{s=1}^r\sum _{-\infty<k<\infty } \textrm{Res}_{t=c_s^{-1}q^{1+k}}F(t)\,dt \nonumber \\&=\lim _{l\rightarrow \infty }\frac{1}{2\pi \sqrt{-1}}\left( \int _{C_1^{(l)}} + \int _{C_2^{(l)}} \right) F(t)\,dt=0, \end{aligned}$$

(2.7)

if $\left| \,b_1\cdots b_r a_1^{-1}\cdots a_r^{-1}\,\right|<|\,x\,|<1,$ where $C_1^{(l)}$ is in the counterclockwise direction and $C_2^{(l)}$ is in the clockwise direction.

Consequently, combining (2.3), (2.4) with (2.7), we obtain

$$\begin{aligned} 0&= \dfrac{\theta (Ax)\, \prod _{i=1}^{r}\theta (a_i)}{(q\,;q)_{\infty }^2\,\prod _{i=1}^r\theta (c_i)} \sum _{-\infty<k<\infty } \prod _{i=1}^{r}\frac{(b_iq^k\,;q)_{\infty }}{(a_iq^k\,;q)_{\infty }}\,x^k\\&\quad -\sum _{s=1}^r \frac{q}{c_s} \dfrac{\theta (Axc_sq^{-1})\, \prod _{i=1}^{r}\theta (c_sa_i^{-1})}{(q\,;q)_{\infty }^2\,\theta (c_s)\, \prod _{\begin{array}{c} 1\le i\le r\\ i\ne s \end{array}} \theta (c_sc_i^{-1})}\,\sum _{-\infty<k<\infty } \prod _{i=1}^{r} \frac{(qb_ic_s^{-1}q^{k}\,;q)_{\infty }}{(qa_ic_s^{-1}q^{k}\,;q)_{\infty }}\,x^k\nonumber \\&= \dfrac{\theta (Ax)\, \prod _{i=1}^{r}(qa_i^{-1},\,b_i\,;\,q)_{\infty }}{(q\,;q)_{\infty }^2\,\prod _{i=1}^r\theta (c_i)} \sum _{-\infty<k<\infty } \prod _{i=1}^{r}\frac{(a_i\,;q)_k}{(b_i\,;q)_k}\,x^k\\&\quad -\sum _{s=1}^r \frac{q}{c_s} \dfrac{\theta (Axc_sq^{-1})\, \prod _{i=1}^{r}(c_sa_i^{-1},\,qb_ic_s^{-1}\,;q)_{\infty }}{(q\,;q)_{\infty }^2\,\theta (c_s)\, \prod _{\begin{array}{c} 1\le i\le r\\ i\ne s \end{array}} \theta (c_sc_i^{-1})}\,\sum _{-\infty<k<\infty } \prod _{i=1}^{r} \frac{(qa_ic_s^{-1}\,;q)_k}{(qb_ic_s^{-1}\,;q)_k}\,x^k, \end{aligned}$$

if $\left| \,b_1\cdots b_r a_1^{-1}\cdots a_r^{-1}\,\right|<|\,x\,|<1.$ This completes the proof of (2.1). $\square$

3 Very-well-poised-balanced ${}_{2r}\psi _{2r}$ series

Slater’s transformation for very-well-poised-balanced ${}_{2r}\psi _{2r}$ series is given by the following, which is (5.5.2) of [1] ( (1.1) of [2] ):

Theorem 2

Suppose that

$$\begin{aligned} a,\, a_i,\, a_ia^{-1},\, a_ia_j^{-1},\, a_i^2a^{-1},\, a_ia_ja^{-1}\, \notin q^{{\mathbb {Z}}},\;1\le i\ne j\le r-2, \end{aligned}$$

for $r\ge 3$, and $\left| \,a^{r-1}q^{r-2}\,\right| <\left| \,\prod _{i=1}^{2(r-1)}b_i\,\right| .$ Then we have

