\def\WhWaAA{21}
\def\WatsAA{20}
\def\VeJaAH{19}
\def\ThoeAA{18}
\def\SrivAN{17}
\def\SlatAC{16}
\def\SlatAD{15}
\def\MoakAB{14}
\def\MoakAA{13}
\def\MaTuAA{12}
\def\LukeAA{11}
\def\LavoAB{10}
\def\LavoAA{9}
\def\KociAA{8}
\def\JackAA{7}
\def\IsLMAA{6}
\def\GaRaAA{5}
\def\ErdeAA{4}
\def\BailAA{3}
\def\AskeAB{2}
\def\AskeAA{1}

\def\R{{\Bbb R}}
\def\C{{\Bbb C}}
\def\liml{\lim\limits}
\def\prodl{\prod\limits}
\def\({\left(}
\def\){\right)}
\def\[{\left[}
\def\]{\right]}
\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\de{\delta}
\def\ep{\varepsilon}
\def\ze{\zeta}
\def\et{\eta}
\def\th{\theta}
\def\vt{\vartheta}
\def\io{\iota}
\def\ka{\kappa}
\def\la{\lambda}
\def\rh{\rho}
\def\si{\sigma}
\def\ta{\tau}
\def\ph{\varphi}
\def\ch{\chi}
\def\ps{\psi}
\def\om{\omega}
\def\Ga{\Gamma}
\def\De{\Delta}
\def\Th{\Theta}
\def\La{\Lambda}
\def\Si{\Sigma}
\def\Ph{\Phi}
\def\Ps{\Psi}
\def\Om{\Omega}


\catcode`\@=11
\def\subjclass{\nofrills@{{\rm1991 {\it Mathematics Subject
   Classification.}\enspace}}\subjclass@
 \DNii@##1\endsubjclass{\def\thesubjclass@{\def\usualspace
  {{\rm\enspace}}\eightpoint\subjclass@\ignorespaces##1\unskip.}}%
 \FN@\next@}
\outer\def\proclaim{\runaway@{proclaim}\def\envir@{proclaim}%
  \nofrills@{.\enspace}\proclaim@
 \DNii@##1{\penaltyandskip@{-100}\medskipamount\varindent@
   \usualspace@{{\proclaimheadfont@\enspace}}\proclaimheadfont@
   \ignorespaces##1\unskip\proclaim@
  \it\ignorespaces}% 
 \FN@\next@}
\catcode`\@=13

\input format
\NoRunningHeads
\TagsOnRight
\def\poq#1#2{(#1;q)_#2}
\def\po#1#2{(#1)_#2}
\def\v#1{\vert #1\vert}
\def\psq{{\tilde \psi_q}}
\def\Ai{(A)}
\def\Bi{(B)}
\def\qAi{(qA)}
\def\qBi{(qB)}
\def\Aiq{(A/q)}
\def\Biq{(B/q)}
\def\ai{A_i}
\def\bi{B_i}

\topmatter 
\title Summations for basic hypergeometric series involving a
$q$-analogue of the digamma function 
\endtitle 
\author C.~Krattenthaler\\
H. M. Srivastava
\endauthor 
%\affil 
%Institut f\"ur Mathematik der Universit\"at Wien,\\
%Strudlhofgasse 4, A-1090 Wien, Austria.\\
%E-mail: KRATT\@PAP.UNIVIE.AC.AT\\\vskip6pt
%University of Victoria, Victoria,\\ British Columbia, Canada V8W
%3P4.\\E-mail: HARIMS\@UVVM.UVIC.CA
%\endaffil 
\address Institut f\"ur Mathematik der Universit\"at Wien,
Strudlhofgasse 4, A-1090 Wien, Austria.
E-mail: KRATT\@PAP.UNIVIE.AC.AT.
\endaddress
\address Department of Mathematics and Statistics,
University of Victoria, Victoria,
British Columbia V8W 3P4, Canada.\newline
E-mail: HMSRI\@UVVM.UVIC.CA or HARIMS\@UVVM.UVIC.CA.
\endaddress
%\dedicatory \enddedicatory
%\date \enddate
%\thanks \endthanks
\subjclass Primary 33D20;
 Secondary 33B15, 33C20, 33D05
\endsubjclass
\keywords basic hypergeometric series, $q$-digamma function,
$q$-polygamma function,
summation formulas
\endkeywords
\abstract Using a simple method, numerous summation formulas for
hypergeometric and
basic hypergeometric series are derived. Among these summation
formulas are nonterminating extensions and $q$-extensions of
identities recorded by Lavoie, Luke, Watson, and Srivastava. At the result
side of the basic hypergeometric 
summations there appears a $q$-analogue of the digamma
function. Some of its properties are also studied.
\endabstract
\endtopmatter
\document

\subhead 1. Introduction and description of method\endsubhead
In \cite{\LukeAA, \LavoAA, \LavoAB, \SrivAN} a simple identity for
hypergeometric series is combined with a limiting process to obtain 
summation formulas for hypergeometric series in which the digamma
function appears at the result side. In this paper we start with the
$q$-analogue of this identity. Again by limiting processes, we derive
numerous identities for basic hypergeometric series, among which are
nonterminating extensions and $q$-analogues of the identities in
\cite{\LukeAA, \LavoAA, \LavoAB, \SrivAN}. Not surprisingly,
at the result side of our identities there
appears a $q$-analogue of the digamma function. Letting $q$ tend to
$1$, also several new summations for hypergeometric series are
obtained.

We adopt the usual conventions of writing basic hypergeometric
series (cf\. \cite{\GaRaAA, sec.~1.2}), $(a;q)_n=\prod _{i=0} ^{n-1}(1-aq^{i})$, 
$(a;q)_\infty=\prod _{i=0} ^{\infty}(1-aq^{i})$,
$${}_r\Phi_s\!\[\matrix a_1,\dots,a_r;\\b_1,\dots,b_s;\endmatrix q,
z\]=\sum _{n=0} ^{\infty}\frac {\poq{a_1}{n}\dotsb\poq{a_r}{n}}
{\poq{q}{n}\poq{b_1}{n}\dotsb\poq{b_s}{n}}\((-1)^nq^{\binom
n2}\)^{s-r+1}z^n\ ,$$
and the short notation (cf\. \cite{\GaRaAA, sec.~2.1})
$$_{r+1}W_r(a;a_1,\dots,a_{ r-2};q,z):=
{}_{r+1}\Phi_r\!\[\matrix a,q\sqrt a,-q\sqrt a,a_1,\dots,a_{r-2};\\
\sqrt a,-\sqrt a,aq/a_1,\dots,aq/a_{r-2};\endmatrix q, z\]$$
for the very well-poised basic hypergeometric series. 

The hypergeometric notations (cf\. \cite{\BailAA, sec.~2.1;
\SlatAC, sec.~2.1}) are $(a)_n=\prod
_{i=0} ^{n-1}(a+i)$, 
$${}_r F_s\!\[\matrix a_1,\dots,a_r;\\b_1,\dots,b_s;\endmatrix  
z\]=\sum _{n=0} ^{\infty}\frac {\po{a_1}{n}\dotsb\po{a_r}{n}}
{n!\,\po{b_1}{n}\dotsb\po{b_s}{n}} z^n\ ,$$
and the short notation 
$$V(a;a_1,\dots,a_{ r-1};z):=
{}_{r+1}F_r\!\[\matrix a,1+a/2,a_1,\dots,a_{r-1};\\
 a/2,1+a-a_1,\dots,1+a-a_{r-2};\endmatrix  z\]$$
for the very well-poised hypergeometric series. 

All identities in our paper are subject to suitable conditions on the
parameters such that the involved hypergeometric or basic
hypergeometric series converge. We shall not state these conditions
for each identity. The reader should consult \cite{\GaRaAA,
pp.~4--5}.

We shall frequently use the short notations
$$\align (a_1,\dots,a_m;q)_\be&:=\prod _{i=1} ^{m}\poq {a_i}\be\ ,\\
(a_1,\dots,a_m)_\be&:=\prod _{i=1} ^{m}\po {a_i}\be\ ,
\endalign$$
and
$$\Gamma \bmatrix a_1,\dots,a_m\\b_1,\dots,b_n\endbmatrix:=
\frac {\prod _{i=1} ^{m}\Gamma(a_i)} {\prod _{i=1} ^{n}\Gamma(b_i)},$$
where $\Gamma(z)$ is the Gamma function.

\medskip
Now let us describe our method.

The aforementioned $q$-analogue of the identity utilized in 
\cite{\LukeAA, \LavoAA, \LavoAB, \SrivAN} is
$$\multline {}_{r+1}\Phi_{s+1}\!\[\matrix
qa_1,\dots,qa_r,q;\\qb_1,\dots,qb_s,q^2;\endmatrix q, z\]\\
=\frac {(1-b_1)\dotsb(1-b_s)(1-q)} {z\,(1-a_1)\dotsb(1-a_r)}
\({}_r\Phi_s\!\[\matrix a_1,\dots,a_r;\\b_1,\dots,b_s;\endmatrix q,
z\]-1\).
\endmultline\tag1.1$$
It is straight-forward to verify this identity. 
Suppose that the parameters in this identity are chosen such that
the resulting $_r\Phi_s$-series at the right-hand side of (1.1) can
be summed. By letting $a_1$, say, approach to $1$ and by computing
the limit at the right-hand side by application of l'H\^ospital's
Rule, we then obtain a summation formula for the resulting basic
hypergeometric series at the left-hand side.

Because of the application of l'H\^ospital's Rule, we have to deal
with derivations of infinite expressions $\poq a\infty$. This results
into the appearance of a $q$-analogue of the digamma function. Though
this $q$-analogue previously appeared in the literature \cite{\AskeAA,
\MoakAA, \MoakAB}, it does not seem to have been studied so far.
Therefore, as a preparation, in the next section we derive some of
its properties. In the subsequent sections we list the identities
that are obtained by our method; in section~3 those that are derived
by summing the $_r\Phi_s$ at the right-hand side of (1.1) with the
help of the nonterminating $q$-Pfaff--Saalsch\"utz sum \cite{\GaRaAA,
(2.10.12)}, in section~4 those
that are derived by using Bailey's nonterminating extension of
Jackson's $_8\Phi_7$-sum \cite{\GaRaAA, (2.11.7)}, in section~5 those
that are derived by using Verma and Jain's \cite{\VeJaAH, (1.6)}
$q$-analogue of Whipple's $_3F_2$-sum, and in section~6 those that are
derived by using Verma and Jain's \cite{\VeJaAH, (1.5)} $q$-analogue of
Watson's $_3F_2$-sum.

\subhead 2. The $q$-digamma function\endsubhead
Thomae \cite{\ThoeAA} and F.~H.~Jackson \cite{\JackAA} introduced the
following $q$-analogue of the gamma function (cf\. \cite{\GaRaAA, sec.~1.10,
Ex.~1.23}),
$$\Gamma_q(x)=\cases \dfrac {\poq q\infty} {\poq
{q^x}\infty}(1-q)^{1-x}& \v q<1\\
\dfrac {(q^{-1};q^{-1})_\infty} {(q^{-x};q^{-1})_\infty}(q-1)^{1-x}
q^{\binom x2} &1<q.\endcases\tag2.1$$
For properties of the $q$-gamma function see \cite{\AskeAA,
\AskeAB,
\GaRaAA, sec.~?, \IsLMAA, \JackAA, \KociAA, \MaTuAA, \MoakAA, \MoakAB,
\ThoeAA}.
It was shown by Askey \cite{\AskeAA} and Moak \cite{\MoakAB} that for
$x\in \R$ (set of real numbers),\break $x\ne0,-1,-2,\dots$, 
the $q$-gamma function
$\Gamma_q(x)$ indeed tends to the gamma function $\Gamma(x)$ when
letting $q\to1$. 

The digamma function $\psi(x)$ (see e.g\. \cite{\ErdeAA, sec.~1.7}) 
by definition is the
logarithmic derivative of $\Gamma(x)$,
$$\psi(x)=\frac {\frac {d} {dx}\Gamma(x)}{\Gamma(x)}.$$
In view of this it is natural to define a $q$-digamma function
$\psi_q(x)$ by
$$\psi_q(x):=\frac {\frac {d} {dx}\Ga_q(x)}{\Ga_q(x)},\tag2.2$$
or more explicitly,
$$\ps_q(x):=\cases \dsize-\log(1-q)+\log q\sum _{n=0} ^{\infty}\dfrac
{q^{n+x}} {1-q^{n+x}}&\v q<1\\
\dsize-\log(q-1)+(x-\frac {1} {2})\log q+\log q\sum _{n=0} ^{\infty}\dfrac
{q^{-n-x}} {q^{-n-x}-1}&1<q.\endcases\tag2.3$$
However, before being allowed to call $\ps_q(x)$ a $q$-analogue of
the digamma function, we have to show that $\ps_q(x)$ tends to
$\ps(x)$ when letting $q\to1$. In Theorem~1 below we prove this fact
for real $q$, which suffices for our purposes. It should not be too
difficult to prove the same for complex $q$, $\vert q\vert\ne 1$.
\proclaim{Theorem 1}Let $q\in\R$. For $x\in\C$ (set of complex
numbers), $x\ne0,-1,-2,\dots$, there holds
$$\lim_{q\to1}\ps_q(x)=\ps(x).$$
\endproclaim
\demo{Proof}First let $0<q<1$. Write $\ps_q(x)$ in the form
$$\ps_q(x)=-\log(1-q)+\log q\sum _{n=0} ^{\infty}\frac {q^{n+1}}
{1-q^{n+1}}-\log q\sum _{n=0} ^{\infty}\frac {q^{n+1}(1-q^{x-1})}
{(1-q^{n+1})(1-q^{n+x})}.\tag2.4$$
We shall show that 
$$\lim_{q\uparrow1}\(\log(1-q)-\log q\sum _{n=0} ^{\infty}\frac
{q^{n+1}} {1-q^{n+1}}\)=\ga\ ,\tag2.5$$
where $\ga$ is the Euler gamma (cf\. \cite{\ErdeAA, 1.1.(4); 
\WhWaAA, sec.~12.1}), and
$$\lim_{q\uparrow1}\( -\log q\sum _{n=0} ^{\infty}\frac {q^{n+1}(1-q^{x-1})}
{(1-q^{n+1})(1-q^{n+x})}\)=\sum _{n=0} ^{\infty}\frac {x-1}
{(n+1)(n+x)}.\tag2.6$$
Supposing that this would be done, we would have
$$\lim_{q\uparrow1}\ps_q(x)=-\ga+\sum _{n=0} ^{\infty}\frac {x-1}
{(n+1)(n+x)}\ .$$
The expression at the right-hand side of this equation is just one of
the many ways of writing the digamma function $\ps(x)$ (cf\.
\cite{\ErdeAA, 1.7.(3)}). This would prove $\lim_{q\uparrow1}\ps_q(x)=\ps(x)$.

We start with proving (2.5). In the first step we rewrite the
expression at the left-hand side of (2.5). We have
$$\align \sum _{n=1} ^{\infty}\( -\frac {q^n\log q}{1-q^n}\right.&\left.-\log\frac
{1-q^{n+1}} {1-q^n}\)\\
&=-\log q\sum _{n=0} ^{\infty}\frac {q^{n+1}} {1-q^{n+1}}-\sum _{n=1}
^{\infty}\big(\log(1-q^{n+1})-\log(1-q^n)\big)\\
&=-\log q\sum _{n=0} ^{\infty}\frac {q^{n+1}} {1-q^{n+1}}+\log (1-q).
\endalign$$
Thus (2.5) is equivalent with
$$\lim_{q\uparrow1}\sum _{n=1} ^{\infty}\( -\frac {q^n\log q} {1-q^n}-\log\frac
{1-q^{n+1}} {1-q^n}\)=\ga.\tag2.7$$
Suppose that we are allowed to interchange limit and summation at the
left-hand side of (2.7). This would give
$$\align
\lim_{q\uparrow1}\sum _{n=1} ^{\infty}\( -\frac {q^n\log q} {1-q^n}-\log\frac
{1-q^{n+1}} {1-q^n}\)&=\sum _{n=1} ^{\infty}\(\frac {1} {n}-\log\frac
{n+1} {n}\)\\
&=\lim_{N\to\infty} \(\sum _{n=1} ^{N}\frac {1} {n}-\log N\).
\endalign$$
But the right-hand side of this equation by definition equals $\ga$ (cf\.
\cite{\ErdeAA, 1.1.(4); \WhWaAA, p.~235}), 
hence (2.7) would be established, and hence also
(2.5). 