$$\begin{aligned}&\frac{\prod _{i=1}^{2(r-1)}(qb_i^{-1},\,qab_i^{-1}\,;\,q)_{\infty }}{(qa,\,qa^{-1}\,;\,q)_{\infty }\, \prod _{i=1}^{r-2}\theta (a_ia^{-1},\,a_i)}\,\nonumber \\&\quad \times {}_{2r}\psi _{2r}\left( \begin{array}{c}{qa^{\frac{1}{2}},\;\;-qa^{\frac{1}{2}}, \;\;\;b_1,\; b_2,\;\ldots ,\;b_{2(r-1)}\;}\\ {\;a^{\frac{1}{2}},\;-a^{\frac{1}{2}},\;qab_1^{-1}, \,qab_2^{-1},\,\ldots ,\,qab_{2(r-1)}^{-1}}\end{array};\,{q,\; \dfrac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}}\, \right) \nonumber \\&=\sum _{s=1}^{r-2} \frac{\prod _{i=1}^{2(r-1)} (qaa_s^{-1}b_i^{-1},\,qa_sb_i^{-1}\,;\,q)_{\infty }}{(qa_s^2a^{-1},\,qaa_s^{-2}\,;\,q)_{\infty }\, \theta (a_sa^{-1},\,a_s)\, \prod _{\begin{array}{c} 1\le i\le r-2\\ i \ne s \end{array}} \theta (a_ia_s^{-1},\,a_sa_ia^{-1})}\,\nonumber \\&\quad \times {}_{2r}\psi _{2r}\left( \begin{array}{c}{qa_sa^{-\frac{1}{2}},\; -qa_sa^{-\frac{1}{2}}, \;\;b_1a_sa^{-1},\; b_2a_sa^{-1},\;\ldots ,\; b_{2(r-1)}a_sa^{-1}}\\ {\;\;a_sa^{-\frac{1}{2}},\;\;-a_sa^{-\frac{1}{2}},\; qa_sb_1^{-1}, \,qa_sb_2^{-1},\,\ldots ,\,qa_sb_{2(r-1)}^{-1}}\end{array};\,{q,\; \dfrac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}}\, \right) . \end{aligned}$$

(3.1)

Proof

Let F(t) be a function defined by

$$\begin{aligned}&F(t)=\dfrac{\prod _{i=1}^{2(r-1)} (qb_i^{-1}t^{-1},\,qab_i^{-1}t\,;q)_{\infty }}{\theta (t^{-1},\,at)\, \prod _{i=1}^{r-2}\theta (a_ia^{-1}t^{-1},\,a_it)} \nonumber \\&=\dfrac{\prod _{i=1}^{2(r-1)} (qb_i^{-1}t^{-1},\,qab_i^{-1}t\,;q)_{\infty }}{(t^{-1},\,qt,\,at,\,qa^{-1}t^{-1}\,;q)_{\infty } \prod _{i=1}^{r-2}(a_ia^{-1}t^{-1},\,qa_i^{-1}at,\, a_it,\,qa_i^{-1}t^{-1}\,;q)_{\infty }}, \end{aligned}$$

(3.2)

which satisfies $F(t)=F(a^{-1}t^{-1}).$

First, for $k\in {\mathbb {Z}}$, we have

$$\begin{aligned}&\textrm{Res}_{t=q^k}F(t)\,dt =\lim _{t\rightarrow q^k }(t-q^k)F(t)\nonumber \\&= \dfrac{q^k\times \prod _{i=1}^{2(r-1)} \theta (b_i)}{(q\,;q)_{\infty }^2\,\theta (a)\, \prod _{i=1}^{r-2}\theta (a_ia^{-1},\,a_i)_{\infty }} \prod _{i=1}^{2(r-1)} \dfrac{(qab_i^{-1}q^k\,;q)_{\infty }}{(b_iq^k\,;q)_{\infty }} \left( \,\frac{(aq)^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k, \end{aligned}$$

(3.3)

and

$$\begin{aligned}&\textrm{Res}_{t=a_sa^{-1}q^k}F(t)\,dt= \lim _{t\rightarrow a_sa^{-1}q^k}(t-a_sa^{-1}q^k)F(t)\nonumber \\&= \dfrac{a_sa^{-1}q^k\times \prod _{i=1}^{2(r-1)} \theta (qaa_s^{-1}b_i^{-1})}{(q\,;q)_{\infty }^2\,\theta (aa_s^{-1},\, a_s,\, a_s^2a^{-1}) \prod _{\begin{array}{c} 1\le i\le r-2\\ i\ne s \end{array}} \theta (a_ia_s^{-1},\,a_sa_ia^{-1})}\, \nonumber \\&\quad \times \prod _{i=1}^{2(r-1)} \dfrac{(qa_sb_i^{-1}q^k\,;q)_{\infty }}{(a_sb_ia^{-1}q^k\,;q)_{\infty }} \left( \,\frac{(qa)^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k\,. \end{aligned}$$

(3.4)

Secondly, for $k\in {\mathbb {Z}}$, we have

$$\begin{aligned}&\textrm{Res}_{t=a^{-1}q^{-k}}F(t)\,dt =-a^{-1}\textrm{Res}_{u=q^k}F(a^{-1}u^{-1})\,u^{-2}\,\textrm{d}u\nonumber \\&=-a^{-1}\textrm{Res}_{u=q^k}F(u)\,u^{-2}\,\textrm{d}u =-a^{-1}q^{-2k}\textrm{Res}_{u=q^k}F(u)\,\textrm{d}u \end{aligned}$$

(3.5)

and

$$\begin{aligned}&\textrm{Res}_{t=a_s^{-1}q^{-k}}F(t)\,dt =-a^{-1}\textrm{Res}_{u=a^{-1}a_sq^k}F(a^{-1}u^{-1})\,u^{-2}\,\textrm{d}u \nonumber \\&=-a^{-1}\textrm{Res}_{u=a^{-1}a_sq^k}F(u)\,u^{-2}\,\textrm{d}u =-aa_s^{-2}q^{-2k}\textrm{Res}_{u=a^{-1}a_sq^k}F(u)\,\textrm{d}u \end{aligned}$$