In order to justify this interchange of limit and
sum, we claim that for $0<q<1$ we have
$$0\le -\frac {q^n\log q} {1-q^n}-\log \frac {1-q^{n+1}} {1-q^n}\le \frac
{1} {n}-\log \frac {n+1} {n}\ .\tag2.8$$
Because of $\frac {1} {n}-\log\frac {n+1} {n}\le \frac {1} {n^2}$
(see \cite{\WhWaAA, p.~235}) this would imply that the sum at the
left-hand side of (2.7) is absolutely convergent, uniformly in $q$.
Therefore we are allowed to interchange limit and sum.

For the proof of (2.8) let us abbreviate
$$h(q):=-\frac {q^n\log q} {1-q^n}-\log \frac {1-q^{n+1}} {1-q^n}.$$
Clearly we have $\lim_{q\downarrow0}h(q)=0$ and
$\lim_{q\uparrow1}h(q)=\frac {1} {n}-\log \frac {n+1} {n}$. Thus (2.8)
would be established if we can show that $h(q)$ is an increasing
function for $0<q<1$. 

Let us compute the derivative of $h(q)$,
$$\frac {d} {dq}h(q)=-\frac {n\,q^{n-1}} {(1-q^n)^2}\(\frac {n+1} {n}\frac
{(1-q)(1-q^n)} {(1-q^{n+1})}+\log q\).$$
In order to show that $h(q)$ is increasing we should have $\frac {d}
{dq}h(q)\ge0$, i.e.
$$\frac {n+1} {n}\frac
{(1-q)(1-q^n)} {(1-q^{n+1})}+\log q\le0,\quad \quad 0<q<1.\tag2.9$$
Again, let us abbreviate
$$g(q):=\frac {n+1} {n}\frac
{(1-q)(1-q^n)} {(1-q^{n+1})}+\log q.$$
The derivation of $g(q)$ is
$$\frac {d} {dq}g(q)=\frac {(1-q)^2q^n}
{n\,q\,(1-q^{n+1})^2}\big(n-(q+q^2+\dots+q^n)\big)
\big((q^{-1}+q^{-2}+\dots+q^{-n})-n\big)\ ,$$
which is greater than 0 for $0<q<1$. Together with
$\lim_{q\uparrow1}g(q)=0$ this implies that $g(q)<0$ for $0<q<1$.
That is, the inequality (2.9) holds, hence also (2.8). This finishes
the proof of (2.5).

\medskip
For the proof of (2.6), we proceed as follows,
$$\align \lim_{q\uparrow1}\bigg(-\log q\sum _{n=0} ^{\infty}&\frac
{q^{n+1}(1-q^{x-1})} {(1-q^{n+1})(1-q^{n+x})}\bigg)\\
&=\lim_{q\uparrow1}\(-\frac {\log q} {1-q}\)\lim_{q\uparrow1} 
\sum _{n=0} ^{\infty}q^{n+1}\frac
{(1-q)(1-q^{x-1})} {(1-q^{n+1})(1-q^{n+x})}\\
&=1\cdot\sum _{n=0} ^{\infty}\frac {x-1} {(n+1)(n+x)}\ .
\endalign$$
The interchange of limit and summation can be justified in a similar
manner as above in the proof of (2.5). We omit the details.

\medskip
Finally we show $\lim_{q\downarrow1}\ps_q(x)=\ps(x)$. If $q>1$ we can
rewrite (2.3) by
$$\align
\ps_q(x)&=(x-\frac {3} {2})\log q-\log(1-\frac {1} {q})+\log \frac
{1} {q}\sum _{n=0} ^{\infty}\frac {q^{-n-x}} {1-q^{-n-x}}\\
&=(x-\frac {3} {2})\log q+\ps_{{1}/{q}}(x).
\endalign$$
From this and the fact that we have already shown 
$\lim_{q\uparrow1}\ps_q(x)=\ps(x)$ we get
$$\align
\lim_{q\downarrow1}\ps_q(x)&=\lim_{q\downarrow1}\big((x-\frac {3}
{2})\log q\big)+\lim_{q\uparrow1}\psi_q(x)\\
&=0+\ps(x)=\ps(x).
\endalign$$
This completes the proof of Theorem~1.\quad \quad \qed
\enddemo

From now on we always assume $q$ to be a complex number with $\v q<1$.

For our purposes, because of notational convenience, it will be
useful to consider a ``logarithmic-free" variant of $\ps_q(x)$,
namely
$$\psq(x):=-(1-q)\sum _{n=0} ^{\infty}\frac {xq^n} {1-xq^n}\
.\tag2.10$$
This definition makes sense for complex $q,x$, $\v q<1$, and
$x\ne1,q^{-1},q^{-2},\dots$.
$\psq(x)$ is related to $\ps(x)$ by
$$\psq(q^x)=-\frac {1-q} {\log q}\ps_q(x)-(1-q)\frac {\log(1-q)}
{\log q}\ .\tag2.11$$
So the limit $\lim_{q\uparrow1}\psq(q^x)$ does not even exist, but at
least, because of (2.11) and Theorem~1, we have
$$\lim_{q\uparrow1}\(\psq(q^x)-\psq(q^y)\)=\ps(x)-\ps(y)\ , \tag2.12$$
which suffices for our purposes.

The next Theorem lists miscellaneous properties of the
$q$-digamma function $\psq(x)$.
These will be of relevance in the subsequent sections when
manipulating $q$-digamma functions and when taking limits of
$q$-digamma functions.
\proclaim{Theorem 2}Let $N$ denote a nonnegative integer. There hold the
following identities.
$$\align  \psq(x)-\psq(y)&=\sum _{n=0} ^{\infty}\frac {q^n(y-x)(1-q)}
{(1-xq^n)(1-yq^n)}\tag2.13a\\
&=\frac {(1-q)(y-x)} {(1-x)(1-y)}\,{}_3\Phi_2\!\[\matrix 
x,y,q;\\qx,qy;\endmatrix q, q\]\ ,\tag2.13b
\endalign$$
\vskip-8pt
$$\gather  \psq(x)-\psq(x\,q^N)=-(1-q)\sum _{n=0} ^{N-1}\frac {xq^n}
{1-xq^n}\ ,\tag2.14\\
\tilde\ps_{q^2}(x)+\tilde\ps_{q^2}(qx)=(1+q)\psq(x)\ ,\tag2.15\\
\tilde \ps_{q^2}(x^2)=\frac {(1+q)} {2}\big(\psq(x)+\psq(-x)\big)\
,\tag2.16\\
\lim_{q\uparrow1}\(\psq(q^x)-\psq(q^y)\)=\ps(x)-\ps(y)\ , \tag2.17\\
\lim_{q\uparrow1}\(\psq(-q^{x})-\psq(-q^{y})\)=0\ ,\tag2.18\\
\lim_{N\to\infty}\(\psq(x/q^N)-\psq(y/q^N)\)=\psq(x)-\psq(y)+\psq(q/y)-
\psq(q/x)\ .\tag2.19\\
\endgather$$
\endproclaim
\demo{Proof}The identities (2.13)--(2.16) follow directly from the
definition (2.10). Identity (2.17) is a restatement of (2.12).

In order to prove (2.18), we use (2.13a) to get
$$\psq(-q^{x})-\psq(-q^{y})=\sum _{n=0} ^{\infty}\frac
{q^n(q^x-q^y)(1-q)} {(1+q^{n+x})(1+q^{n+y})}\ .$$
We may choose $q_0$, $0<q_0<1$, such that 
$\v{(1+q^{n+x})^{-1}(1+q^{n+y})^{-1}}\le 1$ for all $n$ and $q_0<q<1$. 
This implies
$$\align \left\vert \sum _{n=0} ^{\infty}\frac {q^n(q^x-q^y)(1-q)}
{(1+q^{n+x})(1+q^{n+y})}\right\vert&\le 
\sum _{n=0} ^{\infty} {q}^n\v{q^x-q^y}({1-q})\\
&\le \v{q^x-q^y}\ ,\tag2.20
\endalign$$
provided $q_0<q<1$.
For $q\uparrow1$ the right-hand side in (2.20) tends to 0, hence so
does the left-hand side. Thus (2.14) is established.

For the proof of (2.19) we note that by (2.14) we may write
$$\psq(x/q^N)-\psq(x)-\big(\psq(y/q^N)-\psq(y)\big)=
\sum _{n=0} ^{N-1}\frac {q^n(1-q)(q/x-q/y)}
{(1-q^{n+1}/x)(1-q^{n+1}/y)}\ .$$
Therefore, by appealing to (2.13a), we obtain
$$\lim_{N\to\infty}\(\psq(x/q^N)-\psq(x)-\psq(y/q^N)+\psq(y)\)=\psq(q/y)-
\psq(q/x)\ ,$$
which is equivalent with (2.19).\quad \quad \qed
\enddemo

In our identities also the following $q$-analogue of the first
polygamma function (cf\.
\cite{\ErdeAA, 1.16}) will appear,
$$\psq^{(1)}(x):=(1-q)^2\sum _{n=0} ^{\infty}\frac {xq^n}
{(1-xq^n)^2}\ .\tag2.21$$
The following Theorem lists some relevant properties of $\psq^{(1)}(x)$.
\proclaim{Theorem 3}There hold the following identities.
$$ \lim_{q\uparrow1}\psq^{(1)}(q^x)=\ps^{(1)}(x)\ ,\tag2.22$$
where $\ps^{(1)}(x)$ denotes the first polygamma function,
$$\gather \lim_{q\uparrow1}\psq^{(1)}(-q^x)=0\ ,\tag2.23\\
\lim_{N\to\infty}\psq^{(1)}(xq^{-N})=\psq^{(1)}(x)+\psq^{(1)}(q/x)\ ,
\tag2.24\\
\tilde\ps_{q^2}^{(1)}(x)+\tilde\ps_{q^2}^{(1)}(qx)=(1+q)^2\psq^{(1)}(x)\
.\tag2.25
\endgather$$
\endproclaim
\demo{Sketch of Proof} We have
$$\lim_{q\uparrow1}\psq^{(1)}(q^x)=\sum _{n=0} ^{\infty}\frac {1}
{(n+x)^2}\ .$$
The right-hand side in this equation equals $\ps^{(1)}(x)$ (cf\.
\cite{\ErdeAA, 1.16.(9)}). The proof of (2.23) is analogous to that of (2.18) and is
therefore omitted. Identity (2.25) follows directly from the
definition (2.21).

The identity (2.24) is established by the following sequence of
manipulations,
$$\align \lim_{N\to\infty}\psq^{(1)}(xq^{-N})&=\lim_{N\to\infty}
(1-q)^2\(\sum _{n=0} ^{N-1}\frac {xq^{-N+n}} {(1-xq^{-N+n})^2}+
\sum _{n=0} ^{\infty}\frac {xq^n} {(1-xq^n)^2}\)\\
&=\lim_{N\to\infty}(1-q)^2\sum _{n=1} ^{N}\frac {xq^{-n}}
{(1-xq^{-n})^2}+\psq^{(1)}(x)\\
&=\lim_{N\to\infty}(1-q)^2\sum _{n=1} ^{N}\frac {q^n/x}
{(1-q^n/x)^2}\\
&=\psq^{(1)}(x)+\psq^{(1)}(q/x).\quad \quad \qed
\endalign$$

\enddemo

\subhead 3. Identities related to the nonterminating $q$-Pfaff--Saalsch\"utz
sum\endsubhead
In (1.1) we choose $r=3$, $s=2$, $a_1=a$, $a_2=b$, $a_3=c$, $b_1=e$,
$b_2=qabc/e$, $z=q$, and apply the nonterminating extension of the
$q$-Pfaff--Saalsch\"utz sum \cite{\GaRaAA, (II.24)} to the resulting
$_3\Phi_2$ at the right-hand side. This gives
$$\align 
{} _{4} \Phi _{3} \!&\left [ \matrix \let \over / a q, b q, c q, q;\\ \let \over
  / e q, {{a b c {q^2}}\over e}, {q^2};\endmatrix q, {\displaystyle q} \right
  ] \\
&=\frac {1} {q}\frac {(1-e)(1-qabc/e)(1-q)} {(1-a)(1-b)(1-c)}
\bigg({\frac {{(\let \over / {q\over e}, {{b c q}\over e}, {{a c
   q}\over e}, {{a b q}\over e};q)_\infty }} {{(\let \over / {{a q}\over e}, {{b q}\over
   e}, {{c q}\over e}, {{a b c q}\over e};q)_\infty }} }  - 1\\
&\hskip1cm
-  {\frac {{(\let \over / {q\over e}, a, b, c, {{a b c
    {q^2}}\over {{e^2}}};q)_\infty }} {{(\let \over / {e\over q}, {{a q}\over e}, {{b
    q}\over e}, {{c q}\over e}, {{a b c q}\over e};q)_\infty }} }  
   {} _{3} \Phi _{2} \!\left [ \matrix \let \over / {{a q}\over e}, {{b
    q}\over e}, {{c q}\over e};\\ \let \over / {{{q^2}}\over e}, {{a b c
    {q^2}}\over {{e^2}}};\endmatrix q, {\displaystyle q} \right ] 
\bigg).
\endalign$$
Now let $c\to1$. 
The limit $\liml_{c\to 1}\frac {1} {1-c}(\frac {\po\dots\infty}
{\po\dots\infty}-1)$, which has
to be computed at the right-hand side, is obtained by using 
l'H\^ospital's rule, while the limit of the other expressions can be
obtained directly. Thus we obtain
$$\align 
{} _{4} \Phi _{3} \!&\left [ \matrix \let \over / a q, c q, q, q;\\ \let \over /
  b q, {{a c {q^2}}\over b}, {q^2};\endmatrix q, {\displaystyle q} \right ]
\\
&=\frac {1} {q}\frac {(1-b)(1-qac/b)} {(1-a)(1-c)}
\bigg(\psq(q/b)-\psq(cq/b)-\psq(aq/b)+\psq(acq/b)\\
&\hskip1cm
-  {\frac {\left( 1 - q \right)\,{(\let \over / a, c, q,
     {{a c {q^2}}\over {{b^2}}};q)_\infty }} {{(\let \over / {b\over q}, {{a q}\over b},
     {{c q}\over b}, {{a c q}\over b};q)_\infty }} }  
    {} _{3} \Phi _{2} \!\left [ \matrix \let \over / {{a q}\over b}, {{c
     q}\over b}, {q\over b};\\ \let \over / {{{q^2}}\over b}, {{a c {q^2}}\over
     {{b^2}}};\endmatrix q, {\displaystyle q} \right ]\bigg),\tag3.1
\endalign$$
where we replaced $b$ by $c$ and then $e$ by $b$.
Replacing $a$ by $q^a$, $b$ by $q^b$, $c$ by $q^c$ and then 
letting $q\uparrow1$, with the help of (2.17) 
we get the following nonterminating extension of
\cite{\SrivAN, (3.7)--(3.10)}
\NoBlackBoxes
$$\align 
{} &_{4} F _{3} \!\left [ \matrix \let \over / a+1, c+1, 1, 1;\\ \let \over /
  b+1, {{2+a+ c }- b}, {2};\endmatrix  1 \right ]
\\
&\hskip0.2cm=\frac {b(1+a+c-b)} {ac}
\bigg(\ps(1-b)-\ps(1+c-b)-\ps(1+a-b)+\ps(1+a+c-b)\\
&\hskip0.4cm
-  {\Gamma\bmatrix {{ {b- 1}, {{1+a}- b},
     {{1+c}- b}, {{1+a+ c}- b} }} \\
{{ a, c, 
     {{2+a +c {}}- {{2b}}} }} \endbmatrix}  
    {} _{3} F _{2} \!\left [ \matrix \let \over / {{1+a}- b}, {{1+c
     }- b}, {1- b};\\ \let \over / {{{2}}- b}, {{2+a+ c}-
     {{2b}}};\endmatrix  1\right ]\bigg).\\
\tag3.2
\endalign$$

\BlackBoxes
Setting $c=0$ in (3.1) gives a $q$-analogue for \cite{\SrivAN, (3.11)},
$$\align 
{} _{3} \Phi _{2} \!&\left [ \matrix \let \over / a q, q, q;\\ \let \over / b q,
  {q^2};\endmatrix q, {\displaystyle q} \right ] 
=\frac {1} {q}\frac {(1-b)} {(1-a)}
\bigg(\psq(q/b)-\psq(aq/b)\\
&\hskip1cm
-{\frac { \left( 1 - q \right)  \,{(\let \over / a, q;q)_\infty }} {{(\let \over / {b\over q}, {{a q}\over b};q)_\infty }} }  
    {} _{2} \Phi _{1} \!\left [ \matrix \let \over / {{a q}\over b}, {q\over
     b};\\ \let \over / {{{q^2}}\over b};\endmatrix q, {\displaystyle q} \right
     ]  \bigg).\tag 3.3
\endalign$$
In order to obtain a nicer $q$-analogue of \cite{\SrivAN, (3.11)},
in (3.1) set $c=q^{-N}$ and then let $N\to\infty$. By (2.19) this yields
$${}_3\Phi_2\!\[\matrix aq,q,q;\\bq,q^2;\endmatrix q, \frac {b}
{a}\]=\frac {(1-1/b)} {(1-1/a)}\,
\big(\psq(b)-\psq(b/a)\big).\tag3.4$$
Writing this in the form
$$\psq(b)-\psq(b/a)=\frac {\po a\infty} {\po b\infty}\sum _{n=1}
^{\infty}\frac {(1-q)} {(1-q^n)}\frac {\po{bq^n}\infty}
{\po{aq^n}\infty}\(\frac {b} {a}\)^n,\tag3.5$$
we also have a perfect $q$-analogue for \cite{\SrivAN, (3.12)}.

It might be observed that (3.3) could also be obtained by combining
a special
case of a three-term relation for $_3\Phi_2$-series \cite{\GaRaAA,
Appendix, (III.34), $a\to q$, $b\to q/b$, $c\to aq/b$, $d\to q^2/b$,
$e\to aq^2/b$} with (2.13b), while (3.4) could be obtained by
combining the $_3\Phi_2$-transformation \cite{\GaRaAA, Appendix,
(III.9), $a\to q$, $b\to aq$, $c\to q$, $d\to bq$, $e\to q^2$} with
(2.13b).

\subhead 4. Identities related to Bailey's nonterminating Jackson sum\endsubhead
In (1.1) choose $r=8$, $s=7$, $a_1=a$, $a_2=q\sqrt a$,
$a_3=-q\sqrt a$, $a_4=b$, $a_5=c$, $a_6=d$, $a_7=e$,
$a_8=qa^2/bcde$, $b_1=\sqrt a$, $b_2=-\sqrt a$, $b_3=aq/b$,
$b_4=aq/c$, $b_5=aq/d$, $b_6=aq/e$, $b_7=bcde/a$, $z=q$. Then the
resulting $_8\Phi_7$ at the right-hand side can be summed using
Bailey's nonterminating extension of Jackson's sum \cite{\GaRaAA,
(II.25)}. Thus we obtain
$$\align 
{} _{9} \Phi _{8} \!&\left [ \matrix \let \over / a q, {\sqrt{a}} {q^2},
  - {\sqrt{a}} {q^2} , b q, c q, d q, e q, {{{a^2} {q^2}}\over
  {b c d e}}, q;\\ \let \over / {\sqrt{a}} q, - {\sqrt{a}} q  ,
  {{a {q^2}}\over b}, {{a {q^2}}\over c}, {{a {q^2}}\over d}, {{a {q^2}}\over
  e}, {{b c d e q}\over a}, {q^2};\endmatrix q, {\displaystyle q} \right ]
\\
&=\frac {1} {q}\frac {(1-aq/b)(1-aq/c)(1-aq/d)(1-aq/e)(1-bcde/a)(1-q)}
{(1-q^2a)(1-b)(1-c)(1-d)(1-e)(1-qa^2/bcde)}\\
&\hskip1cm
\times\bigg({\frac {{(\let \over / a q, {b\over a}, {{a q}\over {c d}},
   {{a q}\over {c e}}, {{b d e}\over a}, {{a q}\over {d e}}, {{b c e}\over a},
   {{b c d}\over a};q)_\infty }} {{(\let \over / {{a q}\over c}, {{a q}\over d}, {{a
   q}\over e}, {{b c d e}\over a}, {{b c}\over a}, {{b d}\over a}, {{b e}\over
   a}, {{a q}\over {c d e}};q)_\infty }} }  -1
\\
&\hskip1.5cm
-  {\frac {{(\let \over / a q, c, d, e, {{{a^2} q}\over {b c d
    e}}, {b\over a}, {{b q}\over c}, {{b q}\over d}, {{b q}\over e}, {{{b^2} c
    d e}\over {{a^2}}};q)_\infty }} {{(\let \over / {a\over b}, {{a q}\over c}, {{a
    q}\over d}, {{a q}\over e}, {{b c d e}\over a}, {{b c}\over a}, {{b
    d}\over a}, {{b e}\over a}, {{a q}\over {c d e}}, {{{b^2} q}\over
    a};q)_\infty }} }\\
&\hskip2.5cm
  {} _{8} W _{7} ({\displaystyle {{{b^2}}\over
    a}; b, {{b c}\over a}, {{b d}\over a}, {{b e}\over a}, {{a q}\over {c d
    e}}}; q, {\displaystyle q})
\bigg)
\endalign$$
Next we let $e\to1$.
The limit $\liml_{e\to 1}\frac {1} {1-e}(\frac {\po\dots\infty}
{\po\dots\infty}-1)$, which has
to be computed at the right-hand side, is obtained by using 
l'H\^ospital's rule, while the limit of the other expressions can be
obtained directly. This results into
$$\align 
{} _{9} \Phi _{8} \!&\left [ \matrix \let \over /  {\sqrt{a}} {q^2},
  - {\sqrt{a}} {q^2}  , b q, c q, d q, {{{a^2} {q^2}}\over {b c
  d}}, a q, q, q;\\ \let \over / {\sqrt{a}} q, - {\sqrt{a}} q  , {{a
  {q^2}}\over b}, {{a {q^2}}\over c}, {{a {q^2}}\over d}, {{b c d q}\over a},
  {q^2}, a {q^2};\endmatrix q, {\displaystyle q} \right ] 
\\
&=\frac {1} {q}\frac {(1-aq/b)(1-aq/c)(1-aq/d)(1-aq)(1-bcd/a)}
{(1-q^2a)(1-b)(1-c)(1-d)(1-qa^2/bcd)}\\
&\hskip0.5cm
\times\bigg(\psq(aq/d)-\psq(aq)+\psq(aq/c)-\psq(aq/cd)
\\
&\hskip2cm
+\psq(b/a)-\psq(bd/a)+\psq(bcd/a)-\psq(bc/a)
\\
&\hskip1cm
- {\frac { \left( 1 - q \right)\,{(\let \over / c, d, q,
     {{{a^2} q}\over {b c d}}, {{b q}\over c}, {{b q}\over d}, b q, {{{b^2} c
     d}\over {{a^2}}};q)_\infty }} {{(\let \over / {a\over b}, {{a q}\over c}, {{a q}\over
     d}, {{b c d}\over a}, {{b c}\over a}, {{b d}\over a}, {{a q}\over {c d}},
     {{{b^2} q}\over a};q)_\infty }} }  \\
&\hskip2.5cm
    {} _{8} W _{7} ({\displaystyle {{{b^2}}\over a}; b, {{b c}\over a}, {{b
     d}\over a}, {b\over a}, {{a q}\over {c d}}}; q, {\displaystyle q})
\bigg)\ .\tag4.1
\endalign$$
Replacing $a$ by $q^a$, $b$ by $q^b$, $c$ by $q^c$, $d$ by $q^d$, and
letting $q\uparrow1$, it is seen that 
identity (4.1) is a nonterminating $q$-analogue of \cite{\SrivAN,
(1.3)}. In fact, by use of (2.17),
we thus get 
$$\align 
{}& _{8} F _{7} \!\left [ \matrix \let \over / 2+a/2,
 b+1, c +1, d +1, 2+2a- {b- c-
  d}, a+1,  1, 1;\\ \let \over /1+a/2, 2+{a
 - b}, 2+{a- c}, {{2+a }- d}, {{1+b+ c+ d}- a},
 2, 2+a;\endmatrix  1 \right ] 
\\
&\hskip0.2cm=\frac {(1+a-b)(1+a-c)(1+a-d)(1+a)(b+c+d-a)}
{(2+a)\,b\,c\,d\,(1+2a-b-c-d)}\\
&\hskip0.4cm
\times\bigg(\psi(1+a-d)-\psi(1+a)+\psi(1+a-c)-\psi(1+a-c-d)
\\
&\hskip1.5cm
+\psi(b-a)-\psi(b+d-a)+\psi(b+c+d-a)-\psi(b+c-a)
\\
&\hskip0.6cm
- {\Gamma\bigg[\matrix  {{\let \over / {a- b}, {{1+a}- c}, {{1+a}-
     d}, {{b+ c+ d}- a}, {{b +c}- a}, {{b+ d}- a} }},{{1+a}- {c- d}} \\
{ {\let \over / c, d, 
     {{{1+2a}}- {b- c- d}}, {{1+b}- c}, {{1+b}- d}, 1+b }}\endmatrix}  \\
&\hskip1cm
\matrix      {{{1+2b}}- a}\\ {{{2b}+ c
     +d}- {{2a}}}\endmatrix\bigg]
    {} _{7} V _{6} ({\displaystyle {{{2b}}- a}; b, {{b+ c}- a}, {{b+
     d}- a}, {b- a}, {{1+a}- {c- d}}}; 1)
\bigg),\\
\tag4.2
\endalign$$
a nonterminating extension of \cite{\SrivAN, (1.3)}.

Next we find $q$-analogues for \cite{\SrivAN,
(3.1)--(3.6)}. First in (4.1) set $d=q^{-N}$ and then let $N\to\infty$.
Observing that with $d=q^{-N}$ the last term in (4.1) vanishes, by
means of (2.19) we get
$$\align
{}_7\Phi_6&\!\[\matrix aq,\sqrt{a}q^2, -\sqrt aq^2, bq,cq,q,q;\\
\sqrt aq,-\sqrt aq,aq^2/b,aq^2/c,aq^2,q^2;\endmatrix q, \frac {qa}
{bc}\]\\
&=\frac {bc} {qa}\frac {(1-aq/b)(1-aq/c)(1-aq)}
{(1-q^2a)(1-b)(1-c)}\big(\psq(aq/c)-\psq(aq)+\psq(aq/b)
-\psq(aq/bc)\big).
\tag4.3
\endalign$$
This is a $q$-analogue of \cite{\SrivAN, (3.2)}, of which
\cite{\SrivAN, (3.1)} is a terminating special case. Letting $c\to
\infty$ in (4.3), we obtain a $q$-analogue of \cite{\SrivAN, (3.3)},
$$
\align
{}_6\Phi_6\!&\[\matrix aq,\sqrt{a}q^2, -\sqrt aq^2, bq,q,q;\\
\sqrt aq,-\sqrt aq,aq^2/b,aq^2,q^2,0;\endmatrix q, \frac {q^2a}
{b}\]\\
&\hskip3cm=\frac {b} {qa}\frac {(1-aq/b)(1-aq)}
{(1-q^2a)(1-b)}\big(\psq(aq)-\psq(aq/b)\big).
\tag4.4
\endalign
$$

For $c=\sqrt a$, (4.3) reduces to a $q$-analogue of \cite{\SrivAN,
(3.4)},
$$\multline
{}_5\Phi_4\!\[\matrix aq, -\sqrt aq^2, bq,q,q;\\
-\sqrt aq,aq^2/b,aq^2,q^2;\endmatrix q, \frac {q\sqrt a}
{b}\]\\
=\frac {b} {q\sqrt a}\frac {(1-aq/b)(1-aq)(1+\sqrt a)}
{(1-b)(1-a)(1+q\sqrt a)}\big(\psq(\sqrt aq)-\psq(aq)+\psq(aq/b)
-\psq(q\sqrt a/b)\big).
\endmultline
\tag4.5$$
Similarly, setting $b=\sqrt a$ in (4.5) yields
$${}_4\Phi_4\!\[\matrix aq, -\sqrt aq^2, q,q;\\
-\sqrt aq,aq^2,q^2,0;\endmatrix q,  {q^2\sqrt a}
\]
=\frac {1} {q\sqrt a}\frac {(1-aq)(1+\sqrt a)}
{(1-a)(1+q\sqrt a)}\big(\psq(aq)-\psq(q\sqrt a)\big),
\tag4.6$$
a $q$-analogue for \cite{\SrivAN, (3.5)}. A $q$-analogue for
\cite{\SrivAN, (3.6)} could be obtained by starting with the
duplication formula for the $q$-gamma function \cite{\GaRaAA,
(1.10.9)}.

As in the previous section, it should be observed that some
identities could have been obtained in an alternative way. For
instance, (4.3) follows when applying the nonterminating Watson
transformation \cite{\GaRaAA, Appendix, (III.36)} to 
$$_8W_7(aq^2;cq,q,bq,q,aq;q,\frac {aq} {bc})$$
and, with the help of (2.13b), writing the resulting $_3\Phi_2$'s in
terms of $q$-digamma functions. Also we remark that all the series in
this sections are very well-poised series with cancelling leading
parameter. Thus the $_9\Phi_8$ in (4.1) actually is the very
well-poised series
$$_{10}W_9(aq^2;aq,bq,cq,dq,a^2q^2/bcd;q,q).$$