(3.6)

by the change of integration variable from t to u by $t=a^{-1}u^{-1}$ with the equality $F(t)=F(a^{-1}t^{-1}).$

Thirdly, for real positive numbers $R_1$ and $R_2$ satisfying

$$\begin{aligned} R_j,\,aR_j,\,a_iR_j,\,aa_i^{-1}R_j\notin q^{{\mathbb {Z}}}, \quad 1\le i\le r-2,\;\;j=1,\,2, \end{aligned}$$

we have

$$\begin{aligned}&F\left( R_1\,e^{\sqrt{-1}\,\theta }q^{-l}\right) \nonumber \\&= \dfrac{\prod _{i=1}^{2(r-1)} (qb_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l,\, qab_i^{-1}R_1\,e^{\sqrt{-1}\theta }\,;q)_{\infty }}{ (R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l,\, qR_1\,e^{\sqrt{-1}\theta },\, aR_1\,e^{\sqrt{-1}\theta },\, qa^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l \,;q)_{\infty }}\nonumber \\&\quad \times \dfrac{1}{\prod _{i=1}^{r-2} (a_ia^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l,\, qaa_i^{-1}R_1\,e^{\sqrt{-1}\theta },\, a_iR_1\,e^{\sqrt{-1}\theta },\, qa_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l \,;q)_{\infty }}\nonumber \\&\quad \times \dfrac{\prod _{i=1}^{2(r-1)} (b_ia^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta }\,;q)_l}{(R_1^{-1}\,e^{-\sqrt{-1}\theta }q^l,\, a^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta } \,;q)_{l} \prod _{i=1}^{r-2} (a_ia^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta },\, a_i^{-1}R_1^{-1}\,e^{-\sqrt{-1}\theta } \,;q)_l}\nonumber \\&\quad \times \left( \frac{(qa)^{r-1}}{\prod _{i=1}^{2(r-1)}b_i} \right) ^l,\nonumber \\&F(R_2\,e^{\sqrt{-1}\,\theta }q^{l}) \nonumber \\&= \dfrac{\prod _{i=1}^{2(r-1)} (qb_i^{-1}R_2^{-1}\,e^{-\sqrt{-1}\theta },\, qab_i^{-1}R_2\,e^{\sqrt{-1}\theta }q^l\,;q)_{\infty }}{ (R_2^{-1}\,e^{-\sqrt{-1}\theta },\, qR_2\,e^{\sqrt{-1}\theta }q^l,\, aR_2\,e^{\sqrt{-1}\theta }q^l,\, qa^{-1}R_2^{-1}\,e^{-\sqrt{-1}\theta } \,;q)_{\infty }}\nonumber \\&\quad \times \dfrac{1}{\prod _{i=1}^{r-2} (a_ia^{-1}R_2^{-1}\,e^{-\sqrt{-1}\theta },\, qaa_i^{-1}R_2\,e^{\sqrt{-1}\theta }q^l,\, a_iR_2\,e^{\sqrt{-1}\theta }q^l,\, qa_i^{-1}R_2^{-1}\,e^{-\sqrt{-1}\theta } \,;q)_{\infty }}\nonumber \\&\quad \times \dfrac{\prod _{i=1}^{2(r-1)} (b_iR_2\,e^{\sqrt{-1}\theta }\,;q)_l}{(qR_2\,e^{\sqrt{-1}\theta },\, aR_2\,e^{\sqrt{-1}\theta } \,;q)_{l} \prod _{i=1}^{r-2} (qaa_i^{-1}R_2\,e^{\sqrt{-1}\theta },\, a_iR_2\,e^{\sqrt{-1}\theta },\, \,;q)_l} \left( \frac{(qa)^{r-1}}{\prod _{i=1}^{2(r-1)}b_i} \right) ^l, \end{aligned}$$

and thus

$$\begin{aligned} \left| \,\int _{C_1^{(l)}} F(t)\,dt\,\right|&=\left| \,\int _0^{2\pi }F(R_1\,e^{\sqrt{-1}\theta }q^{-l}) \,R_1\,\sqrt{-1}\,e^{\sqrt{-1}\theta }q^{-l} \,\textrm{d}\theta \,\right| \le M_1 \left| \,\frac{a^{r-1}q^{r-2}}{\prod _{i=1}^{2(r-1)}b_i}\, \right| ^l, \end{aligned}$$

(3.7)

$$\begin{aligned} \left| \,\int _{C_2^{(l)}}F(t)\,dt\,\right|&=\left| \,\int _0^{2\pi }F(R_2\,e^{\sqrt{-1}\theta }q^l)\, R_2\,\sqrt{-1}\,e^{\sqrt{-1}\theta }q^l\,\textrm{d}\theta \,\right| \le M_2\, \left| \,\frac{a^{r-1}q^{r}}{\prod _{i=1}^{2(r-1)}b_i}\, \right| ^l, \end{aligned}$$

(3.8)

where $l\in {\mathbb {Z}}_{\ge 0},$ $C_1^{(l)}= \{\,R_1\,e^{\sqrt{-1}\theta }q^{-l}\in {\mathbb {C}}\mid 0\le \theta \le 2\pi \}, C_2^{(l)}= \{\,R_2\,e^{\sqrt{-1}\theta }q^l\in {\mathbb {C}}\mid 0\le \theta \le 2\pi \},$ and $M_1,\,M_2$ are positive numbers independent of l.