\subhead 5. Identities related to Verma and Jain's $q$-analogue of
Whipple's $_3F_2$-sum\endsubhead
In this section we rely on Verma and Jain's 
summation \cite{\VeJaAH, (1.6)} for a very well-poised $_8\Phi_7$,
$$\multline 
{}_8\Phi_7\!\[\matrix
-c,i\,q\,\sqrt{c},-i\,q\,\sqrt{c},a,q/a,c,-d,-q/d;\\
i\,\sqrt{c},-i\,\sqrt{c},-c\,q/a,-a\,c,-q,c\,q/d,c\,d;\endmatrix q, c\]\\
=\frac {(-c,-c\,q;q)_\infty
(a\,c\,d,a\,c\,q/d,c\,d\,q/a,c\,q^2/a\,d;q^2)_\infty}
{(c\,d,c\,q/d,-a\,c,-c\,q/a;q)_\infty}.
\endmultline\tag5.1$$
We also use two summations contiguous to (5.1), namely
$$\multline
{} _{8} \Phi _{7} \! \left [ \matrix \let \over / -c, i\,q\,{\sqrt{c}},
   -i\,q\,\sqrt{c}, a, {1\over a}, c, -d, -{q\over d};\\ \let \over /
   i\,\sqrt{c}, -i\,\sqrt{c}, -{{c\,q}\over a}, - a\,c\,q   ,
   -q, {{c\,q}\over d}, c\,d;\endmatrix q, {\displaystyle c\,q} \right ] 
\\= 
  \frac {1} {1+a}{{{ {(\let \over / - c\,q   , - c\,q    ;q) _\infty
    ({\let \over / a\,c\,d, {{a\,c\,q}\over d}, {{c\,d\,q}\over a},
        {{c\,{q^2}}\over {a\,d}}}; {q^2}) _{\infty}}
        \over {(\let \over / c\,d, {{c\,q}\over d}, - a\,c\,q   ,
        -{{c\,q}\over a} ;q) _\infty} }\,
        }} 
\\+ 
   \frac {a} {1+a}{{{ {(\let \over / - c\,q   , - c\,q    ;q)
        _\infty  
     ({\let \over / a\,c\,d\,q, {{a\,c\,{q^2}}\over d}, {{c\,d}\over a},
        {{c\,q}\over {a\,d}}}; {q^2}) _{\infty}} 
     \over {(\let \over / c\,d, {{c\,q}\over d}, - a\,c\,q
          , -{{c\,q}\over a} ;q) _\infty} }\,
        }}
\endmultline\tag5.2$$
and
$$\align 
{} _{8}& \Phi _{7} \! \left [ \matrix \let \over / - c\,q   ,
   i\,\sqrt{c}\,{q^{{3\over 2}}}, -i\,\sqrt{c}\,{q^{{3\over 2}}}, a\,q,
   {q\over a}, c, - d\,q   , -{{{q^2}}\over d};\\ \let \over /
   i\,{\sqrt{c\,q}}, -i\,{\sqrt{c\,q}}, -{{c\,q}\over a}, - a\,c\,q
     , -{q^2}, {{c\,q}\over d}, c\,d;\endmatrix q, {\displaystyle
   {c\over q}} \right ] 
\\&= {{a\,d\,\left( 1 + {c\over {a\,q}} \right) \,\left( 1 + q \right) 
\over {c\,\left( 1 - a \right) \,\left( 1 - {c\over q} \right) \,
     \left( 1 + a \right) \,\left( 1 + d \right) \,
      \left( 1 + {d\over q} \right) } }} 
\\&\hskip2cm  {{ { {(\let \over / -c, - c\,{q^2}    ;q) _\infty
     ({\let \over / {{a\,c\,d}\over q}, {{a\,c}\over d}, {{c\,d}\over a},
        {{c\,q}\over {a\,d}}}; {q^2}) _{\infty} } \over
        {(\let \over / c\,d, {{c\,q}\over d}, - a\,c\,q   ,
        -{{c\,q}\over a} ;q) _\infty} }\,
       }
     }
\\&\quad -{{a\,d\,\left( 1 + {{a\,c}\over q} \right)\,\left( 1 + q \right)} \over
    {c\,\left( 1 - a \right) \,\left( 1 - {c\over q} \right) \,
        \left( 1 + a \right) \,   \left( 1 + d \right) \,
        \left( 1 + {d\over q} \right) }}
\\&\hskip2cm\quad   {{  { {(\let \over / -c, - c\,{q^2}   
         ;q) _\infty
     }({\let \over / a\,c\,d, {{a\,c\,q}\over d}, {{c\,d}\over {a\,q}},
         {c\over {a\,d}}}; {q^2}) _{\infty} 
      \over {(\let \over / c\,d, {{c\,q}\over d}, -
         a\,c\,q   , -{{c\,q}\over a} ;q) _\infty} }\,
         }
      }.\tag5.3
\endalign$$
Identity (5.2) is a $q$-analogue of \cite{\LavoAB, (3)}. It follows
from the contiguous relation
$$\multline
{} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x, y, \Ai;\\ \let \over /
   {u\over x}, {u\over y}, \Bi;\endmatrix q, {\displaystyle z} \right ] = 
    {{q\,x\,y\,\left( 1 - x \right)  \left( 1 - {u\over {x q}} \right)  
       }\over 
     {u\,\left( x - y \right)  \left( 1 - {{x y q}\over u} \right) }}
\,\,{} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x q, y, \Ai;\\ \let
        \over / {u\over {x q}}, {u\over y}, \Bi;\endmatrix q, {\displaystyle
        {z\over q}} \right ]\\
  -{{q\,x\,y \left( 1 - y \right)   \left( 1 - {u\over {y q}} \right)  
        }\over 
      {u\,\left( x - y \right)  \left( 1 - {{x y q}\over u} \right) }}
\,\,{} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x, y q, \Ai;\\ \let
         \over / {u\over x}, {u\over {y q}}, \Bi;\endmatrix q, {\displaystyle
         {z\over q}} \right ]\ .
\endmultline\tag5.4$$
In fact, by setting $r=7$, $x=a$, $y=1/a$, $u=-c$, and $\Ai$, $\Bi$,
$z$ such
that the left-hand side of (5.4) becomes the $_8\Phi_7$ in (5.2), at
the right-hand side of (5.4) we obtain two $_8\Phi_7$'s which can be
evaluated by means of (5.1). 

Setting $c=1$ in (5.2) and replacing $a$
by $\sqrt {ABq}$ and $d$ by $\sqrt{Aq/B}$ leads to a three-product
relation,
$$\multline 
   ({\let \over / A\,q}; {q^2}) _{\infty} \,
    ({\let \over / B\,q}; {q^2}) _{\infty} \,
    ({\let \over / {{{q}}\over A}}; {q^2}) _{\infty} \,
    ({\let \over / {{{q}}\over B}}; {q^2}) _{\infty} \\
\hskip1cm+  {{{\sqrt{A\,B\,q}}\,({\let \over / {1\over A}}; {q^2}) _{\infty} \,
       ({\let \over / A\,q^2}; {q^2}) _{\infty} \,
       ({\let \over / {1\over B}}; {q^2}) _{\infty} \,
       ({\let \over / B\,q^2}; {q^2}) _{\infty} }} \\=
{{({\let \over / -{{{\sqrt{A\,B\,q}}} }}; q) _{\infty} \,
      ({\let \over / -{ {{\sqrt{q/AB}}}}}; q) _{\infty}
       \,({\let \over / {{{\sqrt{A\,q}}}\over {{\sqrt{B}}}}}; q) _{\infty} \,
      ({\let \over / {{{\sqrt{B\,q}}}\over {{\sqrt{A}}}}}; q) _{\infty} }\over
      {{{({\let \over / -q}; q) _{\infty} }^2}}} \ ,
\endmultline\tag5.5$$
which will be used later. Written in the form
$$\multline 
(Aq,q/A,Bq,q/B,-1,-q^2,-q,-q;q^2)_\infty\\
\hskip3cm+\sqrt{ABq}(Aq^2,1/A,Bq^2,1/B,-1,-q^2,-q,-q;q^2)_\infty\\
=2\cdot(-q^{\frac {1} {2}}\sqrt{AB}, -q^{\frac {3} {2}}/\sqrt{AB},
-q^{\frac {3} {2}}\sqrt{AB}, -q^{\frac {1} {2}}/\sqrt{AB},\\
q^{\frac {1} {2}}\sqrt{A/B},q^{\frac {3} {2}}\sqrt{B/A},
q^{\frac {3} {2}}\sqrt{A/B},q^{\frac {1} {2}}\sqrt{B/A};q^2)_\infty,
\endmultline \tag5.6$$
it turns out to be
a special case of a relation connecting four infinite products
due to Slater \cite{\SlatAD, (3); \GaRaAA, Ex.~5.22}. In fact,
replacing $q$ by $q^2$, $a$ by $\sqrt{ABq^3}$, $b$ by $\sqrt{ABq}$,
$c$ by $-q$, $d$ by $-q^2$, $e$ by $Aq^2$, $f$ by $Bq^2$, $g$ by 1,
and $h$ by $q$ in Slater's formula gives (5.6) after little
manipulation.


Identity (5.3) is a $q$-analogue of \cite{\LavoAB, (4)}. It is
derived from the contiguous relation
$$\multline
{} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x, y, {u\over q},
\Ai;\\ \let
   \over / {u\over x}, {u\over y}, {u\over {{q^2}}}, \Bi;\endmatrix q,
   {\displaystyle z} \right ] \\= 
    {{    \left( 1 - {u\over {x {q^2}}} \right)  
       \left( 1 - {u\over {y {q^2}}} \right)  
       \left( 1 - {u\over {x q}} \right)  \left( 1 - {u\over {y q}} \right)  
         }\over 
     {\left( 1 - {u\over {{q^2}}} \right)  
       \left( -{x\over q} + {y\over q} \right)  
       \left( 1 - {u\over {x y q}} \right)  z 
        }}
{{\prodl_{i = 1}^{r-3} (1 - {{\bi}\over q} )}\over
{\prodl_{i = 1}^{r-2} (1 - {{\ai}\over q} )}}\\
\times{} _{r} \Phi _{r-1} \! \left [ \matrix \let \over / {x\over
        q}, y, \Aiq;\\ \let \over / {u\over {x q}}, {u\over {y {q^2}}},
        \Biq;\endmatrix q, {\displaystyle  z} \right]\\
\hskip1cm  -{{   \left( 1 - {u\over {x {q^2}}} \right)  
        \left( 1 - {u\over {y {q^2}}} \right)  
        \left( 1 - {u\over {x q}} \right)  \left( 1 - {u\over {y q}} \right)  
            }\over 
      {\left( 1 - {u\over {{q^2}}} \right)  
        \left( -{x\over q} + {y\over q} \right)  
        \left( 1 - {u\over {x y q}} \right)  z 
         }} {{\prodl_{i = 1}^{r-3} (1 - {{\bi}\over q}
)}\over
{\prodl_{i = 1}^{r-2} (1 - {{\ai}\over q} )}}\\
\times{} _{r} \Phi _{r-1} \! \left [ \matrix \let \over / x,
         {y\over q}, \Aiq;\\ \let \over / {u\over {x {q^2}}}, {u\over {y q}},
         \Biq;\endmatrix q, {\displaystyle  z} \right ].
\endmultline\tag5.7$$
In order to see this, set $x=aq$, $y=q/a$, $u=-cq^2$,
$\Ai=(-c,i\sqrt{c}q^{3\over 2},-i\sqrt{c}q^{3\over 2},
c,-dq,\mathbreak -q^2/d)$, $\Bi=(i\sqrt{cq},-i\sqrt{cq},-q^2,cq/d,cd)$ in (5.7).
Thus, since the parameter $-c$ is cancelling,
at the left-hand side we just obtain the $_8\Phi_7$ of (5.3). At
the right-hand side we obtain two $_8\Phi_7$'s which can be summed by
means of (5.1).

As first application of our method in this section, 
in (1.1) choose $a_1=-c$,
$a_2=i\,\sqrt c\,q$, $a_3=-i\,\sqrt c \,q$, $a_4=a$, $a_5=q/a$,
$a_6=c$, $a_7=-d$, $a_8=-q/d$, $b_1=i\,\sqrt{c}$, $b_2=-i\,\sqrt{c}$,
$b_3=-c\,q/a$, $b_4=-a\,c$, $b_5=-q$, $b_6=c\,q/d$, $b_7=c\,d$, $z=c$. Then
the resulting $_8\Phi_7$ at the right-hand side can be summed using
(5.1). Letting $a\to1$ with the help of l'H\^ospital's Rule we
obtain the identity
$$\align
{} _{7} &\Phi _{6} \! \left [ \matrix \let \over /
   i\,{\sqrt{c}}\,{q^2}, -i\,{\sqrt{c}}\,{q^2}, q,  c\,q,
   - d\,q   , -{{{q^2}}\over d}, q;\\ \let \over /
   i\,{\sqrt{c}}\,q, -i\,{\sqrt{c}}\,q, -{{c\,{q^2}}}
     , -{q^2}, {{c\,{q^2}}\over d}, c\,d\,q;\endmatrix q,
   {\displaystyle c} \right ] \\
&\hskip1cm=  {{\left( 1 - c\,d \right) \,\left( 1 - {{c\,q}\over d} \right)
  \,\left( 1 + c \right) \,\left( 1 + q \right) \,
      \left( 1 + c\,q \right)  }
\over {c\,\left( 1 - c \right)\,\left( 1 - q \right) \,\left( 1 + d \right)  \,
      \left( 1 + {q\over d} \right) \,\left( 1 + c\,{q^2} \right) }}\\
&\hskip2cm{{ \left( \dfrac {c(1-q)} {(1+c)} + 
        {{\tilde\psi_{q^2} ({\tsize c\,d\,q})-\tilde\psi_{q^2} ({\tsize c\,d}) 
     +\tilde\psi_{q^2} ({\tsize {{c\,{q^2}}\over d}})- \tilde\psi_{q^2} ({\tsize {{c\,q}\over d}}) 
             }
       \over {1 + q}} \right) }.
    }\tag5.8
\endalign$$
Replacing $c$ by $q^c$, $d$ by $q^d$ and letting $q\uparrow1$, by
(2.17) we see that this is a $q$-analogue of \cite{\LavoAA, (1)}. 
On the other hand, by replacing $c$ by
$-q^c$, $d$ by $-q^d$ and letting $q\uparrow1$, by (2.17) we arrive at the
$_5F_4$-evaluation
$$\multline
{} _{5} F _{4} \!\left [ \matrix {\let \over / 2 + {c\over 2}, 1, 1 + d, 2 -
   d, 1};\\ {\let \over / 1 + {c\over 2}, 2 + c, 2 + c - d, 1 + c +
   d};\endmatrix   {\displaystyle -1}\right ]  
= {{c \left( 1 + c \right)  \left( 1 + c - d \right)  \left( c + d \right)  
\over  {\left( 2 + c \right)  \left( 1 - d \right)  d}}}\\
\times \left( {\tsize {1\over c}} + \tfrac {1} {2}\left({{\psi (\tsize{{1 + c - d}\over 2}) -
             \psi (\tsize{{2 + c - d}\over 2}) + \psi (\tsize{{c + d}\over 2}) -
             \psi (\tsize{{1 + c + d}\over 2})}} \right)\right) .
\endmultline\tag5.9$$
The series in (5.8) actually is the very well-poised series
$$_8W_7(-c\,q^2;q,c\,q,-d\,q,-q^2/d,q;q,c),$$
where the parameter $-cq^2$ is cancelling. Applying the
transformation \cite{\GaRaAA, Appendix, (III.24)} to this very
well-poised series we find 
$$\multline
{} _{8} \Phi _{7} \! \left [ \matrix \let \over / - {c^2} q   , i
  c {q^{{3\over 2}}}, -i c {q^{{3\over 2}}}, -q, {{c q}\over d}, c d, -
  c q   , c;\\ \let \over / i c {\sqrt{q}}, -i c {\sqrt{q}}, {c^2} q,
  - c d q   , -{{c {q^2}}\over d}, c q, - c {q^2}  
  ;\endmatrix q, {\displaystyle q} \right ]\\
=  {{\left( 1 + c \right) \, \left( 1 + c q \right)  }\over
  {c\, \left( 1 - q \right)\,\left( 1 + d \right)    
      \left( 1 + {q\over d} \right) }}
    { {(\let \over / c d, {{c q}\over d},-q,  - {c^2} {q^2}   
       ;q) _\infty} \over {(\let \over / q, {c^2} q, - c d q   ,
       -{{c {q^2}}\over d} ;q) _\infty} } \\
      \left( {{c \left( 1 - q \right) }\over {1 + c}} + 
        {{ \tilde\psi_ {{q^2}} ( {\textstyle c d q}) -
          \tilde\psi_ {{q^2}} ( {\textstyle c d}) +
           \tilde\psi_ {{q^2}} ( {\textstyle {{c {q^2}}\over d}})- 
            \tilde\psi_ {{q^2}} ( {\textstyle {{c q}\over d}}) 
           }\over {1 + q}} 
        \right) .
\endmultline\tag5.10$$
This identity is a $q$-analogue of an identity of Watson
\cite{\WatsAA; \LavoAA, (6)}.


Next in (1.1) choose $a_1=-c$, $a_2= i\,q\,{\sqrt{c}}$, $a_3=
-i\,q\,\sqrt{c}$, $a_4= a$, $a_5= {1\over a}$, $a_6= c$, $a_7= -d$,
$a_8= -{q\over d}$, $b_1=i\,\sqrt{c}$, $b_2= -i\,\sqrt{c}$, $b_3= -{{c\,q}\over a}$,
$b_4= - a\,c\,q   $, $b_5=-q$, $b_6= {{c\,q}\over d}$, $b_7= c\,d$,
$z=cq$.
After applying (5.2) at the right-hand side this yields
$$\align 
{} _{9} \Phi _{8}& \! \left [ \matrix \let \over /  - c\,q   ,
   i\,{\sqrt{c}}\,{q^2}, -i\,{\sqrt{c}}\,{q^2}, {q\over a}, a\,q, c\,q,
   - d\,q   , -{{{q^2}}\over d},q;\\ \let \over / 
   i\,{\sqrt{c}}\,q, -i\,{\sqrt{c}}\,q, - a\,c\,{q^2}   ,
   -{{c\,{q^2}}\over a}, -{q^2}, {{c\,{q^2}}\over d}, c\,d\,q,{q^2};\endmatrix q,
   {\displaystyle c\,q} \right ]  \\
&=  c\,{{\left( 1 - c\,d \right) \,\left( 1-d/ cq \right) \,
     \left( 1 - q \right) \,\left( 1 + q \right) \,
      \left(1+ a/ c q \right) \,
      \left( 1 + a\,c\,q \right)\over
   {{{\left( 1 - a \right) }^2}\,\left( 1 - c \right) \,
      \left( 1 + d \right) \,\left( 1+d/ q \right) \,
      \left( 1 + c\,{q^2} \right) }}}\\
& \quad   \left( \frac {a} {1+a} {{{ {(\let \over / - c\,q   , - c\,q
                ;q) _\infty  
      ({\let \over / {{c\,d}\over a}, {{c\,q}\over {a\,d}}, a\,c\,d\,q,
             {{a\,c\,{q^2}}\over d}}; {q^2}) _{\infty}} 
       \over {(\let \over / c\,d, -{{c\,q}\over
             a}, - a\,c\,q   , {{c\,q}\over d} ;q) _\infty} }\,
             }} 
\right.\\
  &\hskip1cm\left. + \frac {1} {1+a}{{{ {(\let \over / - c\,q   , - c\,q    ;q)
             _\infty  
      ({\let \over / a\,c\,d, {{a\,c\,q}\over d}, {{c\,d\,q}\over a},
             {{c\,{q^2}}\over {a\,d}}}; {q^2}) _{\infty}} 
         \over {(\let \over / c\,d, -{{c\,q}\over a}, -
             a\,c\,q   , {{c\,q}\over d} ;q) _\infty} }\,
             }} -1\right). \tag5.11
\endalign$$
Replacing $d$ by $cq/d$ and 
letting $c\to1$, we get a $q$-analogue of \cite{\LavoAA, (2)}, which
after some manipulation can be brought into the form
$$\align 
{} _{9} \Phi _{8}& \! \left [ \matrix \let \over /  i\,{q^2}, -i\,{q^2},
   {q\over a}, a\,q, q, -{{{q^2}}\over d}, - d\,q ,-q,q  ;\\ \let \over
   /  i\,q, -i\,q, - a\,{q^2}   , -{{{q^2}}\over a}, -{q^2},
   d\,q, {{{q^2}}\over d},{q^2};\endmatrix q, {\displaystyle q} \right ] 
\\&=  {{\left( 1 - {d} \right) \,\left( 1 - {d\over q} \right)
      \,\left( 1 + {q} \right) \,
      \left( 1 + {a\over q} \right) \,\left( 1 + a\,q \right) 
       \over
    {{{\left( 1 - a \right) }^2}\,\left( 1 + {d} \right) \,
      \left( 1 + {d\over q} \right) \,\left( 1 + {q^2} \right) }}}\\
&\quad  \bigg( 2\,\psq ({\tsize {q\over d}}) + \psq ({\tsize -{q\over a}}) + 
              \psq ({\tsize - a\,q   }) -2\,\psq ({\tsize -q}) -
  2 { {(\let \over / -q, -q^2;q) _\infty} \over {(\let \over /
             d,{q\over d}, -{q\over a}, - a  ;q) _\infty} } \\
&\hskip1cm
      \Big(({\let \over / {{a\,q}\over d}, a\,d, {{{q^2}}\over {a\,d}},
             {{d\,q}\over a}}; {q^2}) _{\infty}
              {{\big( \tilde\psi_{q^2} ({\tsize {{a\,q}\over d}}) + 
                    \tilde\psi_{q^2} ({\tsize {{{q^2}}\over {a\,d}}}) \big) }
                }  \\
&\hskip1.5cm  + 
            a\,({\let \over / {q\over {a\,d}}, {d\over a}, {{a\,{q^2}}\over d},
             a\,d\,q}; {q^2}) _{\infty} 
      {{\big( \tilde\psi_{q^2} ({\tsize {q\over {a\,d}}}) +
                    \tilde\psi_{q^2} ({\tsize {{a\,{q^2}}\over d}}) \big) }
                } \Big)  \bigg) .\tag5.12
\endalign$$
In the computations which finally lead to (5.11)
we also used the three-product relation (5.5).

An interesting special case of (5.12) arises when setting $d=aq^{-2N}$
and then letting $N\to\infty$. By this procedure the infinite products at
the right-hand side approach to a definite limit while the
$\tilde\psi_{q^2}$'s vanish. Hence what remains is
$$\multline
{} _{7} \Phi _{6} \! \left [ \matrix \let \over / i\,{q^2}, -i\,{q^2}, {q\over
   a}, a\,q, q, -q, q;\\ \let \over / i\,q, -i\,q, - a\,{q^2}   ,
   -{{{q^2}}\over a}, -{q^2}, {q^2};\endmatrix q, {\displaystyle -q} \right ] 
\\   = {{\left( 1 + {a\over q} \right) \,\left( 1 + q \right) \,
      \left( 1 + a\,q \right)}\over
    {{{\left( 1 - a \right) }^2}\,\left( 1 + {q^2} \right) } }
   {{      \left(   \psq ( {\textstyle -{q\over a}}) + 
        \psq ( {\textstyle - a\,q   }) 
      -2\,\psq ( {\textstyle -q}) \right) }
    }\ ,
\endmultline\tag5.13$$
an identity which has no reasonable counterpart for $q\uparrow1$.


If in (5.11) we let $a\to1$, by using 
l'H\^ospital's Rule twice and employing (2.25)
we obtain 
$$\align 
{} _{9} \Phi _{8} &\! \left [ \matrix \let \over /  i
   {\sqrt{c}} {q^2}, -i {\sqrt{c}} {q^2}, q, q, c q, - d q   ,
   -{{{q^2}}\over d},-c\,q,q;\\ \let \over / i {\sqrt{c}} q, -i {\sqrt{c}} q,
   - c {q^2}   , - c {q^2}   , -{q^2}, {{c {q^2}}\over
   d}, c d q,{q^2};\endmatrix q, {\displaystyle c q} \right ] \\
&= {{\left( 1 - c d \right)  \left( 1 - {d\over {c q}} \right)  
      \left( 1 + q \right)  {{\left( 1 + c q \right) }^2} 
     \over
   {2\, q \left( 1 - c \right)  (1-q)\left( 1 + d \right)  
      \left( 1 + {d\over q} \right) 
      \left( 1 + c {q^2} \right) }}}\\
& \quad    \bigg(      \frac {1} {{\left( 1 + q \right) }^2} 
        {{\left( \tilde\psi_{q^2} ({\tsize c d}) + \tilde\psi_{q^2} ({\tsize {{c q}\over d}}) - 
                 \tilde\psi_{q^2} ({\tsize c d q}) - \tilde\psi_{q^2} ({\tsize {{c {q^2}}\over d}}) 
                 \right) }^2}- \psq^{(1)} (c d) -
            \psq^{(1)} ({\tsize{{c q}\over d}}) \\
&   \hskip1cm       -{{{{\left( 1 - q \right)} \over {1 + q}}
     \left( \tilde\psi_{q^2} ({\tsize c d}) + \tilde\psi_{q^2} ({\tsize {{c q}\over d}}) - 
       \tilde\psi_{q^2} ({\tsize c d q}) - \tilde\psi_{q^2} ({\tsize {{c {q^2}}\over d}})\right
               ) }} +
        2 \psq^{(1)} (- c q   )\bigg) .\\
\tag5.14
\endalign$$
This is a $q$-analogue of \cite{\LavoAA,
(3)}, which is seen by replacing $c$ by $q^c$, $d$ by $q^d$, and
letting $q\uparrow1$. For performing the limit we have to take into
account (2.17), (2.22), and (2.23).

On the other hand, 
replacing $c$ by $-q^c$ and $d$ by $-q^d$ in (5.14) gives
$$\multline
{} _{7} F _{6} \!\left [ \matrix {  2 + {c\over 2}, 1, 1, 1 + d, 2 -
   d},1+c,1;\\ { 1 + {c\over 2}, 2 + c, 2 + c, 2 + c - d, 1 + c +
d,2};\endmatrix  
   {\displaystyle -1}\right ] \\
 = {{{{\left( 1 + c \right) }^2} \left( d - c -1 \right)  
      \left( c + d \right)}\over
    {2\,d\, \left( 2 + c \right)  \left(d -1  \right)  }}
    \bigg( \tfrac {1} {4}{{{{\left( \psi (\tsize{{1 + c - d}\over 2}) - 
                \psi (\tsize{{2 + c - d}\over 2}) + \psi (\tsize{{c + d}\over 2}) - 
                \psi (\tsize{{1 + c + d}\over 2}) \right) }^2}}} \\
 +       2 \psi^{(1)} (1 + c) - \psi^{(1)} (1 + c - d) - \psi^{(1)} (c + d)
        \bigg) .
\endmultline\tag5.15$$
An interesting special case of (5.14) is obtained by setting $d=q$,
$$\multline
{} _{7} \Phi _{6} \! \left [ \matrix \let \over /  i
   {\sqrt{c}} {q^2}, -i {\sqrt{c}} {q^2},-q, -c\,q, q, q, q;\\ \let \over /  i
   {\sqrt{c}} q, -i {\sqrt{c}} q, c {q^2}, q^2, -c {q^2} ,  -c {q^2}
   ;\endmatrix q, {\displaystyle c q} \right ] \\= 
  {{\left( 1 - c q \right)  {{\left( 1 + c q \right) }^2}} \over
     {2\, c \left( 1 - q \right)  q \left( 1 + c {q^2} \right) } }
   {{\big(\psq^{(1)} (c q) - 
        \psq^{(1)} (- c q ) \big)
             }}.
\endmultline\tag5.16$$
This identity is a $q$-analogue of \cite{\LavoAA, second identity on
p.~216}. To obtain another identity from (5.14), we replace $c$ by
$-c$, set $d=-q^{-N}$ and let $N\to \infty$. By using (2.19) and (2.24)
we find that 
$$\multline
{} _{7} \Phi _{6} \! \left [ \matrix \let \over / {\sqrt{c}}\,{q^2},
   -{\sqrt{c}}\,{q^2}, q, q, -c\,q, c\,q   , q;\\ \let \over /
   {\sqrt{c}}\,q, -{\sqrt{c}}\,q, c\,{q^2}   , 
   c\,{q^2}   , -{q^2}, {q^2};\endmatrix q, {\displaystyle -q} \right ] 
   \\= {{\left( 1 + q \right) \,{{\left( 1 -c\,q \right) }^2}
       }\over
    {2\,q\,\left( 1 +c \right) \,\left( 1 - q \right) \,
      \left( 1 -c\,{q^2} \right) }}
           \bigg(   {{{{\left( \tilde\psi_{q^2} ( {\textstyle c}) + 
                \tilde\psi_{q^2} ( {\textstyle {q\over c}}) - 
                \tilde\psi_{q^2} ( {\textstyle c\,q   }) - 
                \tilde\psi_{q^2} ( {\textstyle {{{q^2}}\over c}}) \right) }^2}}
           \over {{{\left( 1 + q \right) }^2}}} \\
    +    \tilde\psi^{(1)}_ {q} ( {\textstyle c\,q }) -
     \tilde\psi^{(1)}_ {q} ( {\textstyle { 1\over c}    })
-{{\left( 1 - q \right)}\over{1 + q}}
        {{ \left( \tilde\psi_{q^2} ( {\textstyle c}) + 
               \tilde\psi_{q^2} ( {\textstyle {q\over c}}) - 
               \tilde\psi_{q^2} ( {\textstyle c\,q   }) - 
               \tilde\psi_{q^2} ( {\textstyle {{{q^2}}\over c}}) \right) }
           }    \bigg) .
\endmultline\tag5.17$$
Replacing $c$ by $q^c$ and letting $q\uparrow1$ gives an evaluation
for a $_5F_4$,
$$\multline
{} _{5} F _{4} \!\left [ \matrix { 2 + {c\over 2},  1, 1, 1 + c,1};\\ { 1 +
   {c\over 2}, 2 + c, 2 + c, 2};\endmatrix   {\displaystyle -1}\right ]
\\ = 
{{ {{\left(  c+1 \right) }^2}}\over{2\,\left(  c+2 \right) }}  
        \bigg( \tfrac {1} {4}{{{{\left( \psi ({ \textstyle {{1 - c}\over 2}}) - 
                  \psi ({ \textstyle 1 - {c\over 2}}) + 
                  \psi ({ \textstyle {c\over 2}}) - 
                  \psi ({ \textstyle {{1 + c}\over 2}}) \right) }^2}}} 
          + \psi^{(1)} ({ \textstyle 1 + c})  - \psi^{(1)} ({ \textstyle -c}) 
           \bigg) \ .\\
\quad 
\endmultline\tag5.18$$



Finally, in (1.1) choose $a_1=- c\,q   $, $a_2= 
   i\,\sqrt{c}\,{q^{{3\over 2}}}$, $a_3=  -i\,\sqrt{c}\,{q^{{3\over 2}}}$,
$a_4=  a\,q$, $a_5= 
   {q\over a}$, $a_6=  c$, $a_7=  - d\,q   $, $a_8=  -{{{q^2}}\over
d}$, $b_1=i\,{\sqrt{c\,q}}$, $b_2=  -i\,{\sqrt{c\,q}}$, $b_3=  -{{c\,q}\over a}$,
$b_4=  - a\,c\,q
     $, $b_5=  -{q^2}$, $b_6=  {{c\,q}\over d}$, $b_7=  c\,d$, $z=c/q$. Then
the resulting $_8\Phi_7$ at the right-hand side can be summed using
(5.3). Upon letting $a\to q$ we obtain the following $q$-analogue of
\cite{\LavoAB, third identity on p.~272}
$$\align 
{} _{9}& \Phi _{8} \! \left [ \matrix \let \over / 
   i\,{\sqrt{c}}\,{q^{{5\over 2}}}, -i\,{\sqrt{c}}\,{q^{{5\over 2}}},
    c\,q, - d\,{q^2}   , -{{{q^3}}\over
   d},- c\,{q^2}   ,{q^3}, {{{q}}}, q;\\ 
    \let \over / i\,{\sqrt{c}}\,{q^{{3\over 2}}},
   -i\,{\sqrt{c}}\,{q^{{3\over 2}}}
     , -{q^3}, {{c\,{q^2}}\over d}, c\,d\,q, {q^2}
  , -{{c\,{q}}}, - c\,{q^3};\endmatrix q,
   {\displaystyle {c\over q}} \right ] 
\\&={{q\,\left( 1 - c d \right)   \left( 1 - {d\over {c q}} \right) 
      \left( 1 + { c} \right)   
         \left( 1 + {q^2} \right)  \left( 1 + c {q^2} \right) }\over 
   { (1-q)\left( 1 - c \right)  \left( 1 + {d\over {{q^2}}} \right)  
     \left( 1 + d q \right)  \left( 1 + {q} \right)  
     \left( 1 + c {q^3} \right) }}\\
&\quad \bigg(-{{\left( 1 + {c\over {{q^2}}} \right)  
      \left( (1 - d) + \frac {d} {q}{{\left( 1 - c \right)  }} + 
       \frac {d^2} {q} {{\left( 1 - {c\over d} \right)  }} \right) }\over 
    { \left( 1 - {c\over q} \right) \left( 1 + d \right)   
      \left( 1 + {d\over q} \right)  }} + {{1-q}\over {1 + q}} - 
  {{c\,(1-q)}\over {q\left( 1 + c \right)  }} -
      {{c\,(1-q)}\over {1 + c q}}
   \\
&\hskip1.5cm- \frac {1} {{  q \left( 1 + q \right) }} 
\left( {{2 \,d\,q\,\left( 1 - {c\over d} \right)   \left( 1 - {{c d}\over q} \right)
                   \left( 1 + {c\over {{q^2}}} \right)  
                  }\over 
                 {c \left( 1 - {c\over q} \right) \left( 1 - q \right)
     \left( 1 + d \right)      \left( 1 + {d\over q} \right)   }} 
      -1    \right)\\
&\hskip3cm  (\tilde\psi_{q^2} ({\tsize c\,d})-\tilde\psi_{q^2} ({\tsize c\,d\,q})+\tilde\psi_{q^2} ({\tsize c\,q/d})
   -\tilde\psi_{q^2} ({\tsize c\,q^2/d}))
   \bigg).\tag5.19
\endalign$$
Replacing $c$ by $-q^c$, $d$ by $-q^d$ and letting $q\uparrow1$ leads to
an evaluation for a $_7F_6$,
$$\align 
{} _{7} F _{6}& \!\left [ \matrix {  {5\over 2} + {c\over 2},  2 +
   d, 3 - d,2+c,3, 1, 1};\\ { {3\over 2} + {c\over 2},  2 + c - d, 1 + c +
   d, 2,1 + c, 3 + c};\endmatrix   {\displaystyle -1}\right ]  \\
&= {{ c \left( 2 + c \right)  \left(  d-c-1 \right)  
        \left( c + d \right) }\over
   {2 \left( 3 + c \right)  \left(  d-2 \right)  
      \left( 1 + d \right) } }
   \bigg( {1\over c} + {1\over {1 + c}} +
          {{\left( c-2 \right)  
              \left(2d-c-1\right) }\over 
            {\left( d-1 \right)  d}} \\
&\hskip0.5cm- 
          \left(  {{\left( c-2 \right)  \left( c - d \right)  
                    \left(  c + d-1 \right) }\over 
                  {2\left( d -1\right)  d}} -1\right)  \\
&\hskip1cm \times
              \left( \psi (\tsize{{1 + c - d}\over 2}) - 
                \psi (\tsize{{2 + c - d}\over 2}) + \psi (\tsize{{c + d}\over 2}) - 
                \psi (\tsize{{1 + c + d}\over 2}) \right)  \bigg)  .
\tag5.20
\endalign$$