The residue theorem combined with inequalities (3.7) and (3.8) leads to

$$\begin{aligned}&\sum _{-\infty<k<\infty } \Bigl \{\textrm{Res}_{t=q^k}F(t)\,dt+ \textrm{Res}_{t=a^{-1}q^{-k}}F(t)\,dt\Bigr \} \nonumber \\&+\sum _{s=1}^{r-2}\sum _{-\infty<k<\infty } \Bigl \{\textrm{Res}_{t=a_sa^{-1}q^k}F(t)\,dt+ \textrm{Res}_{t=a_s^{-1}q^{-k}}F(t)\,dt\Bigr \} \nonumber \\&=\lim _{l\rightarrow \infty }\frac{1}{2\pi \sqrt{-1}}\left( \int _{C_1^{(l)}} + \int _{C_2^{(l)}} \right) F(t)\,dt=0, \end{aligned}$$

(3.9)

if $\left| \,a^{r-1}q^{r-2}\,\right| <\left| \,\prod _{i=1}^{2(r-1)}b_i\,\right| ,$ where $C_1^{(l)}$ is in the counterclockwise direction and $C_2^{(l)}$ is in the clockwise direction.

On the other hand, for $k\in {\mathbb {Z}}$, we have

$$\begin{aligned}&\textrm{Res}_{t=q^k}F(t)\,dt+ \textrm{Res}_{t=a^{-1}q^{-k}}F(t)\,dt\nonumber \\&= \bigl (\,q^k-a^{-1}q^{-k}\,\bigr )\times \dfrac{\prod _{i=1}^{2(r-1)} \theta (b_i)}{(q\,;q)_{\infty }^2\,\theta (a) \,\prod _{i=1}^{r-2}\theta (a_ia^{-1},\,a_i)} \nonumber \\&\quad \times \prod _{i=1}^{2(r-1)}\; \dfrac{(qab_i^{-1}q^k\,;\,q)_{\infty }}{(b_i\,q^k\,;\,q)_{\infty }} \left( \,\frac{(aq)^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k \nonumber \\&= -a^{-1}\times \dfrac{ \prod _{i=1}^{2(r-1)} \theta (b_i)}{(q\,;q)_{\infty }^2\,(qa,\,qa^{-1}\,;\,q)_{\infty } \,\prod _{i=1}^{r-2}\theta (a_ia^{-1},\,a_i)} \nonumber \\&\quad \times \frac{1-aq^{2k}}{1-a}\prod _{i=1}^{2(r-1)}\; \dfrac{(qab_i^{-1}q^k\,;\,q)_{\infty }}{(b_i\,q^k\,;\,q)_{\infty }} \left( \,\frac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k \end{aligned}$$

(3.10)

from (3.3) and (3.5), and

$$\begin{aligned}&\textrm{Res}_{t=a_sa^{-1}q^k}F(t)\,dt+ \textrm{Res}_{t=a_s^{-1}q^{-k}}F(t)\,dt\nonumber \\&= \bigl (\,a_sa^{-1}q^k-a_s^{-1}q^{-k}\,\bigr )\times \dfrac{\prod _{i=1}^{2(r-1)} \theta (a_sb_ia^{-1})}{(q\,;q)_{\infty }^2\,\theta (aa_s^{-1},\,a_s,\,a_s^2a^{-1}) \,\prod _{\begin{array}{c} 1\le i\le r-2\\ i\ne s \end{array}} \theta (a_ia_s^{-1},\,a_ia_sa^{-1})} \nonumber \\&\quad \times \prod _{i=1}^{2(r-1)}\; \dfrac{(qa_sb_i^{-1}q^k\,;\,q)_{\infty }}{(a_sb_ia^{-1}\,q^k\,;\,q)_{\infty }} \left( \,\frac{(qa)^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k \nonumber \\&= -a_s^{-1}\times \dfrac{ \prod _{i=1}^{2(r-1)} \theta (a_sb_ia^{-1})}{(q\,;q)_{\infty }^2\,(qa_s^2a^{-1},\,qaa_s^{-2}\,;\,q)_{\infty } \,\theta (aa_s^{-1},\,a_s) \,\prod _{\begin{array}{c} 1\le i\le r-2\\ i\ne s \end{array}}\theta (a_ia_s^{-1},\,a_ia_sa^{-1})} \nonumber \\&\quad \times \frac{1-a_s^2a^{-1}q^{2k}}{1-a_s^2a^{-1}}\prod _{i=1}^{2(r-1)}\; \dfrac{(qa_sb_i^{-1}\,q^k\,;\,q)_{\infty }}{(a_sb_ia^{-1}\,q^k\,;\,q)_{\infty }} \left( \,\frac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k \end{aligned}$$

(3.11)

from (3.4) and (3.6).