As before in (5.14), it is of particular interest to set $d=q$ in
(5.19). With this choice of $d$, (5.19) eventually reduces to
$$\multline
{} _{7} \Phi _{6} \! \left [ \matrix \let \over /  i
   {\sqrt{c}} {q^{{5\over 2}}}, -i {\sqrt{c}} {q^{{5\over 2}}}, {q^3}, q,
   -{q^2}, - c {q^2}   ,q;\\ \let \over / i {\sqrt{c}} {q^{{3\over 2}}}, -i {\sqrt{c}}
   {q^{{3\over 2}}}, - c q   , - c {q^3}   , c {q^2},
   {q^2};\endmatrix q, {\displaystyle {c\over q}} \right ] \\= 
  {{\left( 1 - c q \right)  \left( 1 - 2 c - {c^2} + q + 2 c q - {c^2} q 
        \right)  \left( 1 + c {q^2} \right) }\over 
    {\left( 1 - c \right)  \left( 1 - {c\over q} \right)  
      \left( 1 + q \right)  \left( 1 + c q \right)  
      \left( 1 + c {q^3} \right) }}.
\endmultline\tag5.21$$
Replacing $c$ by $q^c$ and $-q^c$, and then letting
$q$ tend to $1^-$ in (5.21), produces the identities
$$
{} _{3} F _{2} \!\left [ \matrix { 3, 1, 1};\\ { 2 + c, 2};\endmatrix  
   {\displaystyle 1}\right ]  = 
  {{\left( 1 + c \right)  \left( 2 c -1\right) }\over 
    {2 \left(  c-1 \right)  c}}\tag5.22
$$
and
$$
{} _{5} F _{4} \!\left [ \matrix { 2 + c, {5\over 2} + {c\over 2}, 3, 1,
1};\\
   { {3\over 2} + {c\over 2}, 1 + c, 3 + c, 2};\endmatrix   {\displaystyle
   -1}\right ]  = {{\left( 2 + c \right)  \left( 1 + 2 c \right) }\over 
    {2 \left( 1 + c \right)  \left( 3 + c \right) }}\ ,\tag5.23
$$
respectively.

\medskip
The last remark at the end of section~4 also applies here. All the
series appearing in (5.8)--(5.23) with the exception of (5.22) 
are very well-poised series with cancelling leading parameter.




\subhead 6. Identities related to Verma and Jain's $q$-analogue of
Watson's $_3F_2$-sum \endsubhead
In this section we rely on Verma and Jain's identity \cite{\VeJaAH, (1.5)}
$$\align 
{} _{8}& \Phi _{7} \! \left [ \matrix \let \over / -{{{\sqrt{a\,b}}\,c}\over
   {{\sqrt{q}}}}, i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}, -i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}},  a, b, 
   \\ \let \over / {{i\,{{\left( a\,b \right)
   }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}, {{-i\,{{\left( a\,b
   \right) }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}},
    -{{c\,{\sqrt{b\,q}}}\over
   {{\sqrt{a}}}},  -{{c\,{\sqrt{a\,q}}}\over {{\sqrt{b}}}}, \endmatrix\right.\\
&\hskip4cm   \left.\matrix  \let \over /  c,-c, -{{{\sqrt{a\,b\,q}}}\over
   {c}} ;\\   \let \over /
   -{{{\sqrt{a\,b\,q}}}},{\sqrt{a\,b\,q}}, {c^2}
    ;\endmatrix q, {\displaystyle {{c\,{\sqrt{q}}}\over
   {{\sqrt{a\,b}}}}} \right ] \\
&=   { {(\let \over / - c\,{\sqrt{a\,b\,q}}   ,
     -{{c\,{\sqrt{q}}}\over {{\sqrt{a\,b}}}} ;q) _\infty} \over {(\let \over /
     -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}}}\over
     {{\sqrt{b}}}} ;q) _\infty} }\,
    { {(\let \over / a\,q, b\,q, {{{c^2}\,q}\over a}, {{{c^2}\,q}\over b}
     ;{q^2}) _\infty} \over {(\let \over / q, a\,b\,q, {c^2}\,q,
     {{{c^2}\,q}\over {a\,b}} ;{q^2}) _\infty} }\ .\tag6.1
\endalign$$
We start by recording two identities contiguous to (6.1),
$$\align 
{} _{8}& \Phi _{7} \! \left [ \matrix \let \over / -{{{\sqrt{a\,b}}\,c}\over
   {{\sqrt{q}}}}, i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}, -i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}},  a, b, 
   \\ \let \over / {{i\,{{\left( a\,b \right)
   }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}, {{-i\,{{\left( a\,b
   \right) }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}},
    -{{c\,{\sqrt{b\,q}}}\over
   {{\sqrt{a}}}},  -{{c\,{\sqrt{a\,q}}}\over {{\sqrt{b}}}}, \endmatrix\right.\\
&\hskip4cm   \left.\matrix  \let \over / c\,q,-c, -{{{\sqrt{a\,b}}}\over
   {c{\sqrt{q}}}} ;\\   \let \over / 
   -{{{\sqrt{a\,b}}}\over {{\sqrt{q}}}},{\sqrt{a\,b\,q}},{c^2}\,q
    ;\endmatrix q, {\displaystyle {{c\,{\sqrt{q}}}\over
   {{\sqrt{a\,b}}}}} \right ] \\
&= {{\left( 1 + {{{c^2}\,{\sqrt{q}}}\over {{\sqrt{a\,b}}}} \right)  \over
    {\left( 1 + {{{\sqrt{a\,b}}}\over {{\sqrt{q}}}} \right) }}}
  {{       { {(\let \over / -c\,{\sqrt{a\,b\,q}}  ,
        -{c\sqrt {q}\over {{\sqrt{a\,b}}}} ;q) _\infty} \over {(\let \over /
        -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}}}\over
        {{\sqrt{b}}}} ;q) _\infty} }\,
       { {(\let \over / a, b, {{{c^2}\,{q^2}}\over a}, {{{c^2}\,{q^2}}\over b}
        ;{q^2}) _\infty} \over {(\let \over / q, a\,b\,q, {c^2}\,q,
        {{{c^2}\,q}\over {a\,b}} ;{q^2}) _\infty} }}
     }
\\&\quad  + 
   { {(\let \over / - c\,{\sqrt{a\,b\,q}}   ,
     -{{c\,{\sqrt{q}}}\over {{\sqrt{a\,b}}}} ;q) _\infty} \over {(\let \over /
     -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}}}\over
     {{\sqrt{b}}}} ;q) _\infty} }\,
    { {(\let \over / a\,q, b\,q, {{{c^2}\,q}\over a}, {{{c^2}\,q}\over b}
     ;{q^2}) _\infty} \over {(\let \over / q, a\,b\,q, {c^2}\,q,
     {{{c^2}\,q}\over {a\,b}} ;{q^2}) _\infty} }\tag6.2
\endalign$$
and
$$\align 
{} _{8} &\Phi _{7} \! \left [ \matrix \let \over / -{{{\sqrt{a\,b}}\,c}\over
   {{\sqrt{q}}}}, i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}, -i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}, a, b, 
   \\ \let \over / {{i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}, {{-i\,{{\left( a\,b \right)
   }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}},
   -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}}}\over
   {{\sqrt{b}}}}, \endmatrix 
   \right.\\
&\hskip4cm \left.\matrix  \let \over / c, -c, -{{{\sqrt{a\,b}}}\over
   {c\,{\sqrt{q}}}};\\  \let \over / 
   -{\sqrt{a\,b\,q}}, {\sqrt{a\,b\,q}}, {c^2}\,q;\endmatrix
   q,  {\displaystyle {{c\,{q^{{3\over 2}}}}\over {{\sqrt{a\,b}}}}} \right ] 
\\&= 
   { {(\let \over / - c\,{\sqrt{a\,b\,q}}   ,
     -{{c\,{\sqrt{q}}}\over {{\sqrt{a\,b}}}} ;q) _\infty} \over {(\let \over /
     -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}}}\over
     {{\sqrt{b}}}} ;q) _\infty} }\,
    { {(\let \over / a\,q, b\,q, {{{c^2}\,q}\over a}, {{{c^2}\,q}\over b}
     ;{q^2}) _\infty} \over {(\let \over / q, a\,b\,q, {c^2}\,q,
     {{{c^2}\,q}\over {a\,b}} ;{q^2}) _\infty} }
\\&\quad - {{c\sqrt q \over
    {\sqrt {a\,b}}}}
  {{ { {(\let \over / - c\,{\sqrt{a\,b\,q}}   ,
         -{c\sqrt q\over {{\sqrt{a\,b}}}} ;q) _\infty} \over {(\let \over /
         -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}}}\over
         {{\sqrt{b}}}} ;q) _\infty} }\,
        { {(\let \over / a, b, {{{c^2}\,{q^2}}\over a}, {{{c^2}\,{q^2}}\over
         b} ;{q^2}) _\infty} \over {(\let \over / q, a\,b\,q, {c^2}\,q,
         {{{c^2}\,q}\over {a\,b}} ;{q^2}) _\infty} }}
      }\ . \tag6.3
\endalign$$
Identity (6.2) is a $q$-analogue for \cite{\LavoAB, (1)}. It is 
proved by using the contiguous relation 
\NoBlackBoxes
$$\multline
{} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x, y, \Ai;\\ \let \over /
   {u\over x}, {u\over y}, \Bi;\endmatrix q, {\displaystyle z} \right ] = 
  {} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x q, {y\over q},
\Ai;\\
    \let \over / {u\over {x q}}, {{u q}\over y}, \Bi;\endmatrix q,
    {\displaystyle z} \right ] 
\\+    {{ \left( 1 - u \right)  
       \left( 1 - {u\over {x y}} \right)  \left( -x + {y\over q} \right)  z 
       }\over 
     {\left( 1 - {u\over x} \right)  \left( 1 - {u\over y} \right)  
       \left( 1 - {u\over {x q}} \right)  \left( 1 - {{u q}\over y} \right)  
        }}
{{\prodl_{i = 1}^{r-1} (1 - \ai )}\over
{\prodl_{i = 1}^{r-2} (1 - \bi )}}
{} _{r+2} \Phi _{r+1} \! \left [ \matrix \let \over / x q, y, u q,
        \qAi;\\ \let \over / {{u q}\over x}, {{u {q^2}}\over y}, u,
        \qBi;\endmatrix q, {\displaystyle  z} \right ]\\
\quad 
\endmultline\tag6.4$$
with $r=7$, $x=-\sqrt{a\,b}/c\,\sqrt q$, $y=c\,q$,
$u=-c\,\sqrt{a\,b\,q}$, and $\Ai$, $\Bi$ such that the left-hand side
becomes the $_8\Phi_7$ in (6.2). Subsequently, the resulting
$_8\Phi_7$'s at the right-hand side can be evaluated by (6.1). The
contiguous relation (6.4) is equivalent with (5.7).


\BlackBoxes
Likewise, (6.3) is a $q$-analogue of \cite{\LavoAB, (2)}. It follows
from the contiguous relation
$$\multline
{} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x, \Ai;\\ \let \over / y,
   \Bi;\endmatrix q, {\displaystyle z} \right ] = 
  {} _{r+1} \Phi _{r} \! \left [ \matrix \let \over / x q, \Ai;\\ \let \over /
    {y\over q}, \Bi;\endmatrix q, {\displaystyle {z\over q}} \right ]\\
-    {{ \left( 1 - x y \right)  z 
       }\over 
     {\left( 1 - y \right)  \left( 1 - {y\over q} \right)  q 
        }}
{{\prodl_{i = 1}^{r} (1 - \ai )}\over
{\prodl_{i = 1}^{r-1} (1 - \bi )}}
{} _{r+2} \Phi _{r+1} \! \left [ \matrix \let \over / x q, x y q,
        \qAi;\\ \let \over / y q, x y, \qBi;\endmatrix q, {\displaystyle 
   z\over q} \right ]
\endmultline\tag6.5
$$
when setting $r=7$, $x=-\sqrt{a\,b}/c\,\sqrt q$, $y=c^2\,q$, and the
remaining parameters such that the left-hand side becomes the
$_8\Phi_7$ in (6.3). Once again the resulting $_8\Phi_7$'s at the right-hand
side can be evaluated by (6.1).

{\let\over/ Now in (1.1) choose $a_1=-{{{\sqrt{a\,b}}\,c}\over
   {{\sqrt{q}}}}$, $a_2=  i\,{{\left( a\,b \right) }^{{1\over
4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}$, $a_3=\mathbreak
  -i\,{{\left( a\,b \right) }^{{1\over
4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}$, $a_4=   a$, $a_5=  b$, $a_6=  c$,
$a_7= -c$, 
$a_8=  -{{{\sqrt{a\,b\,q}}}\over   {c}} $, $b_1={{i\,{{\left( a\,b \right)
   }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}$, $b_2=  {{-i\,{{\left( a\,b
   \right) }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}$,
$b_3= 
    -{{c\,{\sqrt{b\,q}}}\over
   {{\sqrt{a}}}}$, $b_4=   -{{c\,{\sqrt{a\,q}}}\over {{\sqrt{b}}}}$,
$b_5=   -{{{\sqrt{a\,b\,q}}}}$, $b_6= {\sqrt{a\,b\,q}}$, $b_7= 
{c^2}$, $z=c\,\sqrt q/\sqrt{a\,b}$. The resulting $_8\Phi_7$ at the
right-hand side can be summed by  (6.1). In the limit $b\to1$ we get}
$$\align
{} _{9}& \Phi _{8} \! \left [ \matrix \let \over / 
      i\,{a^{{1\over 4}}}\,{\sqrt{c}}\,{q^{{7\over 4}}},
   -i\,{a^{{1\over 4}}}\,{\sqrt{c}}\,{q^{{7\over 4}}}, q, a\,q,c\,q,
  \\ \let \over /
   i\,{a^{{1\over 4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}, -i\,{a^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}, - {\sqrt{a}}\,c\,{q^{{3\over 2}}}
    , -{{c\,{q^{{3\over 2}}}}\over {{\sqrt{a}}}},
   -{\sqrt{a}}\,{q^{{3\over 2}}}  ,\endmatrix\right.\\
&\left.\matrix  \let \over /  -
   c\,q , -{{{\sqrt{a}}\,{q^{{3\over 2}}}}\over c},
  -c\,{\sqrt{a\,q}}, q;\\ \let \over / 
   {\sqrt{a}}\,{q^{{3\over 2}}},
   {c^2}\,q, {q^2};\endmatrix q, {\displaystyle {{c\,{\sqrt{q}}}\over
   {{\sqrt{a}}}}} \right ] =
 {{\left( 1 + 
        {{c\,{\sqrt{q}}}\over {{\sqrt{a}}}} \right) \,
      \left( 1 - a\,q \right) \,\left( 1 + c\,{\sqrt{a\,q}} \right) }
\over {q\,\left( 1 - a \right) \,
      \left( 1 + {\sqrt{a}}\,c\,{q^{{3\over 2}}} \right) \,
      \left( 1 + {c\over {{\sqrt{a\,q}}}} \right) }}\\
&{{ \left(  \psq ({\tsize -{{c\,{\sqrt{q}}}\over {{\sqrt{a}}}}})-
        \psq ({\tsize - c\,{\sqrt{a\,q}}  }) + 
        {\tilde\psi_{q^2} ({\tsize a\,q})-{\tilde\psi_{q^2} ({\tsize q}) +
            \tilde\psi_{q^2} ({\tsize {c^2}\,q}) - \tilde\psi_{q^2} ({\tsize {{{c^2}\,q}\over a}})}
      \over {1 + q}}  \right).
      }}\tag6.6
\endalign$$
This identity is a $q$-analogue for \cite{\LavoAB, second identity on
p.~272} and for 
$$\multline
{} _{7} F _{6} \!\left [ \matrix {  {7\over 4} +
   {a\over 4} + {c\over 2}, 1,  1 + a, 1 + c, {3\over 2} + {a\over 2} - c},
  {1\over 2} + {a\over 2} + c,1;\\
   { {3\over 4} + {a\over 4} + {c\over 2}, {3\over 2} + {a\over 2} + c,
   {3\over 2} - {a\over 2} + c, {3\over 2} + {a\over 2}, 1 + 2 c, 2};\endmatrix
     {\displaystyle -1}\right ] \\ 
= 
{{\left( 1 + a \right)  \left( a - 2 c-1 \right)  
      \left( 1 + a + 2 c \right)  }\over
  {a \left( 1 + a - 2 c \right)  \left( 3 + a + 2 c \right) } }  
      \bigg( \psi ({ \textstyle {1\over 2} - {a\over 2} + c})-
 \psi ({ \textstyle {1\over 2} + {a\over 2} + c})    \\
\hskip1cm+ 
       \tfrac {1} {2}\big( \psi ({ \textstyle {{1 + a}\over 2}}) - 
          {{\psi ({ \textstyle {1\over 2}}) +
                       \psi ({ \textstyle {{1 + 2 c}\over 2}}) -
            \psi ({ \textstyle {{1 - a + 2 c}\over 2}})}}\big)\bigg),
\endmultline\tag6.7
$$
which is seen by replacing $a$ by $q^a$, $c$ by $-q^c$,
and letting $q\uparrow1$ in (6.6). 
A nice limiting case results from (6.6) by
letting $c\to0$,
$$
{} _{3} \Phi _{3} \! \left [ \matrix \let \over / q, a q, q;\\ \let \over /
   - {\sqrt{a}} {q^{{3\over 2}}}   , {\sqrt{a}} {q^{{3\over 2}}},
   {q^2};\endmatrix q, {\displaystyle -{q^2}} \right ] = 
  {\left( 1 - a q \right) \over {q\,\left( 1 - a \right) \left( 1 + q \right) }}
  {{ \left( \tilde\psi_{q^2} ( {\textstyle a q})-
  \tilde\psi_{q^2} ( {\textstyle q})    \right)\ . }}\tag6.8
$$
Replacing $a$ by $q^a$ and letting $q\uparrow1$ gives
$$
{} _{3} F _{2} \!\left [ \matrix { 1, 1, 1 + a};\\ { {3\over 2} + {a\over 2},
   2};\endmatrix   {\displaystyle {1\over 2}}\right ]  = 
  {{\left( 1 + a \right)}\over{2 \,a}}
  {{\left(   \psi ({ \textstyle {{1 + a}\over 2}}) -
    \psi ({ \textstyle {1\over 2}}) \right)\ . }}\tag6.9
$$



{\let\over/ Next in (1.1) choose $a_1=-{{{\sqrt{a\,b}}\,c}\over
   {{\sqrt{q}}}}$, $a_2=  i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}$, $a_3=\mathbreak
  -i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}$, $a_4=   a$, $a_5=  b$, $a_6= 
-c$, $a_7=  c\,q$, $a_8=  -{{{\sqrt{a\,b}}}\over
   {c{\sqrt{q}}}} $, $b_1={{i\,{{\left( a\,b \right)
   }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}$, $b_2=  {{-i\,{{\left( a\,b
   \right) }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}$,
$b_3= 
    -{{c\,{\sqrt{b\,q}}}\over
   {{\sqrt{a}}}}$, $b_4=  -{{c\,{\sqrt{a\,q}}}\over {{\sqrt{b}}}}$,
$b_5= 
   {\sqrt{a\,b\,q}}$, $b_6= -{{{\sqrt{a\,b}}}\over {{\sqrt{q}}}}$,
$b_7= {c^2}\,q$, $z=c\sqrt q/\sqrt{a\,b}$. 
The resulting $_8\Phi_7$ at the right-hand side can
be summed by (6.2).}
Letting $b\to1$ leads to a $q$-analogue of \cite{\LavoAB, identity at
the bottom of p.~271},
$$\align 
{} _{9}& \Phi _{8} \! \left [ \matrix \let \over / 
 i\,{a^{{1\over 4}}}\,{\sqrt{c}}\,{q^{{5\over 4}}},
   -i\,{a^{{1\over 4}}}\,{\sqrt{c}}\,{q^{{5\over 4}}}, c\,{q},
   -{{{\sqrt{a}\,q^{{3\over 2}}}}\over c}, a\,q, q, - c   ,
   -c\,{\sqrt{a}}/\sqrt{q}, q;\\ \let \over /
   i\,{(a\,q)^{{1\over 4}}}\,{\sqrt{c}}, -i\,{(a\,q)^{{1\over
   4}}}\,{\sqrt{c}}, -{\sqrt{a\,q}}, {c^2},
   -{{c\,{q^{{1\over 2}}}}\over {{\sqrt{a}}}}, -
   {\sqrt{a}}\,c\,{q^{{1\over 2}}}   , {\sqrt{a}}\,{q^{{3\over 2}}},
   {q^2};\endmatrix\right.\\
& q, {\displaystyle {{c}\over {{\sqrt{a\,q}}}}}
   \Big ] = {{ \left( 1 - q \right) \,
      \left( 1 + {{{c^2}}\over {{\sqrt{a}\,q^{{3\over 2}}}}} \right) \,
      \left( 1 + c \right) \,
       \left( 1 + {c\,{\sqrt{a}}\over{\sqrt{q}}} \right) }\over
   {\left( 1 +{ c\over q} \right) \,\left( 1 + c{\sqrt{a\,q}} \right
         ) \,\left( 1 + {\sqrt{a\,q}} \right) }}
{{    { {(\let \over / a\,{q^2}, {c^2}\,{q^2}, {{{c^2}}\over a}, {q^2}
        ;{q^2}) _\infty} \over {(\let \over / q, a\,{q^3}, {c^2}\,{q},
        {{{c^2}}\over {a\,q}} ;{q^2}) _\infty} }}
     }\\
&\quad  + {{\left( 1 - {{c^2}\over q} \right) \,\left( 1 - {\sqrt{a\,q}} \right) \,
    \left( 1 + {{{\sqrt{a}}}\over {{\sqrt{q}}}} \right) \,
              \left( 1 + {c\,{\sqrt{a}}\over{\sqrt{q}}} \right) }\over
   {\left( 1 - a \right) \, \left( 1 - c \right) \,
      \left( 1 + {c\over q} \right) \,
             \left( 1 + c{\sqrt{a\,q}} \right) }}
    \Big(  \tilde\psi_ {q} ( {\textstyle -{{c}\over
             {{\sqrt{a\,q}}}}}) - \tilde\psi_ {q} ( {\textstyle -
             {c\,{\sqrt{a}}\over{\sqrt{q}}}   })  \\
&  \quad \quad + \frac { \tilde\psi_ {{q^2}} ( {\textstyle a\,q}) 
     -\tilde\psi_ {{q^2}} ( {\textstyle q})       + 
         \tilde\psi_ {{q^2}} ( {\textstyle {{c^2}\over q}}) - 
         \tilde\psi_ {{q^2}} ( {\textstyle {{{c^2}}\over {a\,q}}})}
     {1+q}     \Big) 
     .\tag6.10
\endalign$$
Replacing $a$ by $q^a$, $c$ by $-q^c$, in the limit $q\uparrow1$ we
obtain another $_7F_6$-evaluation,
$$\align
{} _{7} F _{6}& \!\left [ \matrix { {5\over 4} + {a\over 4} + {c\over 2},
   {3\over 2} + {a\over 2} - c, 1 + a,  1, c, -{1\over 2} + {a\over 2} +
   c},1;\\ { {1\over 4} + {a\over 4} + {c\over 2}, 2 c, {1\over 2} - {a\over 2} +
   c, {1\over 2} + {a\over 2} + c, {3\over 2} + {a\over 2}, 2};\endmatrix  
   {\displaystyle -1}\right ] \\
& = {{c \left( 2 c+a-1 \right)  }\over
     {2 \left(  c -1\right)  \left(  2c+a+1 \right) }}
  {{  \Gamma \left [ \matrix {1\over 2}, {{ a+3}\over 2}, {{ 2 c+1}\over
        2}, {{2 c-a-1}\over 2}\\  {{a+2}\over 2}, {{2 c+2}\over 2},
        {{2 c-a}\over 2}\endmatrix \right ]}} \\
&\quad -
   {{\left( 1 + a \right)  \left(  2 c-1 \right)  
       \left( 2 c+a-1 \right)  }\over
    {2\, a\, \left( c-1 \right)  \left(  2 c+a+1 \right) }}  
       \Big(  \psi ({ \textstyle c + {a\over 2} -{1\over 2}})- 
          \psi ({ \textstyle c- {a\over 2}-{1\over 2} }) \\
&\hskip2cm+\tfrac {1} {2}
        \big( \psi ({ \textstyle {1\over 2}}) -
         \psi ({ \textstyle {{a+1}\over 2}})  - 
         \psi ({ \textstyle {{ 2 c-1}\over 2}}) + 
         \psi ({ \textstyle {{ 2 c-a-1}\over 2}})\big) \Big) \ .
\tag6.11
\endalign$$
Also letting $c\to0$ in (6.10) is of interest. We thus obtain an
identity related to (6.8),
$$\multline
{} _{3} \Phi _{3} \! \left [ \matrix \let \over / a\,q, q, q;\\ \let \over /
   -{\sqrt{a\,q}}, {\sqrt{a}}\,{q^{{3\over 2}}}, {q^2};\endmatrix q,
   {\displaystyle -q} \right ] \\= 
{{\left( 1 - q \right) }\over{(1 + {\sqrt{a\,q}}) }}  
   {{{ {(\let \over / a\,{q^2}, {q^2} ;{q^2}) _\infty}
        \over {(\let \over / q, a\,{q^3} ;{q^2}) _\infty} }}}
 +{{\left( 1 - {\sqrt{a\,q}} \right)\,
   \left( 1 + {{{\sqrt{a}}}\over {{\sqrt{q}}}} \right)   }\over
  {\left( 1 - a \right) \,\left( 1 + q \right) }} 
  {{       \left(       \tilde\psi_{q^2} ( {\textstyle a\,q}) 
     -\tilde\psi_{q^2} ( {\textstyle q})   \right) }
     }
\endmultline\tag6.12$$


{\let\over/ Finally, if in (1.1) we choose $a_1=-{{{\sqrt{a\,b}}\,c}\over
   {{\sqrt{q}}}}$, $a_2=  i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}$, $a_3=  -i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}\,{q^{{3\over 4}}}$, $a_4=  a$, $a_5=  b$, $a_6=  c$,
$a_7=  -c$, $a_8=  -{{{\sqrt{a\,b}}}\over
   {c\,{\sqrt{q}}}}$, $b_1={{i\,{{\left( a\,b \right) }^{{1\over
   4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}$, $b_2=  {{-i\,{{\left( a\,b \right)
   }^{{1\over 4}}}\,{\sqrt{c}}}\over {{q^{{1\over 4}}}}}$, $b_3= 
   -{{c\,{\sqrt{b\,q}}}\over {{\sqrt{a}}}}$, $b_4=  -{{c\,{\sqrt{a\,q}}}\over
   {{\sqrt{b}}}}$, $b_5= -{\sqrt{a\,b\,q}}$, $b_6=  {\sqrt{a\,b\,q}}$,
$b_7=  {c^2}\,q$, $z=c\,q^{3\over 2}/\sqrt{a\,b}$, by applying (6.3)
at the right-hand side we obtain }
$$\align 
{} _{9} &\Phi _{8} \! \left [ \matrix \let \over /  i {(a\,b)^{{1\over 4}}} 
   {\sqrt{c}} {q^{{7\over 4}}}, -i {(a\,b)^{{1\over 4}}} 
   {\sqrt{c}} {q^{{7\over 4}}}, a q, b q,\\ \let \over / i
   {(a\,b)^{{1\over 4}}}  {\sqrt{c}} {q^{{3\over 4}}}, -i
   {(a\,b)^{{1\over 4}}}  {\sqrt{c}} {q^{{3\over 4}}},
   -{{{\sqrt{b}} c {q^{{3\over 2}}}}\over {{\sqrt{a}}}}, -{{{\sqrt{a}} c
   {q^{{3\over 2}}}}\over {{\sqrt{b}}}}, 
  \endmatrix \right.\\
&\hskip2cm\left.\matrix   \let \over / c q, - c q   ,
   -{{{\sqrt{a\,b\,q}}}\over c},-c\,{\sqrt{a\,b\,q}}  , q;\\  \let \over / 
   - {\sqrt{a\,b}} {q^{{3\over 2}}} 
  , {\sqrt{a\,b}} {q^{{3\over 2}}}, {c^2}
   {q^2}, {q^2};\endmatrix q, {\displaystyle {{c {q^{{3\over 2}}}}\over
   {{\sqrt{a\,b}} }}} \right ] \\
&= {{\left( 1 - q \right)  \left( 1 - a b q \right)  
      \left( 1 - {c^2} q \right)
    \left( 1 + {{c\,{\sqrt{a\,q}}  }\over {{\sqrt{b}}}} \right)  
      \left( 1 + {{c\,{\sqrt{b\,q}}  }\over {{\sqrt{a}}}} \right)  
        }\over
   {q\,\left( 1 - a \right)  \left( 1 - b \right)  \left( 1 - c \right)  
      \left( 1 + c \right)  \left( 1 + 
        {{c {\sqrt{q}}}\over {{\sqrt{a\,b}} }} \right) 
      \left( 1 + {\sqrt{a\,b}}\,  c \,{q^{{3\over 2}}} \right) }}\\
&\quad \quad         \bigg(  - {{c\sqrt q }\over
    {\sqrt{a\,b}}}
    {{  { {(\let \over / - c\, {\sqrt{a\,b\,q}}  
               , -{{c\,\sqrt q}\over {{\sqrt{a\,b}}}} ;q)
             _\infty} \over {(\let \over / -{{c\,{\sqrt{b\,q}}  }\over
             {{\sqrt{a}}}}, -{{c\,{\sqrt{a\,q}} }\over {{\sqrt{b}}}} ;q)
             _\infty} } { {(\let \over / a, b, {{{c^2} {q^2}}\over a}, {{{c^2}
             {q^2}}\over b} ;{q^2}) _\infty} \over {(\let \over / q, a b q,
             {c^2} q, {{{c^2} q}\over {a b}} ;{q^2}) _\infty} }}
          }\\
&\hskip2cm + 
        { {(\let \over / - c\,{\sqrt{a\,b\,q}}     ,
          -{{c {\sqrt{q}}}\over {{\sqrt{a\,b}} }} ;q) _\infty} \over
          {(\let \over / -{{c\,{\sqrt{b\,q}}  }\over {{\sqrt{a}}}},
          -{{c\,{\sqrt{a\,q}}  }\over {{\sqrt{b}}}} ;q) _\infty} } 
         { {(\let \over / a q, b q, {{{c^2} q}\over a}, {{{c^2} q}\over b}
          ;{q^2}) _\infty} \over {(\let \over / q, a b q, {c^2} q, {{{c^2}
          q}\over {a b}} ;{q^2}) _\infty} } -1\bigg) .\\
\tag6.13
\endalign$$
Letting $b\to1$ yields a $q$-analogue of \cite{\LavoAB, identity at
the top of p.~272}
\NoBlackBoxes
$$\align 
{} _{9}& \Phi _{8} \! \left [ \matrix \let \over / 
       i {a^{{1\over 4}}} {\sqrt{c}} {q^{{7\over 4}}}, -i
   {a^{{1\over 4}}} {\sqrt{c}} {q^{{7\over 4}}}, a q, q, c q, - c q
     , -{{{\sqrt{a\,q}} }\over c},\\ \let \over / i {a^{{1\over
   4}}} {\sqrt{c}} {q^{{3\over 4}}}, -i {a^{{1\over 4}}} {\sqrt{c}}
   {q^{{3\over 4}}}, -{{c {q^{{3\over 2}}}}\over {{\sqrt{a}}}}, -
   {\sqrt{a}} c {q^{{3\over 2}}}   , - {\sqrt{a}} {q^{{3\over 2}}}
     , {\sqrt{a}} {q^{{3\over 2}}}, {c^2} {q^2}, \endmatrix \right.\\
&\left.\matrix   \let \over /- c\,{\sqrt{a\,q}} , q;\\  \let \over /
{q^2};\endmatrix q,
   {\displaystyle {{c {q^{{3\over 2}}}}\over {{\sqrt{a}}}}} \right ] =
-{{c  \left( 1 - q \right)
        \left( 1 + c\,{\sqrt{a\,q}}   \right)   }\over
   {\sqrt{a\,q}\left( 1 - c \right)  \left( 1 + c \right)  
        \left( 1 + {\sqrt{a}} c {q^{{3\over 2}}} \right) }}
  {{   { {(\let \over / {q^2}, a {q^2}, {c^2} {q^2}, {{{c^2} {q^2}}\over a}
         ;{q^2}) _\infty} \over {(\let \over / {{{c^2} q}\over a}, q, a {q^3},
         {c^2} {q^3} ;{q^2}) _\infty} }}}\\
& + {{\left( 1 - a q \right)      \left( 1 - {c^2} q \right)  
     \left( 1 + c\,{\sqrt{a\,q}}   \right)          }\over
    {q\,\left( 1 - a \right)  \left( 1 - c \right)  \left( 1 + c \right)  
         \left( 1 + {\sqrt{a}} c {q^{{3\over 2}}} \right) 
       }}
     \Big( \tilde\psi_ {q} ( {\textstyle -{{c {\sqrt{q}}}\over
             {{\sqrt{a}}}}}) - \tilde\psi_ {q} ( {\textstyle -c\,{\sqrt{a\,q}} 
                }) \\
&\hskip2cm+ \frac { \tilde\psi_ {{q^2}} ( {\textstyle a q})  - 
         \tilde\psi_ {{q^2}} ( {\textstyle q}) +
    \tilde\psi_ {{q^2}} ( {\textstyle {c^2} q}) - 
         \tilde\psi_ {{q^2}} ( {\textstyle {{{c^2} q}\over a}}) }
    {1+q}
     \Big) .
\tag6.14
\endalign$$
A $_7F_6$-evaluation related to (6.7) and (6.11) is obtained from
(6.14) by replacing $a$ by $q^a$, $c$ by $-q^c$, and then letting
$q\uparrow1$,
$$\align
{} _{7} F _{6}& \!\left [ \matrix { {7\over 4} + {a\over 4} + {c\over 2}, 1 +
   a, 1, 1 + c, {1\over 2} + {a\over 2} - c, {1\over 2} + {a\over 2} +
c},1;\\
   { {3\over 4} + {a\over 4} + {c\over 2}, {3\over 2} - {a\over 2} + c,
   {3\over 2} + {a\over 2} + c, {3\over 2} + {a\over 2}, 2 + 2 c, 2};\endmatrix
     {\displaystyle -1}\right ] \\
& = {{\left(  2 c+a+1 \right)  }\over{2 \,c\, \left(  2 c+a+3 \right) }}
  {{  \Gamma \left [ \matrix {1\over 2}, {{3 + a}\over 2}, {{3 + 2 c}\over
        2}, {{1 - a + 2 c}\over 2}\\  {{2 + a}\over 2}, {{2 + 2 c}\over 2},
        {{2 - a + 2 c}\over 2}\endmatrix \right ]}
     }\\
&\quad  - {{\left( 1 + a \right)  \left( 1 + 2 c \right)  
       \left(  2 c +a+1\right)  }\over
    {2 a c \left( 2 c+a+3 \right) }}
      \Big(  \psi ({ \textstyle {1\over 2} + {a\over 2} + c})- 
         \psi ({ \textstyle {1\over 2} - {a\over 2} + c})\\
&\hskip1cm+\frac {1} {2}
       \big( \psi ({ \textstyle {1\over 2}}) -
        \psi ({ \textstyle {{1 + a}\over 2}})  -
         \psi ({ \textstyle {{1 + 2 c}\over 2}}) + 
         \psi ({ \textstyle {{1 - a + 2 c}\over 2}})\big) \Big) .
\tag6.15
\endalign$$