Consequently, combining (3.10), (3.11) with (3.9), we obtain

$$\begin{aligned} 0&= -a^{-1}\times \dfrac{ \prod _{i=1}^{2(r-1)} \theta (b_i)}{(q\,;q)_{\infty }^2\,(qa,\,qa^{-1}\,;q)_{\infty } \,\prod _{i=1}^{r-2}\theta (a_ia^{-1},\,a_i)} \nonumber \\&\quad \times \sum _{-\infty<k<\infty } \frac{1-aq^{2k}}{1-a}\prod _{i=1}^{2(r-1)}\; \dfrac{(qab_i^{-1}q^k\,;q)_{\infty }}{(b_i\,q^k\,;q)_{\infty }} \left( \,\frac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k \nonumber \\&-\sum _{s=1}^{r-2} a_s^{-1}\times \dfrac{ \prod _{i=1}^{2(r-1)} \theta (a_sb_ia^{-1})}{(q\,;\,q)_{\infty }^2\,(qa_s^2a^{-1},\,qaa_s^{-2}\,;\,q)_{\infty } \,\theta (aa_s^{-1},\,a_s) \,\prod _{\begin{array}{c} 1\le i\le r-2\\ i\ne s \end{array}} \theta (a_ia_s^{-1},\,a_ia_sa^{-1})} \nonumber \\&\quad \times \sum _{-\infty<k<\infty } \frac{1-a_s^2a^{-1}q^{2k}}{1-a_s^2a^{-1}}\prod _{i=1}^{2(r-1)}\; \dfrac{(qa_sb_i^{-1}\,q^k\,;\,q)_{\infty }}{(a_sb_ia^{-1}\,q^k\,;\,q)_{\infty }} \left( \,\frac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k, \end{aligned}$$

(3.12)

which is equivalent to

$$\begin{aligned}&\dfrac{ \prod _{i=1}^{2(r-1)} (qb_i^{-1},\, qab_i^{-1}\,;\,q)_{\infty }}{(qa,\,qa^{-1}\,;\,q)_{\infty } \,\prod _{i=1}^{r-2}\theta (a_ia^{-1},\,a_i)} \nonumber \\&\quad \times \sum _{-\infty<k<\infty }\frac{1-aq^{2k}}{1-a}\prod _{i=1}^{2(r-1)}\; \dfrac{(b_i\,;q)_k}{(qab_i^{-1}\,;q)_k} \left( \,\frac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k \nonumber \\&=\sum _{s=1}^{r-2} \dfrac{ \prod _{i=1}^{2(r-1)} (qaa_s^{-1}b_i^{-1},\, qa_sb_i^{-1}\,;\,q)_{\infty }}{(qa_s^2a^{-1},\,qaa_s^{-2}\,;\,q)_{\infty } \,\theta (a_sa^{-1},\,a_s) \,\prod _{\begin{array}{c} 1\le i\le r-2\\ i\ne s \end{array}} \theta (a_ia_s^{-1},\,a_ia_sa^{-1})} \nonumber \\&\quad \times \sum _{-\infty<k<\infty }\frac{1-a_s^2a^{-1}q^{2k}}{1-a_s^2a^{-1}}\prod _{i=1}^{2(r-1)}\; \dfrac{(a_sb_ia^{-1}\,;\,q)_k}{(qa_sb_i^{-1}\,;q)_k} \left( \,\frac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}\,\right) ^k. \end{aligned}$$

(3.13)

This completes the proof of (3.1). $\square$

Remark 1

If we change r to $r+2$ and substitute $b_{2r+1}=a^{\frac{1}{2}},\; b_{2r+2}=-a^{\frac{1}{2}}$ in (3.1), we obtain (5.5.1) of [1] ((7) of [6], and (7.2.1.1) of [7]), since

$$\begin{aligned}{} & {} \frac{(a^{\frac{1}{2}},\,-a^{\frac{1}{2}};q)_{\infty }}{\theta (a)}= \frac{(qa^{\frac{1}{2}},\,-qa^{\frac{1}{2}};q)_{\infty }}{(qa,\,qa^{-1};q)_{\infty }},\; a_s\frac{(a^{\frac{1}{2}}a_s^{-1},\,-a^{\frac{1}{2}}a_s^{-1};q)_{\infty }}{\theta (aa_s^{-1},\,a_s^2a^{-1})}\\{} & {} = \frac{(qa^{\frac{1}{2}}a_s^{-1},\,-qa^{\frac{1}{2}}a_s^{-1};q)_{\infty }}{\theta (a_sa^{-1})\,(qa_s^2a^{-1},\,qaa_s^{-2};q)_{\infty }}. \end{aligned}$$

Remark 2

To obtain (5.5.1) of [1] by the same manner as in the proof of Theorem 2, it is enough to consider the function

$$\begin{aligned} F(t)&=\dfrac{\prod _{i=1}^{2r} \left( qb_i^{-1}t^{-1},\,qab_i^{-1}t\,;q\right) _{\infty }}{\left( t^{-1},\,qt,\,at,\,qa^{-1}t^{-1}\,;q\right) _{\infty } \prod _{i=1}^{r}\left( a_ia^{-1}t^{-1},\,qa_i^{-1}at,\, a_it,\,qa_i^{-1}t^{-1}\,;q\right) _{\infty }}. \end{aligned}$$

4 Very-well-poised ${}_{2r}\psi _{2r}$ and ${}_{2r-1}\psi _{2r-1}$ series

If we change r to $r+1$ and substitute $b_{2r-1}=q^{\frac{1}{2}}a^{\frac{1}{2}},\; b_{2r}=-q^{\frac{1}{2}}a^{\frac{1}{2}}$ in (3.1), we obtain the following transformation for very-well-poised ${}_{2r}\psi _{2r}$ series:

Theorem 3

Suppose that

$$\begin{aligned} a,\, a_i,\, a_ia^{-1},\, a_ia_j^{-1},\, a_i^2a^{-1},\, a_ia_ja^{-1}\, \notin q^{{\mathbb {Z}}},\;1\le i\ne j\le r-1, \end{aligned}$$

for $r\ge 2$, and $\left| \,a^{r-2}q^{r-1}\,\right| <\left| \,\prod _{i=1}^{2(r-1)}b_i\,\right| .$ Then we have

$$\begin{aligned}&\frac{\prod _{i=1}^{2(r-1)}(qb_i^{-1},\,qab_i^{-1}\,;\,q)_{\infty }}{(q^2a,\,q^2a^{-1}\,;\,q^2)_{\infty }\, \prod _{i=1}^{r-1}\theta (a_ia^{-1},\,a_i)}\,\nonumber \\&\quad \times {}_{2r}\psi _{2r}\left( \begin{array}{c}{qa^{\frac{1}{2}},\;\;-qa^{\frac{1}{2}}, \;\;\;b_1,\; b_2,\;\ldots ,\;b_{2(r-1)}\;}\\ {\;a^{\frac{1}{2}},\;-a^{\frac{1}{2}},\;qab_1^{-1}, \,qab_2^{-1},\,\ldots ,\,qab_{2(r-1)}^{-1}}\end{array};\,{q,\; -\dfrac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}}\, \right) \nonumber \\&=\sum _{s=1}^{r-1} \frac{\prod _{i=1}^{2(r-1)} (qaa_s^{-1}b_i^{-1},\,qa_sb_i^{-1}\,;\,q)_{\infty }}{(q^2a_s^2a^{-1},\,q^2aa_s^{-2}\,;\,q^2)_{\infty }\, \theta (a_sa^{-1},\,a_s)\, \prod _{\begin{array}{c} 1\le i\le r-1\\ i \ne s \end{array}} \theta (a_ia_s^{-1},\,a_sa_ia^{-1})}\,\nonumber \\&\quad \times {}_{2r}\psi _{2r}\left( \begin{array}{c}{qa_sa^{-\frac{1}{2}},\; -qa_sa^{-\frac{1}{2}}, \;\;b_1a_sa^{-1},\; b_2a_sa^{-1},\;\ldots ,\; b_{2(r-1)}a_sa^{-1}}\\ {\;\;a_sa^{-\frac{1}{2}},\;\;-a_sa^{-\frac{1}{2}},\; qa_sb_1^{-1}, \,qa_sb_2^{-1},\,\ldots ,\,qa_sb_{2(r-1)}^{-1}}\end{array};\,{q,\; -\dfrac{q^{r-2}a^{r-1}}{\prod _{i=1}^{2(r-1)}b_i}}\, \right) . \end{aligned}$$

(4.1)

Remark 3

To obtain (4.1) by the same manner as in the proof of Theorem 2, it is enough to consider the function

$$\begin{aligned} F(t) =\dfrac{(q^{\frac{1}{2}}a^{-\frac{1}{2}}t^{-1},\, q^{\frac{1}{2}}a^{\frac{1}{2}}t,\, -q^{\frac{1}{2}}a^{-\frac{1}{2}}t^{-1},\, -q^{\frac{1}{2}}a^{\frac{1}{2}}t;q)_{\infty } \prod _{i=1}^{2(r-1)} (qb_i^{-1}t^{-1},\,qab_i^{-1}t;q)_{\infty }}{(t^{-1},\,qt,\,at,\,qa^{-1}t^{-1};q)_{\infty } \prod _{i=1}^{r-1}(a_ia^{-1}t^{-1},\,qa_i^{-1}at,\, a_it,\,qa_i^{-1}t^{-1};q)_{\infty }}. \end{aligned}$$

Remark 4

If we change r to $r+2$ and substitute $b_{2r+1}=a^{\frac{1}{2}},\; b_{2r+2}=-a^{\frac{1}{2}}$ in (4.1), we obtain (5.5.4) of [1] ( (8) of [6], and (7.2.1.3) of [7] ), since

$$\begin{aligned}&(qa^{-\frac{1}{2}},\,qa^{\frac{1}{2}},\, -qa^{-\frac{1}{2}},\,-qa^{\frac{1}{2}}\,;q)_{\infty } (q^{\frac{1}{2}}a^{-\frac{1}{2}},\,q^{\frac{1}{2}}a^{\frac{1}{2}},\, -q^{\frac{1}{2}}a^{-\frac{1}{2}},\, -q^{\frac{1}{2}}a^{\frac{1}{2}}\,;q)_{\infty }\\&\quad =(qa,\,qa^{-1}\,;q)_{\infty }, \end{aligned}$$

and

$$\begin{aligned}&(qa^{\frac{1}{2}}a_s^{-1},\,qa_sa^{-\frac{1}{2}},\, -qa^{\frac{1}{2}}a_s^{-1},\,-qa_sa^{-\frac{1}{2}}\,;q)_{\infty }\\&\quad \times (q^{\frac{1}{2}}a^{\frac{1}{2}}a_s^{-1},\,q^{\frac{1}{2}}a^{-\frac{1}{2}}a_s,\, -q^{\frac{1}{2}}a^{\frac{1}{2}}a_s^{-1},\, -q^{\frac{1}{2}}a^{-\frac{1}{2}}a_s\,;q)_{\infty } =(qa_s^2a^{-1},\,qaa_s^{-2}\,;q)_{\infty }. \end{aligned}$$

Remark 5

To obtain (5.5.4) of [1] by the same manner as in the proof of Theorem 2, it is enough to consider the function

$$\begin{aligned} F(t)&=\dfrac{(q^{\frac{1}{2}}a^{-\frac{1}{2}}t^{-1},\, q^{\frac{1}{2}}a^{\frac{1}{2}}t,\, -q^{\frac{1}{2}}a^{-\frac{1}{2}}t^{-1},\, -q^{\frac{1}{2}}a^{\frac{1}{2}}t\,;q)_{\infty }\, (qa^{-\frac{1}{2}}t^{-1},\,qa^{\frac{1}{2}}t,\, -qa^{-\frac{1}{2}}t^{-1},\,-qa^{\frac{1}{2}}t\,;q)_{\infty }}{(t^{-1},\,qt,\,at,\,qa^{-1}t^{-1}\,;q)_{\infty }} \\&\quad \times \dfrac{\prod _{i=1}^{2r} (qb_i^{-1}t^{-1},\,qab_i^{-1}t\,;q)_{\infty }}{\prod _{i=1}^{r+1}(a_ia^{-1}t^{-1},\,qa_i^{-1}at,\, a_it,\,qa_i^{-1}t^{-1}\,;q)_{\infty }}. \end{aligned}$$

The substitution $b_{2(r-1)}=\pm q^{\frac{1}{2}}a^{\frac{1}{2}}$ in (4.1) implies the following transformation for very-well-poised ${}_{2r-1}\psi _{2r-1}$ series:

Theorem 4

Suppose that

$$\begin{aligned} a,\, a_i,\, a_ia^{-1},\, a_ia_j^{-1},\, a_i^2a^{-1},\, a_ia_ja^{-1}\, \notin q^{{\mathbb {Z}}},\;1\le i\ne j\le r-2, \end{aligned}$$

for $r\ge 3$, and $\left| \,a^{r-\frac{5}{2}}q^{r-\frac{3}{2}}\,\right| <\left| \,\prod _{i=1}^{2r-3}b_i\,\right| .$ Then we have

$$\begin{aligned}&\frac{ \big (\pm q^{\frac{1}{2}}a^{-\frac{1}{2}},\, \pm q^{\frac{1}{2}}a^{\frac{1}{2}}\,;q\big )_{\infty }\, \prod _{i=1}^{2r-3}\big (qb_i^{-1},\,qab_i^{-1}\,;\,q)_{\infty }}{\big (qa,\,qa^{-1}\,;\,q\big )_{\infty }\, \prod _{i=1}^{r-2}\theta \big (a_ia^{-1},\,a_i\big )}\,\nonumber \\&\quad \times {}_{2r-1}\psi _{2r-1}\left( \begin{array}{c}{qa^{\frac{1}{2}},\;\;-qa^{\frac{1}{2}}, \;\;\;b_1,\; b_2,\;\ldots ,\;b_{2r-3}\;}\\ {\;a^{\frac{1}{2}},\;-a^{\frac{1}{2}},\;qab_1^{-1}, \,qab_2^{-1},\,\ldots ,\,qab_{2r-3}^{-1}}\end{array};\,{q,\; \pm \dfrac{q^{r-\frac{5}{2}}a^{r-\frac{3}{2}}}{\prod _{i=1}^{2r-3}b_i}}\, \right) \nonumber \\&=\sum _{s=1}^{r-2} \frac{ \big (\pm q^{\frac{1}{2}}a^{\frac{1}{2}}a_s^{-1},\, \pm q^{\frac{1}{2}}a^{-\frac{1}{2}}a_s\,;q\big )_{\infty }\, \prod _{i=1}^{2r-3} \big (qaa_s^{-1}b_i^{-1},\,qa_sb_i^{-1}\,;\,q\big )_{\infty }}{\big (qa_s^2a^{-1},\,qaa_s^{-2}\,;\,q\big )_{\infty }\, \theta (a_sa^{-1},\,a_s)\, \prod _{\begin{array}{c} 1\le i\le r-2\\ i \ne s \end{array}} \theta \big (a_ia_s^{-1},\,a_sa_ia^{-1}\big )}\,\nonumber \\&\quad \times {}_{2r-1}\psi _{2r-1}\left( \begin{array}{c}{qa_sa^{-\frac{1}{2}},\; -qa_sa^{-\frac{1}{2}}, \;\;b_1a_sa^{-1},\; b_2a_sa^{-1},\;\ldots ,\; b_{2r-3}a_sa^{-1}}\\ {\;\;a_sa^{-\frac{1}{2}},\;\;-a_sa^{-\frac{1}{2}},\; qa_sb_1^{-1}, \,qa_sb_2^{-1},\,\ldots ,\,qa_sb_{2r-3}^{-1}}\end{array};\,{q,\;\pm \dfrac{q^{r-\frac{5}{2}}a^{r-\frac{3}{2}}}{\prod _{i=1}^{2r-3}b_i}}\, \right) . \end{aligned}$$

(4.2)

Here either all the upper or all the lower signs are taken throughout.

Remark 6

To obtain (4.2) by the same manner as in the proof of Theorem 2, it is enough to consider the function

$$\begin{aligned} F(t) =\dfrac{\big (\pm q^{\frac{1}{2}}a^{-\frac{1}{2}}t^{-1},\, \pm q^{\frac{1}{2}}a^{\frac{1}{2}}t;q\big )_{\infty }\, \prod _{i=1}^{2r-3} \big (qb_i^{-1}t^{-1},\,qab_i^{-1}t;q\big )_{\infty }}{\big (t^{-1},\,qt,\,at,\,qa^{-1}t^{-1};q\big )_{\infty } \prod _{i=1}^{r-2}\big (a_ia^{-1}t^{-1},\,qa_i^{-1}at,\, a_it,\,qa_i^{-1}t^{-1};q\big )_{\infty }}. \end{aligned}$$

Remark 7

If we change r with $r+2$ and substitute $b_{2r}=a^{\frac{1}{2}},\; b_{2r+1}=-a^{\frac{1}{2}}$ in (4.2), we obtain (5.5.5) of [1] ( (9) of [6], and (7.2.1.4) of [7]), since

$$\begin{aligned}&\big (a_sa^{-\frac{1}{2}},\,-a_sa^{-\frac{1}{2}},\\&\, qa^{\frac{1}{2}}a_s^{-1},\,-qa^{\frac{1}{2}}a_s^{-1}\,;q\big )_{\infty } =-a_s^2a^{-1} \big (qa_sa^{-\frac{1}{2}},\,-qa_sa^{-\frac{1}{2}},\, a^{\frac{1}{2}}a_s^{-1},\,-a^{\frac{1}{2}}a_s^{-1}\,;q\big )_{\infty }, \end{aligned}$$

and

$$\begin{aligned}&\theta (a_sa^{-1})=-a_sa^{-1}\theta (aa_s^{-1}). \end{aligned}$$

Remark 8

To obtain (5.5.5) of [1] by the same manner as in the proof of Theorem 2, it is enough to consider the function

$$\begin{aligned} F(t)&=\dfrac{\big (\pm q^{\frac{1}{2}}a^{-\frac{1}{2}}t^{-1},\, \pm q^{\frac{1}{2}}a^{\frac{1}{2}}t\,;q\big )_{\infty }\, \big (qa^{-\frac{1}{2}}t^{-1},\,qa^{\frac{1}{2}}t,\, -qa^{-\frac{1}{2}}t^{-1},\,-qa^{\frac{1}{2}}t\,;q\big )_{\infty }}{(t^{-1},\,qt,\,at,\,qa^{-1}t^{-1}\,;q)_{\infty }} \\&\quad \times \dfrac{\prod _{i=1}^{2r-1} (qb_i^{-1}t^{-1},\,qab_i^{-1}t\,;q)_{\infty }}{\prod _{i=1}^{r}(a_ia^{-1}t^{-1},\,qa_i^{-1}at,\, a_it,\,qa_i^{-1}t^{-1}\,;q)_{\infty }}. \end{aligned}$$