On the other hand, letting $c\to1$ in (6.13) leads to a $q$-analogue
of \cite{\LavoAB, (7)}
$$\align 
{} _{9}& \Phi _{8} \! \left [ \matrix \let \over /
      i {(a\,b)^{{1\over 4}}} 
   {q^{{7\over 4}}}, -i {(a\,b)^{{1\over 4}}} {q^{{7\over 4}}}, a
   q, b q, q,\\ \let
   \over / i {(a\,b)^{{1\over 4}}}  {q^{{3\over 4}}}, -i
   {(a\,b)^{{1\over 4}}} {q^{{3\over 4}}}, -{{{\sqrt{b}}
   {q^{{3\over 2}}}}\over {{\sqrt{a}}}}, -{{{\sqrt{a}} {q^{{3\over 2}}}}\over
   {{\sqrt{b}}}}, - {\sqrt{a\,b}}  {q^{{3\over 2}}}   ,\endmatrix \right.\\
&\hskip1cm \left.\matrix   \let \over / -q, - {\sqrt{a\,b\,q}} ,
    - {\sqrt{a\,b\,q}}   , q;\\
    \let \over /  {\sqrt{a\,b}} {q^{{3\over 2}}}, {q^2}, {q^2};\endmatrix q,
   {\displaystyle {{{q^{{3\over 2}}}}\over {{\sqrt{a\,b}} }}} \right ]
   = {{\left( 1 - q \right)  \left( 1 - a b q \right) 
     \left( 1 + {{{\sqrt{a\,q}} }\over {{\sqrt{b}}}}   \right)
        \left( 1 + {{{\sqrt{b\,q}} }\over {{\sqrt{a}}}} \right)  
         }\over
     {2 \left( 1 - a \right)  \left( 1 - b \right)  
      \left( 1 + {{{\sqrt{q}}}\over {{\sqrt{a\,b}} }} \right)  q 
      \left( 1 + {\sqrt{a\,b}}  {q^{{3\over 2}}} \right) }}\\
&\quad      \bigg({{\left( 1 - q \right)  {\sqrt{q}} }\over {{\sqrt{a\,b}} }} 
    {{       { {(\let \over / - {\sqrt{a\,b\,q}}  
                  , -{{{\sqrt{q}}}\over {{\sqrt{a\,b}} }} ;q)
                _\infty} \over {(\let \over / -{{{\sqrt{b\,q}} }\over
                {{\sqrt{a}}}}, -{{{\sqrt{a\,q}} }\over {{\sqrt{b}}}}
                ;q) _\infty} } { {(\let \over / a, b, {{{q^2}}\over a},
                {{{q^2}}\over b} ;{q^2}) _\infty} \over {(\let \over / q, a b
                q, q, {q\over {a b}} ;{q^2}) _\infty} }} } \\
&\quad  \tilde\psi {q} ( {\textstyle -{{{\sqrt{a\,q}}
           }\over {{\sqrt{b}}}}})- 
   \tilde\psi_ {q} ( {\textstyle -{{{\sqrt{q}}}\over {{\sqrt{a\,b}}
           }}})  +
          \tilde\psi {q} ( {\textstyle -{{{\sqrt{b\,q}} }\over
           {{\sqrt{a}}}}}) - \tilde\psi {q} ( {\textstyle - {\sqrt{a\,b\,q}}
               }) +
          {2\over {1+q}}{{ \left( \tilde\psi {{q^2}} ( {\textstyle q}) + 
                \tilde\psi {{q^2}} ( {\textstyle {q\over {a b}}}) \right) }}\\
&      \quad   - {2\over{1 + q}} { {(\let \over / - {\sqrt{a\,b\,q}} 
                  , -{{{\sqrt{q}}}\over {{\sqrt{a\,b}}
               }} ;q) _\infty} \over {(\let \over / -{{{\sqrt{b\,q}}
               }\over {{\sqrt{a}}}}, -{{{\sqrt{a\,q}} }\over
               {{\sqrt{b}}}} ;q) _\infty} } 
              \Big( { {(\let \over / a q, b q, {q\over a}, {q\over b} ;{q^2})
                  _\infty} \over {(\let \over / q, a b q, q, {q\over {a b}}
                  ;{q^2}) _\infty} } 
                 \left( \tilde\psi {{q^2}} ( {\textstyle {q\over a}}) + 
                   \tilde\psi {{q^2}} ( {\textstyle {q\over b}}) \right)\\
&\quad   - 
                {{\sqrt{q}}\over {{\sqrt{a\,b}} }}
       {{ { {(\let \over / a, b, {{{q^2}}\over a},
                     {{{q^2}}\over b} ;{q^2}) _\infty} \over {(\let \over / q,
                     a b q, q, {q\over {a b}} ;{q^2}) _\infty} } 
                    \big( \tilde\psi {{q^2}} ( {\textstyle {{{q^2}}\over a}}) + 
                      \tilde\psi {{q^2}} ( {\textstyle {{{q^2}}\over b}})\big) }
                    } \Big) \bigg).
\tag6.16
\endalign$$

Again, the series in (6.6), (6.7), (6.10)--(6.16) are very
well-poised series with cancelling leading parameter.

\remark{Acknowledgements} The present investigation was initiated 
during the first-named author's visit to the University of Victoria
in July 1992. The first author was partially supported by the FWF
Austria, Projekt Nr\. P10191-MAT.
The second author was supported, in part, by the Natural
Sciences and Engineering Research Council of Canada under Grant
OGP0007353.
\endremark

\comment
\head ---\quad ---\quad --- \endhead

\bigskip
\remark{Comments and Questions} 
\roster
\item Should we extend the proof of Theorem~1 such that the theorem
also holds for complex $q$, $\v q\ne1$? (I suppose that this can be done.
However, there might be some complications.)
\item Should $\psi_q(x)$ be studied more extensively (e.g\. as a
function of $q$, such as Askey \cite{\AskeAA} and Moak \cite{\MoakAA,
\MoakAB} did for the $q$-gamma function)? Perhaps then we should 
reserve a separate paper for section~2. The article is very long even
now, I fear.
\item Should the identities of Theorem~2 also be formulated in
terms of $\ps_q(x)$? Should we also introduce (and discuss) the ``actual"
$q$-polygamma function $\ps_q^{(1)}(x):=\frac {d} {dx}\psi_q(x)$?
\item In a remark after (5.5) I wrote that the three-product formula
follows from a formula due to Slater \cite{\SlatAD, (3); \GaRaAA,
Ex.~5.22}. However this is just a suspicion of mine. I could not
manage to actually verify this. Maybe you succeed.

The first thing that I did was to try
to derive it from Bailey's three-product formula \cite{\GaRaAA,
Ex.~5.21}. But it appears that it is impossible. 
What has to be done is to rewrite (5.5) in the form
$$\multline 
(Aq,q/A,Bq,q/B,-1,-q^2,-q,-q;q^2)_\infty\\
\hskip3cm+\sqrt{ABq}(Aq^2,1/A,Bq^2,1/B,-1,-q^2,-q,-q;q^2)_\infty\\
=2\cdot(-q^{\frac {1} {2}}\sqrt{AB}, -q^{\frac {3} {2}}/\sqrt{AB},
-q^{\frac {3} {2}}\sqrt{AB}, -q^{\frac {1} {2}}/\sqrt{AB},\\
q^{\frac {1} {2}}\sqrt{A/B},q^{\frac {3} {2}}\sqrt{B/A},
q^{\frac {3} {2}}\sqrt{A/B},q^{\frac {1} {2}}\sqrt{B/A};q^2)_\infty,
\endmultline\tag$*$ $$
and then somehow choose the parameters in Slater's formula with $q$
replaced by $q^2$ such that
two of the four products become identical, thus giving the right-hand
side of ($*$), and the two other products producing the left-hand
side. I think that it is evident that
(5.5) must be a special case of Slater's formula since the parameter
pairing in ($*$) perfectly fits into Slater's formula.
\item Should we include further identities (other special or limiting
cases, outcomes of transforms, outcomes of other contiguous
relations)?
If I am not wrong, when applying the transformation \cite{\GaRaAA,
(III.24)} to (5.1)--(5.3) and {\it then} applying our method gives rise to
different identities. As another example consider (5.10) which came
out by applying a transformation {\it after} having applied our
method.
\item Do some of the hypergeometric identities (which momentarily
have no reference) already appear in the literature?

\endroster
\endremark

\endcomment
\message{Erdelyi: Vols. I,II!}
\Refs

\ref\no \AskeAA\by R.    Askey \yr 1978 
\paper The $q$-Gamma and $q$-Beta functions
\jour Appl\. Analysis\vol 8
\pages 125--141\endref

\ref\no \AskeAB\by R.    Askey \yr 1980 \paper Ramanujan's extensions of the 
Gamma and Beta functions\jour Amer\. Math\. Monthly\vol 87\pages 
346--359\endref

\ref\no \BailAA\by W. N. Bailey \yr 1935 
\book Generalized hypergeometric series
\publ Cambridge University Press
\publaddr Cambridge\endref

\ref\no \ErdeAA\by A.    Erd\'elyi (ed.) \yr 1953 
\book Higher Transcendental Functions Vol.~I
\publ McGraw--Hill
\publaddr New York\endref

\ref\no \GaRaAA\by G.    Gasper and M. Rahman \yr 1990 
\book Basic hypergeometric series
\publ Encyclopedia of Mathematics And Its Applications~35, Cambridge University Press
\publaddr Cambridge\endref

\ref\no \IsLMAA\by M.E.H.Ismail, L. Lorch and M. E. Muldoon \yr 1988 
\paper Completely monotonic functions associated with the gamma 
function and its $q$-analogues\jour J. Math\. Anal\. Appl\.\vol 
116\pages 1--9\endref

\ref\no \JackAA\by F. H. Jackson \yr 1904 
\paper A generalization of the functions $\Gamma(n)$ and $x^n$
\jour Proc\. Roy\. Soc\. London\vol 74
\pages 64--72\endref

\ref\no \KociAA\by V. L. Koci\'c \yr 1990 \paper A note on $q$-gamma 
function\jour Univ\. Beograd\. Publ\. Elektrotehn\. Fak\. Ser\. Mat\.\vol 
1\pages 31--34\endref

\ref\no \LavoAA\by J. L. Lavoie \yr 1986 
\paper Some evaluations for the generalized hypergeometric series
\jour Math\. Comp\.\vol 46
\pages 215--218\endref

\ref\no \LavoAB\by J. L. Lavoie \yr 1987 
\paper Some summation formulas for the series $_3F_2(1)$
\jour Math\. Comp\.\vol 49
\pages 269--274\endref

\ref\no \LukeAA\by Y. L. Luke \yr 1975 
\book Mathematical functions and their approximations
\publ Academic Press
\publaddr New York\endref

\ref\no \MaTuAA\by O. I. Marichev and V. K. Tuan \yr 1982 \paper Some 
properties of the $q$-gamma function\jour Dokl\. Akad\. Nauk BSSR\vol 
26\pages 488--491, 571\endref

\ref\no \MoakAA\by D. S. Moak \yr 1980 
\paper The $q$-gamma function for $q>1$
\jour Aequationes Math\.\vol 20
\pages 278--285\endref

\ref\no \MoakAB\by D. S. Moak \yr 1980 
\paper The $q$-gamma function for $x<0$
\jour Aequationes Math\.\vol 21
\pages 179--191\endref

\ref\no \SlatAD\by L. J. Slater \yr 1954 
\paper A note on equivalent product theorems
\jour Math\. Gazette\vol 38
\pages 127--128\endref

\ref\no \SlatAC\by L. J. Slater \yr 1966 
\book Generalized hypergeometric functions
\publ Cambridge University Press
\publaddr Cambridge\endref

\ref\no \SrivAN\by H. M. Srivastava \yr 1992 
\paper A simple algorithm for the evaluation of a class of generalized hypergeometric series
\jour Stud\. Appl\. Math\.\vol 86
\pages 79--86\endref

\ref\no \ThoeAA\by J.    Thomae \yr 1869 
\paper Beitr\"age zur Theorie der durch die Heinesche Reihe \dots
\jour J. Reine Angew\. Math\.\vol 70
\pages 258--281\endref

\ref\no \VeJaAH\by A.    Verma and V. K. Jain \yr 1985 
\paper Some summation formulae for nonterminating basic hypergeometric series
\jour SIAM J. Math\. Anal\.\vol 16
\pages 647--655\endref

\ref\no \WatsAA\by G. N. Watson \yr 1917 \paper The integral formula
for generalized Legendre functions\jour Proc\. London Math\. Soc\. (2)\vol
17\pages 241--246\endref

\ref\no \WhWaAA\by E. T. Whittaker and G. N. Watson \yr 1965 
\book A Course of Modern Analysis
\bookinfo 4th edition
\publ Cambridge University Press
\publaddr Cambridge\endref

\endRefs
\enddocument

Bemerkungen:
