Summations for basic hypergeometric series

involving a q-analogue of the digamma function

C. Krattenthaler H. M. Srivastava

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.

1. Introduction and description of method. In [11, 9, 10, 17] 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 [11, 9, 10, 17]. Not surprisingly, at the result side of our identities there appears a qanalogue 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. [5, sec. 1.2]), (a; q)_{n} = ∏_{i=0}^{n−1}(1 − aq^{i}), (a; q)_{∞} = ∏_{i=0}^{∞}(1 − aq^{i}),

rΦ_{s}[ \begin{matrix}a_{1}, …, a_{r}; \\ b_{1}, …, b_{s}; \end{matrix} q, z] = ∑_{n=0}^{∞} \frac{(a_{1}; q)_{n} ⋯ (a_{r}; q)_{n}}{(q; q)_{n}(b_{1}; q)_{n} ⋯ (b_{s}; q)_{n}} ((−1)^{n}q^{\binom{n}{2}})^{s−r+1} z^{n} ,

and the short notation (cf. [5, sec. 2.1])

r+1W_{r}(a; a_{1}, …, a_{r−2}; q, z) := _{r+1}Φ_{r}[ \begin{matrix}a, q\sqrt{a}, −q\sqrt{a}, a_{1}, …, a_{r−2}; \\ \sqrt{a}, −\sqrt{a}, aq/a_{1}, …, aq/a_{r−2}; \end{matrix} q, z]

for the very well-poised basic hypergeometric series.

The hypergeometric notations (cf. [3, sec. 2.1; 16, sec. 2.1]) are (a)_{n} = ∏_{i=0}^{n−1}(a+i),

rF_{s}[ \begin{matrix}a_{1}, …, a_{r}; \\ b_{1}, …, b_{s}; \end{matrix} z] = ∑_{n=0}^{∞} \frac{(a_{1})_{n} ⋯ (a_{r})_{n}}{n! (b_{1})_{n} ⋯ (b_{s})_{n}} z^{n} ,

summation formulas.

1991 Mathematics Subject Classification. Primary 33D20; Secondary 33B15, 33C20, 33D05.

Key words and phrases. basic hypergeometric series, q-digamma function, q-polygamma function,

and the short notation

V (a; a_{1}, …, a_{r−1}; z) := _{r+1}F_{r}[ \begin{matrix}a, 1 + a/2, a_{1}, …, a_{r−1}; \\ a/2, 1 + a − a_{1}, …, 1 + a − a_{r−2}; \end{matrix} 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 [5, pp. 4–5].

We shall frequently use the short notations

(a_{1}, …, a_{m}; q)_{β} := ∏_{i=1}^{m}(a_{i}; q)_{β} ,

(a_{1}, …, a_{m})_{β} := ∏_{i=1}^{m}(a_{i})_{β} ,

and

Γ [ \begin{matrix}a_{1}, …, a_{m} \\ b_{1}, …, b_{n} \end{matrix} ] := \frac{∏_{i=1}^{m} Γ(a_{i})}{∏_{i=1}^{n} Γ(b_{i})},

where Γ(z) is the Gamma function.

Now let us describe our method.

The aforementioned q-analogue of the identity utilized in [11, 9, 10, 17] is

r+1Φ_{s+1}[ \begin{matrix}qa_{1}, …, qa_{r}, q; \\ qb_{1}, …, qb_{s}, q^{2}; \end{matrix} q, z]

= \frac{(1 − b_{1}) ⋯ (1 − b_{s})(1 − q)}{z (1 − a_{1}) ⋯ (1 − a_{r})} (_{r}Φ_{s}[ \begin{matrix}a_{1}, …, a_{r}; \\ b_{1}, …, b_{s}; \end{matrix} q, z] − 1) . (1.1)

It is straight-forward to verify this identity. Suppose that the parameters in this identity are chosen such that the resulting _{r}Φ_{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’Hôspital’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’Hôspital’s Rule, we have to deal with derivations of infinite expressions (a; q)_{∞}. This results into the appearance of a q-analogue of the digamma function. Though this q-analogue previously appeared in the literature [1, 13, 14], 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}Φ_{s} at the right-hand side of (1.1) with the help of the nonterminating q-Pfaff–Saalschütz sum [5, (2.10.12)], in section 4 those that are derived by using Bailey’s nonterminating extension of Jackson’s _{8}Φ_{7}-sum [5, (2.11.7)], in section 5 those that are derived by using Verma and Jain’s [19, (1.6)] q-analogue of Whipple’s 3F_{2}-sum, and in section 6 those that are derived by using Verma and Jain’s [19, (1.5)] q-analogue of Watson’s _{3}F_{2}-sum.

2. The q-digamma function. Thomae [18] and F. H. Jackson [7] introduced the following q-analogue of the gamma function (cf. [5, sec. 1.10, Ex. 1.23]),

Γ_{q}(x) = \{ \begin{matrix}\frac{(q; q)_{∞}}{(q^{x}; q)_{∞}} (1 − q)^{1−x} & |q| < 1 \\ \frac{(q^{−1}; q^{−1})_{∞}}{(q^{−x}; q^{−1})_{∞}} (q − 1)^{1−x}q^{\binom{x}{2}} & 1 < q. \end{matrix}  (2.1)

For properties of the q-gamma function see [1, 2, 5, sec. ?, 6, 7, 8, 12, 13, 14, 18]. It was shown by Askey [1] and Moak [14] that for x ∈ ℝ (set of real numbers), x ≠ 0, −1, −2, …, the q-gamma function Γ_{q}(x) indeed tends to the gamma function Γ(x) when letting q → 1.

The digamma function ψ(x) (see e.g. [4, sec. 1.7]) by definition is the logarithmic derivative of Γ(x),

ψ(x) = \frac{\frac{d}{dx}Γ(x)}{Γ(x)}.

In view of this it is natural to define a q-digamma function ψ_{q}(x) by

ψ_{q}(x) := \frac{\frac{d}{dx}Γ_{q}(x)}{Γ_{q}(x)},  (2.2)

or more explicitly,

ψ_{q}(x) := \{ \begin{matrix}− log(1 − q) + log q ∑_{n=0}^{∞} \frac{q^{n+x}}{1 − q^{n+x}} & |q| < 1 \\ − log(q − 1) + (x − \frac{1}{2}) log q + log q ∑_{n=0}^{∞} \frac{q^{−n−x}}{q^{−n−x} − 1} & 1 < q. \end{matrix}  (2.3)

However, before being allowed to call ψ_{q}(x) a q-analogue of the digamma function, we have to show that ψ_{q}(x) tends to ψ(x) when letting q → 1. 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, |q| ≠ 1.

Theorem 1. Let q ∈ ℝ. For x ∈ ℂ (set of complex numbers), x ≠ 0, −1, −2, …, there holds

lim_{q→1} ψ_{q}(x) = ψ(x).

Proof. First let 0 < q < 1. Write ψ_{q}(x) in the form

ψ_{q}(x) = − log(1 − q) + log q ∑_{n=0}^{∞} \frac{q^{n+1}}{1 − q^{n+1}} − log q ∑_{n=0}^{∞} \frac{q^{n+1}(1 − q^{x−1})}{(1 − q^{n+1})(1 − q^{n+x})}. (2.4)

We shall show that

lim_{q↑1} (log(1 − q) − log q ∑_{n=0}^{∞} \frac{q^{n+1}}{1 − q^{n+1}} ) = γ ,  (2.5)

where γ is the Euler gamma (cf. [4, 1.1.(4); 21, sec. 12.1]), and

lim_{q↑1} (− log q ∑_{n=0}^{∞} \frac{q^{n+1}(1 − q^{x−1})}{(1 − q^{n+1})(1 − q^{n+x})}) = ∑_{n=0}^{∞} \frac{x − 1}{(n + 1)(n + x)}.  (2.6)

Supposing that this would be done, we would have

lim_{q↑1} ψ_{q}(x) = −γ + ∑_{n=0}^{∞} \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 ψ(x) (cf. [4, 1.7.(3)]). This would prove lim_{q↑1} ψ_{q}(x) = ψ(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

∑_{n=1}^{∞} (−\frac{q^{n} log q}{1 − q^{n}} − log \frac{1 − q^{n+1}}{1 − q^{n}} )

= − log q ∑_{n=0}^{∞} \frac{q^{n+1}}{1 − q^{n+1}} − ∑_{n=1}^{∞} ( log(1 − q^{n+1}) − log(1 − q^{n}))

= − log q ∑_{n=0}^{∞} \frac{q^{n+1}}{1 − q^{n+1}} + log(1 − q).

Thus (2.5) is equivalent with

lim_{q↑1} ∑_{n=1}^{∞} (−\frac{q^{n} log q}{1 − q^{n}} − log \frac{1 − q^{n+1}}{1 − q^{n}} ) = γ.  (2.7)

Suppose that we are allowed to interchange limit and summation at the left-hand side of (2.7). This would give

lim_{q↑1} ∑_{n=1}^{∞} (−\frac{q^{n} log q}{1 − q^{n}} − log \frac{1 − q^{n+1}}{1 − q^{n}} ) = ∑_{n=1}^{∞} (\frac{1}{n} − log \frac{n + 1}{n})

= lim_{N→∞} (∑_{n=1}^{N} \frac{1}{n} − log N) .

But the right-hand side of this equation by definition equals γ (cf. [4, 1.1.(4); 21, 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 ≤ −\frac{q^{n} log q}{1 − q^{n}} − log \frac{1 − q^{n+1}}{1 − q^{n}} ≤ \frac{1}{n} − log \frac{n + 1}{n} .  (2.8)

Because of \frac{1}{n} − log \frac{n+1}{n} ≤ \frac{1}{n^{2}} (see [21, 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↓0} h(q) = 0 and lim_{q↑1} 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) ≥ 0, i.e.

\frac{n + 1}{n} \frac{(1 − q)(1 − q^{n})}{(1 − q^{n+1})} + log q ≤ 0,  0 < q < 1.  (2.9)

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)^{2}q^{n}}{n q (1 − q^{n+1})^{2}} (n − (q + q^{2} + ⋯ + q^{n}))((q^{−1} + q^{−2} + ⋯ + q^{−n}) − n) ,

which is greater than 0 for 0 < q < 1. Together with lim_{q↑1} 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).

For the proof of (2.6), we proceed as follows,

lim_{q↑1} ( − log q ∑_{n=0}^{∞} \frac{q^{n+1}(1 − q^{x−1})}{(1 − q^{n+1})(1 − q^{n+x})})

= lim_{q↑1} (−\frac{log q}{1 − q} ) lim_{q↑1} ∑_{n=0}^{∞} q^{n+1} \frac{(1 − q)(1 − q^{x−1})}{(1 − q^{n+1})(1 − q^{n+x})}

= 1 · ∑_{n=0}^{∞} \frac{x − 1}{(n + 1)(n + x)} .

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.

Finally we show lim_{q↓1} ψ_{q}(x) = ψ(x). If q > 1 we can rewrite (2.3) by

ψ_{q}(x) = (x − \frac{3}{2}) log q − log(1 − \frac{1}{q}) + log \frac{1}{q} ∑_{n=0}^{∞} \frac{q^{−n−x}}{1 − q^{−n−x}}

= (x − \frac{3}{2}) log q + ψ_{1/q}(x).

From this and the fact that we have already shown lim_{q↑1} ψ_{q}(x) = ψ(x) we get

lim_{q↓1} ψ_{q}(x) = lim_{q↓1} ((x − \frac{3}{2}) log q) + lim_{q↑1} ψ_{q}(x)
= 0 + ψ(x) = ψ(x).

This completes the proof of Theorem 1. □

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

For our purposes, because of notational convenience, it will be useful to consider a “logarithmic-free” variant of ψ_{q}(x), namely

ψ̃_{q}(x) := −(1 − q) ∑_{n=0}^{∞} \frac{xq^{n}}{1 − xq^{n}} .  (2.10)

This definition makes sense for complex q, x, |q| < 1, and x ≠ 1, q^{−1}, q^{−2}, . . .. ψ̃_{q}(x) is related to ψ(x) by

ψ̃_{q}(q^{x}) = −\frac{1 − q}{log q} ψ_{q}(x) − (1 − q)\frac{log(1 − q)}{log q} .  (2.11)

So the limit lim_{q↑1} ψ̃_{q}(q^{x}) does not even exist, but at least, because of (2.11) and Theorem 1, we have

lim_{q↑1} (ψ̃_{q}(q^{x}) − ψ̃_{q}(q^{y})) = ψ(x) − ψ(y) ,  (2.12)

which suffices for our purposes.

The next Theorem lists miscellaneous properties of the q-digamma function ψ̃_{q}(x). These will be of relevance in the subsequent sections when manipulating q-digamma functions and when taking limits of q-digamma functions.

Theorem 2. Let N denote a nonnegative integer. There hold the following identities.

ψ̃_{q}(x) − ψ̃_{q}(y) = ∑_{n=0}^{∞} \frac{q^{n}(y − x)(1 − q)}{(1 − xq^{n})(1 − yq^{n})}  (2.13a)

= \frac{(1 − q)(y − x)}{(1 − x)(1 − y)} _{3}Φ_{2}[ \begin{matrix}x, y, q; \\ qx, qy; \end{matrix} q, q] ,  (2.13b)

ψ̃_{q}(x) − ψ̃_{q}(x q^{N}) = −(1 − q) ∑_{n=0}^{N−1} \frac{xq^{n}}{1 − xq^{n}} ,  (2.14)
ψ̃_{q^{2}}(x) + ψ̃_{q^{2}}(qx) = (1 + q)ψ̃_{q}(x) ,  (2.15)
ψ̃_{q^{2}}(x^{2}) = \frac{(1 + q)}{2}(ψ̃_{q}(x) + ψ̃_{q}(−x)) ,  (2.16)
lim_{q↑1} (ψ̃_{q}(q^{x}) − ψ̃_{q}(q^{y})) = ψ(x) − ψ(y) ,  (2.17)
lim_{q↑1} (ψ̃_{q}(−q^{x}) − ψ̃_{q}(−q^{y})) = 0 ,  (2.18)
lim_{N→∞} (ψ̃_{q}(x/q^{N}) − ψ̃_{q}(y/q^{N})) = ψ̃_{q}(x) − ψ̃_{q}(y) + ψ̃_{q}(q/y) − ψ̃_{q}(q/x) . (2.19)

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

ψ̃_{q}(−q^{x}) − ψ̃_{q}(−q^{y}) = ∑_{n=0}^{∞} \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 |(1 + q^{n+x})^{−1}(1 + q^{n+y})^{−1}| ≤ 1 for all n and q_{0} < q < 1. This implies

|∑_{n=0}^{∞} \frac{q^{n}(q^{x} − q^{y})(1 − q)}{(1 + q^{n+x})(1 + q^{n+y})}| ≤ ∑_{n=0}^{∞} q^{n}|q^{x} − q^{y}|(1 − q)
≤ |q^{x} − q^{y}| ,  (2.20)

provided q_{0} < q < 1. For q ↑ 1 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

ψ̃_{q}(x/q^{N}) − ψ̃_{q}(x) − (ψ̃_{q}(y/q^{N}) − ψ̃_{q}(y)) = ∑_{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→∞} (ψ̃_{q}(x/q^{N}) − ψ̃_{q}(x) − ψ̃_{q}(y/q^{N}) + ψ̃_{q}(y)) = ψ̃_{q}(q/y) − ψ̃_{q}(q/x) ,

which is equivalent with (2.19). □

In our identities also the following q-analogue of the first polygamma function (cf. [4, 1.16]) will appear,

ψ̃_{q}^{(1)}(x) := (1 − q)^{2} ∑_{n=0}^{∞} \frac{xq^{n}}{(1 − xq^{n})^{2}} .  (2.21)

The following Theorem lists some relevant properties of ψ̃_{q}^{(1)}(x).

Theorem 3. There hold the following identities.

lim_{q↑1} ψ̃_{q}^{(1)}(q^{x}) = ψ^{(1)}(x) ,  (2.22)

where ψ^{(1)}(x) denotes the first polygamma function,

lim_{q↑1} ψ̃_{q}^{(1)}(−q^{x}) = 0 ,  (2.23)
lim_{N→∞} ψ̃_{q}^{(1)}(xq^{−N}) = ψ̃_{q}^{(1)}(x) + ψ̃_{q}^{(1)}(q/x) ,  (2.24)
ψ̃_{q^{2}}^{(1)}(x) + ψ̃_{q^{2}}^{(1)}(qx) = (1 + q)^{2}ψ̃_{q}^{(1)}(x) .  (2.25)

Sketch of Proof. We have

lim_{q↑1} ψ̃_{q}^{(1)}(q^{x}) = ∑_{n=0}^{∞} \frac{1}{(n + x)^{2}} .

The right-hand side in this equation equals ψ^{(1)}(x) (cf. [4, 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,

lim_{N→∞} ψ̃_{q}^{(1)}(xq^{−N}) = lim_{N→∞}(1 − q)^{2} ( ∑_{n=0}^{N−1} \frac{xq^{−N+n}}{(1 − xq^{−N+n})^{2}} + ∑_{n=0}^{∞} \frac{xq^{n}}{(1 − xq^{n})^{2}} )

= lim_{N→∞}(1 − q)^{2} ∑_{n=1}^{N} \frac{xq^{−n}}{(1 − xq^{−n})^{2}} + ψ̃_{q}^{(1)}(x)

= lim_{N→∞}(1 − q)^{2} ∑_{n=1}^{N} \frac{q^{n}/x}{(1 − q^{n}/x)^{2}}

= ψ̃_{q}^{(1)}(x) + ψ̃_{q}^{(1)}(q/x). □

3. Identities related to the nonterminating q-Pfaff–Saalschütz sum. 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–Saalschütz sum [5, (II.24)] to the resulting _{3}Φ_{2} at the right-hand side. This gives

4Φ_{3}[ \begin{matrix}aq, bq, cq, q; \\ eq, abcq^{2}/e, q^{2}; \end{matrix} q, q]

= \frac{1}{q} \frac{(1 − e)(1 − qabc/e)(1 − q)}{(1 − a)(1 − b)(1 − c)}(\frac{(q/e, bcq/e, acq/e, abq/e; q)_{∞}}{(aq/e, bq/e, cq/e, abcq/e; q)_{∞}} − 1

− \frac{(q/e, a, b, c, abcq^{2}/e^{2}; q)_{∞}}{(e/q, aq/e, bq/e, cq/e, abcq/e; q)_{∞}}_{3}Φ_{2}[ \begin{matrix}aq/e, bq/e, cq/e; \\ q^{2}/e, abcq^{2}/e^{2}; \end{matrix} q, q] ).

Now let c → 1. The limit lim_{c→1} \frac{1}{1−c}(\frac{(...)_{∞}}{(...)_{∞}} − 1), which has to be computed at the right-hand side, is obtained by using l’Hôspital’s rule, while the limit of the other expressions can be obtained directly. Thus we obtain

4Φ_{3}[ \begin{matrix}aq, cq, q, q; \\ bq, acq^{2}/b, q^{2}; \end{matrix} q, q]

= \frac{1}{q} \frac{(1 − b)(1 − qac/b)}{(1 − a)(1 − c)}(ψ̃_{q}(q/b) − ψ̃_{q}(cq/b) − ψ̃_{q}(aq/b) + ψ̃_{q}(acq/b)

− \frac{(1 − q) (a, c, q, acq^{2}/b^{2}; q)_{∞}}{(b/q, aq/b, cq/b, acq/b; q)_{∞}}_{3}Φ_{2}[ \begin{matrix}aq/b, cq/b, q/b; \\ q^{2}/b, acq^{2}/b^{2}; \end{matrix} q, q] ),  (3.1)

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 ↑ 1, with the help of (2.17) we get the following nonterminating extension of [17, (3.7)–(3.10)]

4F_{3}[ \begin{matrix}a + 1, c + 1, 1, 1; \\ b + 1, 2 + a + c − b, 2; \end{matrix} 1]

= \frac{b(1 + a + c − b)}{ac}(ψ(1 − b) − ψ(1 + c − b) − ψ(1 + a − b) + ψ(1 + a + c − b)

− Γ [ \begin{matrix}b − 1, 1 + a − b, 1 + c − b, 1 + a + c − b \\ a, c, 2 + a + c − 2b \end{matrix} ]_{3}F_{2}[ \begin{matrix}1 + a − b, 1 + c − b, 1 − b; \\ 2 − b, 2 + a + c − 2b; \end{matrix} 1] ).

(3.2)

Setting c = 0 in (3.1) gives a q-analogue for [17, (3.11)],

3Φ_{2}[ \begin{matrix}aq, q, q; \\ bq, q^{2}; \end{matrix} q, q] = \frac{1}{q} \frac{(1 − b)}{(1 − a)}(ψ̃_{q}(q/b) − ψ̃_{q}(aq/b)
− \frac{(1 − q) (a, q; q)_{∞}}{(b/q, aq/b; q)_{∞}}_{2}Φ_{1}[ \begin{matrix}aq/b, q/b; \\ q^{2}/b; \end{matrix} q, q] ).  (3.3)

In order to obtain a nicer q-analogue of [17, (3.11)], in (3.1) set c = q^{−N} and then let N → ∞. By (2.19) this yields

3Φ_{2}[ \begin{matrix}aq, q, q; \\ bq, q^{2}; \end{matrix} q, \frac{b}{a}] = \frac{(1 − 1/b)}{(1 − 1/a)} (ψ̃_{q}(b) − ψ̃_{q}(b/a)).  (3.4)

Writing this in the form

ψ̃_{q}(b) − ψ̃_{q}(b/a) = \frac{(a)_{∞}}{(b)_{∞}} ∑_{n=1}^{∞} \frac{(1 − q)}{(1 − q^{n})} \frac{(bq^{n})_{∞}}{(aq^{n})_{∞}} (\frac{b}{a})^{n} ,  (3.5)

we also have a perfect q-analogue for [17, (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}Φ_{2}-series [5, Appendix, (III.34), a → q, b → q/b, c → aq/b, d → q^{2}/b, e → aq^{2}/b] with (2.13b), while (3.4) could be obtained by combining the 3Φ_{2}-transformation [5, Appendix, (III.9), a → q, b → aq, c → q, d → bq, e → q^{2}] with (2.13b).

4. Identities related to Bailey’s nonterminating Jackson sum. 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}Φ_{7} at the right-hand side can be summed using Bailey’s nonterminating extension of Jackson’s sum [5, (II.25)]. Thus we obtain

9Φ_{8}[ \begin{matrix}aq, \sqrt{a}q^{2}, −\sqrt{a}q^{2}, bq, cq, dq, eq, a^{2}q^{2}/bcde, q; \\ \sqrt{a}q, −\sqrt{a}q, aq^{2}/b, aq^{2}/c, aq^{2}/d, aq^{2}/e, bcdeq/a, q^{2}; \end{matrix} q, q]

= \frac{1}{q} \frac{(1 − aq/b)(1 − aq/c)(1 − aq/d)(1 − aq/e)(1 − bcde/a)(1 − q)}{(1 − q^{2}a)(1 − b)(1 − c)(1 − d)(1 − e)(1 − qa^{2}/bcde)}

× (\frac{(aq, b/a, aq/cd, aq/ce, bde/a, aq/de, bce/a, bcd/a; q)_{∞}}{(aq/c, aq/d, aq/e, bcde/a, bc/a, bd/a, be/a, aq/cde; q)_{∞}} − 1

− \frac{(aq, c, d, e, a^{2}q/bcde, b/a, bq/c, bq/d, bq/e, b^{2}cde/a^{2}; q)_{∞}}{(a/b, aq/c, aq/d, aq/e, bcde/a, bc/a, bd/a, be/a, aq/cde, b^{2}q/a; q)_{∞}}

8W_{7}(\frac{b^{2}}{a}; b, \frac{bc}{a}, \frac{bd}{a}, \frac{be}{a}, \frac{aq}{cde}; q, q))

Next we let e → 1. The limit lim_{e→1} \frac{1}{1−e}(\frac{(...)_{∞}}{(...)_{∞}} − 1), which has to be computed at the right-hand side, is obtained by using l’Hôspital’s rule, while the limit of the other expressions can be obtained directly. This results into

9Φ_{8}[ \begin{matrix}\sqrt{a}q^{2}, −\sqrt{a}q^{2}, bq, cq, dq, a^{2}q^{2}/bcd, aq, q, q; \\ \sqrt{a}q, −\sqrt{a}q, aq^{2}/b, aq^{2}/c, aq^{2}/d, bcdq/a, q^{2}, aq^{2}; \end{matrix} q, q]

= \frac{1}{q} \frac{(1 − aq/b)(1 − aq/c)(1 − aq/d)(1 − aq)(1 − bcd/a)}{(1 − q^{2}a)(1 − b)(1 − c)(1 − d)(1 − qa^{2}/bcd)}

× (ψ̃_{q}(aq/d) − ψ̃_{q}(aq) + ψ̃_{q}(aq/c) − ψ̃_{q}(aq/cd)

+ ψ̃_{q}(b/a) − ψ̃_{q}(bd/a) + ψ̃_{q}(bcd/a) − ψ̃_{q}(bc/a)

− \frac{(1 − q) (c, d, q, a^{2}q/bcd, bq/c, bq/d, bq, b^{2}cd/a^{2}; q)_{∞}}{(a/b, aq/c, aq/d, bcd/a, bc/a, bd/a, aq/cd, b^{2}q/a; q)_{∞}}

8W_{7}(\frac{b^{2}}{a}; b, \frac{bc}{a}, \frac{bd}{a}, \frac{b}{a}, \frac{aq}{cd}; q, q)) .  (4.1)

Replacing a by q^{a}, b by q^{b}, c by q^{c}, d by q^{d}, and letting q ↑ 1, it is seen that identity (4.1) is a nonterminating q-analogue of [17, (1.3)]. In fact, by use of (2.17), we thus get

8F_{7}[ \begin{matrix}2 + a/2, b + 1, c + 1, d + 1, 2 + 2a − b − c − d, a + 1, 1, 1; \\ 1 + a/2, 2 + a − b, 2 + a − c, 2 + a − d, 1 + b + c + d − a, 2, 2 + a; \end{matrix} 1]

= \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)}

× (ψ(1 + a − d) − ψ(1 + a) + ψ(1 + a − c) − ψ(1 + a − c − d)

+ ψ(b − a) − ψ(b + d − a) + ψ(b + c + d − a) − ψ(b + c − a)

− Γ[ \begin{matrix}a − b, 1 + a − c, 1 + a − d, b + c + d − a, b + c − a, b + d − a, 1 + a − c − d \\ c, d, 1 + 2a − b − c − d, 1 + b − c, 1 + b − d, 1 + b \end{matrix}

\begin{matrix}1 + 2b − a \\ 2b + c + d − 2a \end{matrix} ]_{7}V_{6}(2b − a; b, b + c − a, b + d − a, b − a, 1 + a − c − d; 1)),

(4.2)

a nonterminating extension of [17, (1.3)].

Next we find q-analogues for [17, (3.1)–(3.6)]. First in (4.1) set d = q^{−N} and then let N → ∞. Observing that with d = q^{−N} the last term in (4.1) vanishes, by means of (2.19) we get

7Φ_{6}[ \begin{matrix}aq, \sqrt{a}q^{2}, −\sqrt{a}q^{2}, bq, cq, q, q; \\ \sqrt{a}q, −\sqrt{a}q, aq^{2}/b, aq^{2}/c, aq^{2}, q^{2}; \end{matrix} q, \frac{qa}{bc}]

= \frac{bc}{qa} \frac{(1 − aq/b)(1 − aq/c)(1 − aq)}{(1 − q^{2}a)(1 − b)(1 − c)}(ψ̃_{q}(aq/c) − ψ̃_{q}(aq) + ψ̃_{q}(aq/b) − ψ̃_{q}(aq/bc)). (4.3)

This is a q-analogue of [17, (3.2)], of which [17, (3.1)] is a terminating special case. Letting c → ∞ in (4.3), we obtain a q-analogue of [17, (3.3)],

6Φ_{6}[ \begin{matrix}aq, \sqrt{a}q^{2}, −\sqrt{a}q^{2}, bq, q, q; \\ \sqrt{a}q, −\sqrt{a}q, aq^{2}/b, aq^{2}, q^{2}, 0; \end{matrix} q, \frac{q^{2}a}{b}]
= \frac{b}{qa} \frac{(1 − aq/b)(1 − aq)}{(1 − q^{2}a)(1 − b)}(ψ̃_{q}(aq) − ψ̃_{q}(aq/b)). (4.4)

For c = \sqrt{a}, (4.3) reduces to a q-analogue of [17, (3.4)],

5Φ_{4}[ \begin{matrix}aq, −\sqrt{a}q^{2}, bq, q, q; \\ −\sqrt{a}q, aq^{2}/b, aq^{2}, q^{2}; \end{matrix} 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})}(ψ̃_{q}(\sqrt{a}q) − ψ̃_{q}(aq) + ψ̃_{q}(aq/b) − ψ̃_{q}(q\sqrt{a}/b)). (4.5)

Similarly, setting b = \sqrt{a} in (4.5) yields

4Φ_{4}[ \begin{matrix}aq, −\sqrt{a}q^{2}, q, q; \\ −\sqrt{a}q, aq^{2}, q^{2}, 0; \end{matrix} q, q^{2}\sqrt{a}] = \frac{1}{q\sqrt{a}} \frac{(1 − aq)(1 + \sqrt{a})}{(1 − a)(1 + q\sqrt{a})}(ψ̃_{q}(aq) − ψ̃_{q}(q\sqrt{a})), (4.6)

a q-analogue for [17, (3.5)]. A q-analogue for [17, (3.6)] could be obtained by starting with the duplication formula for the q-gamma function [5, (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 [5, 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}Φ_{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}Φ_{8} in (4.1) actually is the very well-poised series

10W_{9}(aq^{2}; aq, bq, cq, dq, a^{2}q^{2}/bcd; q, q).

5. Identities related to Verma and Jain’s q-analogue of Whipple’s _{3}F_{2}- sum. In this section we rely on Verma and Jain’s summation [19, (1.6)] for a very well-poised _{8}Φ_{7},

8Φ_{7}[ \begin{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; \end{matrix} q, c]

= \frac{(−c, −c q; q)_{∞}(a c d, a c q/d, c d q/a, c q^{2}/a d; q^{2})_{∞}}{(c d, c q/d, −a c, −c q/a; q)_{∞}} . (5.1)

We also use two summations contiguous to (5.1), namely

8Φ_{7}[ \begin{matrix}−c, i q \sqrt{c}, −i q \sqrt{c}, a, 1/a, c, −d, −q/d; \\ i \sqrt{c}, −i \sqrt{c}, −c q/a, −a c q, −q, c q/d, c d; \end{matrix} q, c q]

= \frac{1}{1 + a} \frac{(−c q, −c q; q)_{∞}(a c d, a c q/d, c d q/a, c q^{2}/a d; q^{2})_{∞}}{(c d, c q/d, −a c q, −c q/a; q)_{∞}}

+ \frac{a}{1 + a} \frac{(−c q, −c q; q)_{∞}(a c d q, a c q^{2}/d, c d/a, c q/a d; q^{2})_{∞}}{(c d, c q/d, −a c q, −c q/a; q)_{∞}} (5.2)

and

8Φ_{7}[ \begin{matrix}−c q, i \sqrt{c} q^{3/2}, −i \sqrt{c} q^{3/2}, a q, q/a, c, −d q, −q^{2}/d; \\ i \sqrt{c q}, −i \sqrt{c q}, −c q/a, −a c q, −q^{2}, c q/d, c d; \end{matrix} q, \frac{c}{q} ]

= \frac{a d (1 + \frac{c}{a q} ) (1 + q)}{c (1 − a) (1 − \frac{c}{q} ) (1 + a) (1 + d) (1 + \frac{d}{q})}

\frac{(−c, −c q^{2}; q)_{∞}(a c d/q, a c/d, c d/a, c q/a d; q^{2})_{∞}}{(c d, c q/d, −a c q, −c q/a; q)_{∞}}

− \frac{a d (1 + \frac{a c}{q}) (1 + q)}{c (1 − a) (1 − \frac{c}{q} ) (1 + a) (1 + d) (1 + \frac{d}{q})}

\frac{(−c, −c q^{2}; q)_{∞}(a c d, a c q/d, c d/a q, c/a d; q^{2})_{∞}}{(c d, c q/d, −a c q, −c q/a; q)_{∞}} .  (5.3)

Identity (5.2) is a q-analogue of [10, (3)]. It follows from the contiguous relation

r+1Φ_{r}[ \begin{matrix}x, y, (A); \\ u/x, u/y, (B); \end{matrix} q, z] = \frac{q x y (1 − x) (1 − \frac{u}{xq} )}{u (x − y) (1 − \frac{xyq}{u} )} _{r+1}Φ_{r}[ \begin{matrix}xq, y, (A); \\ u/xq, u/y, (B); \end{matrix} q, \frac{z}{q} ]

− \frac{q x y (1 − y) (1 − \frac{u}{yq} )}{u (x − y) (1 − \frac{xyq}{u} )} _{r+1}Φ_{r}[ \begin{matrix}x, yq, (A); \\ u/x, u/yq, (B); \end{matrix} q, \frac{z}{q} ] . (5.4)

In fact, by setting r = 7, x = a, y = 1/a, u = −c, and (A), (B), z such that the left-hand side of (5.4) becomes the _{8}Φ_{7} in (5.2), at the right-hand side of (5.4) we obtain two _{8}Φ_{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,

(A q; q^{2})_{∞} (B q; q^{2})_{∞} (q/A; q^{2})_{∞} (q/B; q^{2})_{∞}

+ \sqrt{A B q} (1/A; q^{2})_{∞} (A q^{2}; q^{2})_{∞} (1/B; q^{2})_{∞} (B q^{2}; q^{2})_{∞}

= \frac{(−\sqrt{A B q}; q)_{∞} (−\sqrt{q/AB}; q)_{∞} (\sqrt{A q}/\sqrt{B}; q)_{∞} (\sqrt{B q}/\sqrt{A}; q)_{∞}}{(−q; q)_{∞}^{2}} , (5.5)

which will be used later. Written in the form

(Aq, q/A, Bq, q/B, −1, −q^{2}, −q, −q; q^{2})_{∞}

+ \sqrt{ABq}(Aq^{2}, 1/A, Bq^{2}, 1/B, −1, −q^{2}, −q, −q; q^{2})_{∞}
= 2 · (−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})_{∞}, (5.6)

it turns out to be a special case of a relation connecting four infinite products due to Slater [15, (3); 5, 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 [10, (4)]. It is derived from the contiguous relation

r+1Φ_{r}[ \begin{matrix}x, y, u/q, (A); \\ u/x, u/y, u/q^{2}, (B); \end{matrix} q, z]

= \frac{(1 − \frac{u}{xq^{2}} ) (1 − \frac{u}{yq^{2}} ) (1 − \frac{u}{xq} ) (1 − \frac{u}{yq} )}{(1 − \frac{u}{q^{2}} ) (−\frac{x}{q} + \frac{y}{q} ) (1 − \frac{u}{xyq} ) z} \frac{∏_{i=1}^{r−3}(1 − \frac{B_{i}}{q})}{∏_{i=1}^{r−2}(1 − \frac{A_{i}}{q})}

×_{r}Φ_{r−1}[ \begin{matrix}x/q, y, (A/q); \\ u/xq, u/yq^{2}, (B/q); \end{matrix} q, z]

− \frac{(1 − \frac{u}{xq^{2}} ) (1 − \frac{u}{yq^{2}} ) (1 − \frac{u}{xq} ) (1 − \frac{u}{yq} )}{(1 − \frac{u}{q^{2}} ) (−\frac{x}{q} + \frac{y}{q} ) (1 − \frac{u}{xyq} ) z} \frac{∏_{i=1}^{r−3}(1 − \frac{B_{i}}{q})}{∏_{i=1}^{r−2}(1 − \frac{A_{i}}{q})}

×_{r}Φ_{r−1}[ \begin{matrix}x, y/q, (A/q); \\ u/xq^{2}, u/yq, (B/q); \end{matrix} q, z] . (5.7)

In order to see this, set x = aq, y = q/a, u = −cq^{2}, (A) = (−c, i\sqrt{c}q^{\frac{3}{2}} , −i\sqrt{c}q^{\frac{3}{2}} , c, −dq, −q^{2}/d), (B) = (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}Φ_{7} of (5.3). At the right-hand side we obtain two _{8}Φ_{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}Φ_{7} at the right-hand side can be summed using (5.1). Letting a → 1 with the help of l’Hôspital’s Rule we obtain the identity

7Φ_{6}[ \begin{matrix}i \sqrt{c} q^{2}, −i \sqrt{c} q^{2}, q, c q, −d q, −q^{2}/d, q; \\ i \sqrt{c} q, −i \sqrt{c} q, −c q^{2}, −q^{2}, c q^{2}/d, c d q; \end{matrix} q, c]

= \frac{(1 − c d) (1 − \frac{c q}{d} ) (1 + c) (1 + q) (1 + c q)}{c (1 − c) (1 − q) (1 + d) (1 + \frac{q}{d}) (1 + c q^{2})}

(\frac{c(1 − q)}{(1 + c)} + \frac{ψ̃_{q^{2}}(c d q) − ψ̃_{q^{2}}(c d) + ψ̃_{q^{2}}(\frac{c q^{2}}{d}) − ψ̃_{q^{2}}(\frac{c q}{d})}{1 + q}). (5.8)

Replacing c by q^{c}, d by q^{d} and letting q ↑ 1, by (2.17) we see that this is a q-analogue of [9, (1)]. On the other hand, by replacing c by −q^{c}, d by −q^{d} and letting q ↑ 1, by (2.17) we arrive at the _{5}F_{4}-evaluation

5F_{4}[ \begin{matrix}2 + c/2, 1, 1 + d, 2 − d, 1; \\ 1 + c/2, 2 + c, 2 + c − d, 1 + c + d; \end{matrix} −1] = \frac{c (1 + c) (1 + c − d) (c + d)}{(2 + c) (1 − d) d}

× (\frac{1}{c} + \frac{1}{2} (ψ(\frac{1+c−d}{2}) − ψ(\frac{2+c−d}{2}) + ψ(\frac{c+d}{2}) − ψ(\frac{1+c+d}{2}))) . (5.9)

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 [5, Appendix, (III.24)] to this very well-poised series we find

8Φ_{7}[ \begin{matrix}−c^{2}q, icq^{3/2}, −icq^{3/2}, −q, cq/d, cd, −cq, c; \\ ic\sqrt{q}, −ic\sqrt{q}, c^{2}q, −cdq, −cq^{2}/d, cq, −cq^{2}; \end{matrix} q, q]

= \frac{(1 + c) (1 + cq)}{c (1 − q) (1 + d) (1 + \frac{q}{d})} \frac{(cd, cq/d, −q, −c^{2}q^{2}; q)_{∞}}{(q, c^{2}q, −cdq, −cq^{2}/d; q)_{∞}}

(\frac{c (1 − q)}{1 + c} + \frac{ψ̃_{q^{2}}(cdq) − ψ̃_{q^{2}}(cd) + ψ̃_{q^{2}}(\frac{cq^{2}}{d}) − ψ̃_{q^{2}}(\frac{cq}{d})}{1 + q}) . (5.10)

This identity is a q-analogue of an identity of Watson [20; 9, (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} = \frac{1}{a}, a_{6} = c, a_{7} = −d, a_{8} = −\frac{q}{d}, b_{1} = i \sqrt{c}, b_{2} = −i \sqrt{c}, b_{3} = −\frac{c q}{a}, b_{4} = −a c q, b_{5} = −q, b_{6} = \frac{c q}{d}, b_{7} = c d, z = cq. After applying (5.2) at the right-hand side this yields

9Φ_{8}[ \begin{matrix}−c q, i \sqrt{c} q^{2}, −i \sqrt{c} q^{2}, q/a, a q, c q, −d q, −q^{2}/d, q; \\ i \sqrt{c} q, −i \sqrt{c} q, −a c q^{2}, −c q^{2}/a, −q^{2}, c q^{2}/d, c d q, q^{2}; \end{matrix} q, c q]

= c \begin{matrix}(1 − c d) & (1 − d/cq) & (1 − q) & (1 + q) & (1 + a/cq) & (1 + a c q) \\ (1 − a)^{2} & (1 − c) & (1 + d) & (1 + d/q) & (1 + c q^{2}) \end{matrix}

(\frac{a}{1 + a} \frac{(−c q, −c q; q)_{∞}(c d/a, c q/a d, a c d q, a c q^{2}/d; q^{2})_{∞}}{(c d, −c q/a, −a c q, c q/d; q)_{∞}}

+\frac{1}{1 + a} \frac{(−c q, −c q; q)_{∞}(a c d, a c q/d, c d q/a, c q^{2}/a d; q^{2})_{∞}}{(c d, −c q/a, −a c q, c q/d; q)_{∞}} − 1) . (5.11)

Replacing d by cq/d and letting c → 1, we get a q-analogue of [9, (2)], which after some manipulation can be brought into the form

9Φ_{8}[ \begin{matrix}i q^{2}, −i q^{2}, q/a, a q, q, −q^{2}/d, −d q, −q, q; \\ i q, −i q, −a q^{2}, −q^{2}/a, −q^{2}, d q, q^{2}/d, q^{2}; \end{matrix} q, q]

= \begin{matrix}(1 − d) & (1 − \frac{d}{q}) & (1 + q) & (1 + \frac{a}{q}) & (1 + a q) \\ (1 − a)^{2} & (1 + d) & (1 + \frac{d}{q}) & (1 + q^{2}) \end{matrix}

(2 ψ̃_{q}(\frac{q}{d}) + ψ̃_{q}(−\frac{q}{a}) + ψ̃_{q}(−a q) − 2 ψ̃_{q}(−q) − 2\frac{(−q, −q^{2}; q)_{∞}}{(d, q/d, −q/a, −a; q)_{∞}}

((a q/d, a d, q^{2}/a d, d q/a; q^{2})_{∞}(ψ̃_{q^{2}}(\frac{a q}{d}) + ψ̃_{q^{2}}(\frac{q^{2}}{a d}))
+ a (q/a d, d/a, a q^{2}/d, a d q; q^{2})_{∞}(ψ̃_{q^{2}}(\frac{q}{a d}) + ψ̃_{q^{2}}(\frac{a q^{2}}{d})))). (5.12)

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 → ∞. By this procedure the infinite products at the right-hand side approach to a definite limit while the ψ̃_{q^{2}}’s vanish. Hence what remains is

7Φ_{6}[ \begin{matrix}i q^{2}, −i q^{2}, q/a, a q, q, −q, q; \\ i q, −i q, −a q^{2}, −q^{2}/a, −q^{2}, q^{2}; \end{matrix} q, −q]

= \begin{matrix}(1 + \frac{a}{q}) & (1 + q) & (1 + a q) \\ (1 − a)^{2} & (1 + q^{2}) \end{matrix}(ψ̃_{q}(−\frac{q}{a}) + ψ̃_{q}(−a q) − 2 ψ̃_{q}(−q)) , (5.13)

an identity which has no reasonable counterpart for q ↑ 1.

If in (5.11) we let a → 1, by using l’Hôspital’s Rule twice and employing (2.25) we obtain

9Φ_{8}[ \begin{matrix}i\sqrt{c}q^{2}, −i\sqrt{c}q^{2}, q, q, cq, −dq, −q^{2}/d, −c q, q; \\ i\sqrt{c}q, −i\sqrt{c}q, −cq^{2}, −cq^{2}, −q^{2}, cq^{2}/d, cdq, q^{2}; \end{matrix} q, cq]

= \frac{(1 − cd) (1 − \frac{d}{cq} ) (1 + q) (1 + cq)^{2}}{2 q (1 − c) (1 − q) (1 + d) (1 + \frac{d}{q}) (1 + cq^{2})}

(\frac{1}{(1 + q)^{2}} (ψ̃_{q^{2}}(cd) + ψ̃_{q^{2}}(\frac{cq}{d}) − ψ̃_{q^{2}}(cdq) − ψ̃_{q^{2}}(\frac{cq^{2}}{d}))^{2} − ψ̃_{q}^{(1)}(cd) − ψ̃_{q}^{(1)}(\frac{cq}{d})

− \frac{(1 − q)}{1 + q} (ψ̃_{q^{2}}(cd) + ψ̃_{q^{2}}(\frac{cq}{d}) − ψ̃_{q^{2}}(cdq) − ψ̃_{q^{2}}(\frac{cq^{2}}{d})) + 2ψ̃_{q}^{(1)}(−cq)).

(5.14)

This is a q-analogue of [9, (3)], which is seen by replacing c by q^{c}, d by q^{d}, and letting q ↑ 1. 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

7F_{6}[ \begin{matrix}2 + \frac{c}{2}, 1, 1, 1 + d, 2 − d, 1 + c, 1; \\ 1 + \frac{c}{2}, 2 + c, 2 + c, 2 + c − d, 1 + c + d, 2; \end{matrix} −1]

= \frac{(1 + c)^{2} (d − c − 1) (c + d)}{2 d (2 + c) (d − 1)}(\frac{1}{4}(ψ(\frac{1+c−d}{2}) − ψ(\frac{2+c−d}{2}) + ψ(\frac{c+d}{2}) − ψ(\frac{1+c+d}{2}))^{2}

+ 2ψ^{(1)}(1 + c) − ψ^{(1)}(1 + c − d) − ψ^{(1)}(c + d)). (5.15)

An interesting special case of (5.14) is obtained by setting d = q,

7Φ_{6}[ \begin{matrix}i\sqrt{c}q^{2}, −i\sqrt{c}q^{2}, −q, −c q, q, q, q; \\ i\sqrt{c}q, −i\sqrt{c}q, cq^{2}, q^{2}, −cq^{2}, −cq^{2}; \end{matrix} q, cq]

= \frac{(1 − cq) (1 + cq)^{2}}{2 c (1 − q) q (1 + cq^{2})}(ψ̃_{q}^{(1)}(cq) − ψ̃_{q}^{(1)}(−cq)). (5.16)

This identity is a q-analogue of [9, second identity on p. 216]. To obtain another identity from (5.14), we replace c by −c, set d = −q^{−N} and let N → ∞. By using (2.19) and (2.24) we find that

7Φ_{6}[ \begin{matrix}\sqrt{c} q^{2}, −\sqrt{c} q^{2}, q, q, −c q, c q, q; \\ \sqrt{c} q, −\sqrt{c} q, c q^{2}, c q^{2}, −q^{2}, q^{2}; \end{matrix} q, −q]

= \frac{(1 + q) (1 − c q)^{2}}{2 q (1 + c) (1 − q) (1 − c q^{2})}(\begin{matrix}(ψ̃_{q^{2}}(c) + ψ̃_{q^{2}}(\frac{q}{c}) − ψ̃_{q^{2}}(c q) − ψ̃_{q^{2}}(\frac{q^{2}}{c})) & 2 \\ (1 + q) & 2 \end{matrix}

+ ψ̃_{q}^{(1)}(c q) − ψ̃_{q}^{(1)}(\frac{1}{c}) − \frac{(1 − q)}{1 + q}(ψ̃_{q^{2}}(c) + ψ̃_{q^{2}}(\frac{q}{c}) − ψ̃_{q^{2}}(c q) − ψ̃_{q^{2}}(\frac{q^{2}}{c}))). (5.17)

Replacing c by q^{c} and letting q ↑ 1 gives an evaluation for a _{5}F_{4},

5F_{4}[ \begin{matrix}2 + \frac{c}{2}, 1, 1, 1 + c, 1; \\ 1 + \frac{c}{2}, 2 + c, 2 + c, 2; \end{matrix} −1]

= \frac{(c + 1)^{2}}{2 (c + 2)}(\frac{1}{4}(ψ(\frac{1−c}{2}) − ψ(1 − \frac{c}{2}) + ψ(\frac{c}{2}) − ψ(\frac{1+c}{2}))^{2} + ψ^{(1)}(1 + c) − ψ^{(1)}(−c)) .

(5.18)

Finally, in (1.1) choose a_{1} = −c q, a_{2} = i \sqrt{c} q^{\frac{3}{2}} , a_{3} = −i \sqrt{c} q^{\frac{3}{2}} , a_{4} = a q, a_{5} = \frac{q}{a}, a_{6} = c, a_{7} = −d q, a_{8} = −\frac{q^{2}}{d}, b_{1} = i \sqrt{c q}, b_{2} = −i \sqrt{c q}, b_{3} = −\frac{c q}{a}, b_{4} = −a c q, b_{5} = −q^{2}, b_{6} = \frac{c q}{d}, b_{7} = c d, z = c/q. Then the resulting _{8}Φ_{7} at the right-hand side can be summed using (5.3). Upon letting a → q we obtain the following q-analogue of [10, third identity on p. 272]

9Φ_{8}[ \begin{matrix}i \sqrt{c} q^{5/2}, −i \sqrt{c} q^{5/2}, c q, −d q^{2}, −q^{3}/d, −c q^{2}, q^{3}, q, q; \\ i \sqrt{c} q^{3/2}, −i \sqrt{c} q^{3/2}, −q^{3}, c q^{2}/d, c d q, q^{2}, −c q, −c q^{3}; \end{matrix} q, \frac{c}{q} ]

= \frac{q (1 − cd) (1 − \frac{d}{cq} ) (1 + c) (1 + q^{2}) (1 + cq^{2})}{(1 − q) (1 − c) (1 + \frac{d}{q^{2}} ) (1 + dq) (1 + q) (1 + cq^{3})}

( − \begin{matrix}(1 + \frac{c}{q^{2}} ) & ((1 − d) + \frac{d}{q}(1 − c) + \frac{d^{2}}{q} (1 − \frac{c}{d})) \\ (1 − \frac{c}{q} ) & (1 + d) & (1 + \frac{d}{q}) \end{matrix} + \frac{1 − q}{1 + q} − \frac{c (1 − q)}{q (1 + c)} − \frac{c (1 − q)}{1 + cq}

− \frac{1}{q (1 + q)} (\frac{2 d q (1 − \frac{c}{d}) (1 − \frac{cd}{q}) (1 + \frac{c}{q^{2}} )}{c (1 − \frac{c}{q} ) (1 − q) (1 + d) (1 + \frac{d}{q})} − 1)

(ψ̃_{q^{2}}(c d) − ψ̃_{q^{2}}(c d q) + ψ̃_{q^{2}}(c q/d) − ψ̃_{q^{2}}(c q^{2}/d))).  (5.19)

Replacing c by −q^{c}, d by −q^{d} and letting q ↑ 1 leads to an evaluation for a _{7}F_{6},

7F_{6}[ \begin{matrix}\frac{5}{2} + \frac{c}{2}, 2 + d, 3 − d, 2 + c, 3, 1, 1; \\ \frac{3}{2} + \frac{c}{2}, 2 + c − d, 1 + c + d, 2, 1 + c, 3 + c; \end{matrix} −1]

= \frac{c (2 + c) (d − c − 1) (c + d)}{2 (3 + c) (d − 2) (1 + d)}(\frac{1}{c} + \frac{1}{1 + c} + \frac{(c − 2) (2d − c − 1)}{(d − 1) d}

− (\frac{(c − 2) (c − d) (c + d − 1)}{2 (d − 1) d} − 1)

× (ψ(\frac{1+c−d}{2}) − ψ(\frac{2+c−d}{2}) + ψ(\frac{c+d}{2}) − ψ(\frac{1+c+d}{2})) ).  (5.20)

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

7Φ_{6}[ \begin{matrix}i\sqrt{c}q^{5/2}, −i\sqrt{c}q^{5/2}, q^{3}, q, −q^{2}, −cq^{2}, q; \\ i\sqrt{c}q^{3/2}, −i\sqrt{c}q^{3/2}, −cq, −cq^{3}, cq^{2}, q^{2}; \end{matrix} q, \frac{c}{q} ]

= \begin{matrix}(1 − cq) & (1 − 2c − c^{2} + q + 2cq − c^{2}q) & (1 + cq^{2}) \\ (1 − c) & (1 − \frac{c}{q} ) & (1 + q) & (1 + cq) & (1 + cq^{3}) \end{matrix}. (5.21)

Replacing c by q^{c} and −q^{c}, and then letting q tend to 1^{−} in (5.21), produces the identities

3F_{2}[ \begin{matrix}3, 1, 1; \\ 2 + c, 2; \end{matrix} 1] = \frac{(1 + c) (2c − 1)}{2 (c − 1) c}  (5.22)

and

5F_{4}[ \begin{matrix}2 + c, \frac{5}{2} + \frac{c}{2}, 3, 1, 1; \\ \frac{3}{2} + \frac{c}{2}, 1 + c, 3 + c, 2; \end{matrix} −1] = \frac{(2 + c) (1 + 2c)}{2 (1 + c) (3 + c)} ,  (5.23)

respectively.

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.

6. Identities related to Verma and Jain’s q-analogue of Watson’s _{3}F_{2}- sum. In this section we rely on Verma and Jain’s identity [19, (1.5)]

8Φ_{7}[ \begin{matrix}−\sqrt{a b} c/\sqrt{q}, i (a b)^{1/4} \sqrt{c} q^{3/4}, −i (a b)^{1/4} \sqrt{c} q^{3/4}, a, b, \\ i (a b)^{1/4} \sqrt{c}/q^{1/4}, −i (a b)^{1/4} \sqrt{c}/q^{1/4}, −c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}, \end{matrix}

\begin{matrix}c, −c, −\sqrt{a b q}/c; \\ −\sqrt{a b q}, \sqrt{a b q}, c^{2}; \end{matrix} q, \frac{c \sqrt{q}}{\sqrt{a b}}]

= \frac{(−c \sqrt{a b q}, −c \sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a q, b q, c^{2} q/a, c^{2} q/b; q^{2})_{∞}}{(q, a b q, c^{2} q, c^{2} q/a b; q^{2})_{∞}} .  (6.1)

We start by recording two identities contiguous to (6.1),

8Φ_{7}[ \begin{matrix}−\sqrt{a b} c/\sqrt{q}, i (a b)^{1/4} \sqrt{c} q^{3/4}, −i (a b)^{1/4} \sqrt{c} q^{3/4}, a, b, \\ i (a b)^{1/4} \sqrt{c}/q^{1/4}, −i (a b)^{1/4} \sqrt{c}/q^{1/4}, −c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}, \end{matrix}

\begin{matrix}c q, −c, −\sqrt{a b}/c\sqrt{q}; \\ −\sqrt{a b}/\sqrt{q}, \sqrt{a b q}, c^{2} q; \end{matrix} q, \frac{c \sqrt{q}}{\sqrt{a b}}]

= \frac{(1 + \begin{matrix}c^{2} & \sqrt{q} \\ √ & \overline{a b} \end{matrix} )}{(1 + \begin{matrix}√ & \overline{a b} \\ √ & \overline{q} \end{matrix})} \frac{(−c \sqrt{a b q}, −c\sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a, b, c^{2} q^{2}/a, c^{2} q^{2}/b; q^{2})_{∞}}{(q, a b q, c^{2} q, c^{2} q/a b; q^{2})_{∞}}

+ \frac{(−c \sqrt{a b q}, −c \sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a q, b q, c^{2} q/a, c^{2} q/b; q^{2})_{∞}}{(q, a b q, c^{2} q, c^{2} q/a b; q^{2})_{∞}}  (6.2)

and

8Φ_{7}[ \begin{matrix}−\sqrt{a b} c/\sqrt{q}, i (a b)^{1/4} \sqrt{c} q^{3/4}, −i (a b)^{1/4} \sqrt{c} q^{3/4}, a, b, \\ i (a b)^{1/4} \sqrt{c}/q^{1/4}, −i (a b)^{1/4} \sqrt{c}/q^{1/4}, −c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}, \end{matrix}

\begin{matrix}c, −c, −\sqrt{a b}/c \sqrt{q}; \\ −\sqrt{a b q}, \sqrt{a b q}, c^{2} q; \end{matrix} q, \frac{c q^{\frac{3}{2}}}{\sqrt{a b}}]

= \frac{(−c \sqrt{a b q}, −c \sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a q, b q, c^{2} q/a, c^{2} q/b; q^{2})_{∞}}{(q, a b q, c^{2} q, c^{2} q/a b; q^{2})_{∞}}

− \frac{c\sqrt{q}}{\sqrt{a b}} \frac{(−c \sqrt{a b q}, −c\sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a, b, c^{2} q^{2}/a, c^{2} q^{2}/b; q^{2})_{∞}}{(q, a b q, c^{2} q, c^{2} q/a b; q^{2})_{∞}} . (6.3)

Identity (6.2) is a q-analogue for [10, (1)]. It is proved by using the contiguous relation

r+1Φ_{r}[ \begin{matrix}x, y, (A); \\ u/x, u/y, (B); \end{matrix} q, z] = _{r+1}Φ_{r}[ \begin{matrix}xq, y/q, (A); \\ u/xq, uq/y, (B); \end{matrix} q, z]

+\frac{(1 − u) (1 − \frac{u}{xy} ) (−x + \frac{y}{q} ) z}{(1 − \frac{u}{x}) (1 − \frac{u}{y}) (1 − \frac{u}{xq} ) (1 − \frac{uq}{y} )} \frac{∏_{i=1}^{r−1}(1 − A_{i})}{∏_{i=1}^{r−2}(1 − B_{i})}_{r+2}Φ_{r+1}[ \begin{matrix}xq, y, uq, (qA); \\ uq/x, uq^{2}/y, u, (qB); \end{matrix} q, z]

(6.4)

with r = 7, x = −\sqrt{a b}/c \sqrt{q}, y = c q, u = −c \sqrt{a b q}, and (A), (B) such that the left-hand side becomes the _{8}Φ_{7} in (6.2). Subsequently, the resulting _{8}Φ_{7}’s at the right-hand side can be evaluated by (6.1). The contiguous relation (6.4) is equivalent with (5.7).

Likewise, (6.3) is a q-analogue of [10, (2)]. It follows from the contiguous relation

r+1Φ_{r}[ \begin{matrix}x, (A); \\ y, (B); \end{matrix} q, z] = _{r+1}Φ_{r}[ \begin{matrix}xq, (A); \\ y/q, (B); \end{matrix} q, \frac{z}{q} ]

− \frac{(1 − xy) z}{(1 − y) (1 − \frac{y}{q} ) q} \frac{∏_{i=1}^{r}(1 − A_{i})}{∏_{i=1}^{r−1}(1 − B_{i})}_{r+2}Φ_{r+1}[ \begin{matrix}xq, xyq, (qA); \\ yq, xy, (qB); \end{matrix} q, \frac{z}{q} ] (6.5)

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}Φ_{7} in (6.3). Once again the resulting _{8}Φ_{7}’s at the right-hand side can be evaluated by (6.1).

Now in (1.1) choose a_{1} = −\sqrt{a b} c/\sqrt{q}, a_{2} = i (a b)^{1/4} \sqrt{c} q^{3/4}, a_{3} = −i (a b)^{1/4} \sqrt{c} q^{3/4}, a_{4} = a, a_{5} = b, a_{6} = c, a_{7} = −c, a_{8} = −\sqrt{a b q}/c, b_{1} = i (a b)^{1/4} \sqrt{c}/q^{1/4}, b_{2} = −i (a b)^{1/4} \sqrt{c}/q^{1/4}, b_{3} = −c \sqrt{b q}/\sqrt{a}, b_{4} = −c \sqrt{a q}/\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}Φ_{7} at the right-hand side can be summed by (6.1). In the limit b → 1 we get

9Φ_{8}[ \begin{matrix}i a^{1/4} \sqrt{c} q^{7/4}, −i a^{1/4} \sqrt{c} q^{7/4}, q, a q, c q, \\ i a^{1/4} \sqrt{c} q^{3/4}, −i a^{1/4} \sqrt{c} q^{3/4}, −\sqrt{a} c q^{3/2}, −c q^{3/2}/\sqrt{a}, −\sqrt{a} q^{3/2}, \end{matrix}

\begin{matrix}−c q, −\sqrt{a} q^{3/2}/c, −c \sqrt{a q}, q; \\ \sqrt{a} q^{3/2}, c^{2} q, q^{2}; \end{matrix} q, \frac{c \sqrt{q}}{\sqrt{a}} ] = \frac{(1 + \frac{c \sqrt{q}}{\sqrt{a}} ) (1 − a q) (1 + c \sqrt{a q})}{q (1 − a) (1 + \sqrt{a} c q^{\frac{3}{2}} ) (1 + \frac{c}{\sqrt{a q}} )}

(ψ̃_{q}(−\frac{c \sqrt{q}}{\sqrt{a}}) − ψ̃_{q}(−c \sqrt{a q}) + \frac{ψ̃_{q^{2}}(a q) − ψ̃_{q^{2}}(q) + ψ̃_{q^{2}}(c^{2} q) − ψ̃_{q^{2}}(\frac{c^{2} q}{a})}{1 + q}) . (6.6)

This identity is a q-analogue for [10, second identity on p. 272] and for

7F_{6}[ \begin{matrix}\frac{7}{4} + \frac{a}{4} + \frac{c}{2}, 1, 1 + a, 1 + c, \frac{3}{2} + \frac{a}{2} − c, \frac{1}{2} + \frac{a}{2} + c, 1; \\ \frac{3}{4} + \frac{a}{4} + \frac{c}{2}, \frac{3}{2} + \frac{a}{2} + c, \frac{3}{2} − \frac{a}{2} + c, \frac{3}{2} + \frac{a}{2}, 1 + 2c, 2; \end{matrix} −1]

= \frac{(1 + a) (a − 2c − 1) (1 + a + 2c)}{a (1 + a − 2c) (3 + a + 2c)}(ψ(\frac{1}{2} − \frac{a}{2} + c) − ψ(\frac{1}{2} + \frac{a}{2} + c)

+ \frac{1}{2}(ψ(\frac{1+a}{2}) − ψ(\frac{1}{2}) + ψ(\frac{1+2c}{2}) − ψ(\frac{1−a+2c}{2}))), (6.7)

which is seen by replacing a by q^{a}, c by −q^{c}, and letting q ↑ 1 in (6.6). A nice limiting case results from (6.6) by letting c → 0,

3Φ_{3}[ \begin{matrix}q, aq, q; \\ −\sqrt{a}q^{3/2}, \sqrt{a}q^{3/2}, q^{2}; \end{matrix} q, −q^{2}] = \frac{(1 − aq)}{q (1 − a) (1 + q)}(ψ̃_{q^{2}}(aq) − ψ̃_{q^{2}}(q)) . (6.8)

Replacing a by q^{a} and letting q ↑ 1 gives

3F_{2}[ \begin{matrix}1, 1, 1 + a; \\ \frac{3}{2} + \frac{a}{2}, 2; \end{matrix} \frac{1}{2}] = \frac{(1 + a)}{2 a}(ψ(\frac{1+a}{2}) − ψ(\frac{1}{2})) .  (6.9)

Next in (1.1) choose a_{1} = −\sqrt{a b} c/\sqrt{q}, a_{2} = i (a b)^{1/4} \sqrt{c} q^{3/4}, a_{3} = −i (a b)^{1/4} \sqrt{c} q^{3/4}, a_{4} = a, a_{5} = b, a_{6} = −c, a_{7} = c q, a_{8} = −\sqrt{a b}/c\sqrt{q}, b_{1} = i (a b)^{1/4} \sqrt{c}/q^{1/4}, b_{2} = −i (a b)^{1/4} \sqrt{c}/q^{1/4}, b_{3} = −c \sqrt{b q}/\sqrt{a}, b_{4} = −c \sqrt{a q}/\sqrt{b}, b_{5} = \sqrt{a b q}, b_{6} = −\sqrt{a b}/\sqrt{q}, b_{7} = c^{2} q, z = c\sqrt{q}/\sqrt{a b}. The resulting _{8}Φ_{7} at the right-hand side can be summed by (6.2). Letting b → 1 leads to a q-analogue of [10, identity at the bottom of p. 271],

9Φ_{8}[ \begin{matrix}i a^{1/4} \sqrt{c} q^{5/4}, −i a^{1/4} \sqrt{c} q^{5/4}, c q, −\sqrt{a} q^{3/2}/c, a q, q, −c, −c \sqrt{a}/\sqrt{q}, q; \\ i (a q)^{1/4} \sqrt{c}, −i (a q)^{1/4} \sqrt{c}, −\sqrt{a q}, c^{2}, −c q^{1/2}/\sqrt{a}, −\sqrt{a} c q^{1/2}, \sqrt{a} q^{3/2}, q^{2}; \end{matrix}

q, \frac{c}{\sqrt{a q}} ] = \begin{matrix}(1 − q) & (1 + \frac{c^{2}}{\sqrt{a} q^{\frac{3}{2}}} ) & (1 + c) & (1 + \frac{c \sqrt{a}}{\sqrt{q}}) \\ (1 + \frac{c}{q} ) & (1 + c\sqrt{a q}) & (1 + \sqrt{a q}) \end{matrix} \frac{(a q^{2}, c^{2} q^{2}, c^{2}/a, q^{2}; q^{2})_{∞}}{(q, a q^{3}, c^{2} q, c^{2}/a q; q^{2})_{∞}}

+ \begin{matrix}(1 − \frac{c^{2}}{q} ) & (1 − \sqrt{a q}) & (1 + \begin{matrix}√ & \overline{a} \\ √ & \overline{q} \end{matrix}) & (1 + \frac{c \sqrt{a}}{\sqrt{q}}) \\ (1 − a) & (1 − c) & (1 + \frac{c}{q} ) & (1 + c\sqrt{a q}) \end{matrix}(ψ̃_{q}(−\frac{c}{\sqrt{a q}}) − ψ̃_{q}(−\frac{c \sqrt{a}}{\sqrt{q}})

+ \frac{ψ̃_{q^{2}}(a q) − ψ̃_{q^{2}}(q) + ψ̃_{q^{2}}(\frac{c^{2}}{q}) − ψ̃_{q^{2}}(\frac{c^{2}}{a q})}{1 + q}).  (6.10)

Replacing a by q^{a}, c by −q^{c}, in the limit q ↑ 1 we obtain another _{7}F_{6}-evaluation,

7F_{6}[ \begin{matrix}\frac{5}{4} + \frac{a}{4} + \frac{c}{2}, \frac{3}{2} + \frac{a}{2} − c, 1 + a, 1, c, −\frac{1}{2} + \frac{a}{2} + c, 1; \\ \frac{1}{4} + \frac{a}{4} + \frac{c}{2}, 2c, \frac{1}{2} − \frac{a}{2} + c, \frac{1}{2} + \frac{a}{2} + c, \frac{3}{2} + \frac{a}{2}, 2; \end{matrix} −1]

= \frac{c (2c + a − 1)}{2 (c − 1) (2c + a + 1)}Γ [ \begin{matrix}\frac{1}{2}, \frac{a+3}{2}, \frac{2c+1}{2}, \frac{2c−a−1}{2} \\ \frac{a+2}{2}, \frac{2c+2}{2}, \frac{2c−a}{2} \end{matrix} ]

− \frac{(1 + a) (2c − 1) (2c + a − 1)}{2 a (c − 1) (2c + a + 1)}(ψ(c + \frac{a}{2} − \frac{1}{2}) − ψ(c − \frac{a}{2} − \frac{1}{2})

+ \frac{1}{2}(ψ(\frac{1}{2}) − ψ(\frac{a+1}{2}) − ψ(\frac{2c−1}{2}) + ψ(\frac{2c−a−1}{2}))) .  (6.11)

Also letting c → 0 in (6.10) is of interest. We thus obtain an identity related to (6.8),

3Φ_{3}[ \begin{matrix}a q, q, q; \\ −\sqrt{a q}, \sqrt{a} q^{3/2}, q^{2}; \end{matrix} q, −q]

= \frac{(1 − q)}{(1 + \sqrt{a q})} \frac{(a q^{2}, q^{2}; q^{2})_{∞}}{(q, a q^{3}; q^{2})_{∞}} + \begin{matrix}(1 − \sqrt{a q}) & (1 + \begin{matrix}√ & \overline{a} \\ √ & \overline{q} \end{matrix}) \\ (1 − a) & (1 + q) \end{matrix}(ψ̃_{q^{2}}(a q) − ψ̃_{q^{2}}(q)) (6.12)

Finally, if in (1.1) we choose a_{1} = −\sqrt{a b} c/\sqrt{q}, a_{2} = i (a b)^{1/4} \sqrt{c} q^{3/4}, a_{3} = −i (a b)^{1/4} \sqrt{c} q^{3/4}, a_{4} = a, a_{5} = b, a_{6} = c, a_{7} = −c, a_{8} = −\sqrt{a b}/c \sqrt{q}, b_{1} = i (a b)^{1/4} \sqrt{c}/q^{1/4}, b_{2} = −i (a b)^{1/4} \sqrt{c}/q^{1/4}, b_{3} = −c \sqrt{b q}/\sqrt{a}, b_{4} = −c \sqrt{a q}/\sqrt{b}, b_{5} = −\sqrt{a b q}, b_{6} = \sqrt{a b q}, b_{7} = c^{2} q, z = c q^{3/2}/\sqrt{a b}, by applying (6.3) at the right-hand side we obtain

9Φ_{8}[ \begin{matrix}i(a b)^{1/4}\sqrt{c}q^{7/4}, −i(a b)^{1/4}\sqrt{c}q^{7/4}, aq, bq, \\ i(a b)^{1/4}\sqrt{c}q^{3/4}, −i(a b)^{1/4}\sqrt{c}q^{3/4}, −\sqrt{b}cq^{3/2}/\sqrt{a}, −\sqrt{a}cq^{3/2}/\sqrt{b}, \end{matrix}

\begin{matrix}cq, −cq, −\sqrt{a b q}/c, −c \sqrt{a b q}, q; \\ −\sqrt{a b}q^{3/2}, \sqrt{a b}q^{3/2}, c^{2}q^{2}, q^{2}; \end{matrix} q, \frac{cq^{\frac{3}{2}}}{\sqrt{a b}}]

= \frac{(1 − q) (1 − abq) (1 − c^{2}q) (1 + \frac{c \sqrt{a q}}{\sqrt{b}} ) (1 + \frac{c \sqrt{b q}}{\sqrt{a}} )}{q (1 − a) (1 − b) (1 − c) (1 + c) (1 + \frac{c\sqrt{q}}{\sqrt{a b}}) (1 + \sqrt{a b} c q^{\frac{3}{2}} )}

( − \frac{c\sqrt{q}}{\sqrt{a b}} \frac{(−c \sqrt{a b q}, −c \sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a, b, c^{2}q^{2}/a, c^{2}q^{2}/b; q^{2})_{∞}}{(q, abq, c^{2}q, c^{2}q/ab; q^{2})_{∞}}

+ \frac{(−c \sqrt{a b q}, −c\sqrt{q}/\sqrt{a b}; q)_{∞}}{(−c \sqrt{b q}/\sqrt{a}, −c \sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(aq, bq, c^{2}q/a, c^{2}q/b; q^{2})_{∞}}{(q, abq, c^{2}q, c^{2}q/ab; q^{2})_{∞}} − 1).

(6.13)

Letting b → 1 yields a q-analogue of [10, identity at the top of p. 272]

9Φ_{8}[ \begin{matrix}ia^{1/4}\sqrt{c}q^{7/4}, −ia^{1/4}\sqrt{c}q^{7/4}, aq, q, cq, −cq, −\sqrt{a q}/c, \\ ia^{1/4}\sqrt{c}q^{3/4}, −ia^{1/4}\sqrt{c}q^{3/4}, −cq^{3/2}/\sqrt{a}, −\sqrt{a}cq^{3/2}, −\sqrt{a}q^{3/2}, \sqrt{a}q^{3/2}, c^{2}q^{2}, \end{matrix}

\begin{matrix}−c \sqrt{a q}, q; \\ q^{2}; \end{matrix} q, \frac{cq^{\frac{3}{2}}}{\sqrt{a}} ] = −\frac{c (1 − q) (1 + c \sqrt{a q})}{\sqrt{a q} (1 − c) (1 + c) (1 + \sqrt{a}cq^{\frac{3}{2}} )} \frac{(q^{2}, aq^{2}, c^{2}q^{2}, c^{2}q^{2}/a; q^{2})_{∞}}{(c^{2}q/a, q, aq^{3}, c^{2}q^{3}; q^{2})_{∞}}

+ \frac{(1 − aq) (1 − c^{2}q) (1 + c \sqrt{a q})}{q (1 − a) (1 − c) (1 + c) (1 + \sqrt{a}cq^{\frac{3}{2}} )}(ψ̃_{q}(−\frac{c\sqrt{q}}{\sqrt{a}}) − ψ̃_{q}(−c \sqrt{a q})

+ \frac{ψ̃_{q^{2}}(aq) − ψ̃_{q^{2}}(q) + ψ̃_{q^{2}}(c^{2}q) − ψ̃_{q^{2}}(\frac{c^{2}q}{a})}{1 + q}).  (6.14)

A _{7}F_{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 ↑ 1,

7F_{6}[ \begin{matrix}\frac{7}{4} + \frac{a}{4} + \frac{c}{2}, 1 + a, 1, 1 + c, \frac{1}{2} + \frac{a}{2} − c, \frac{1}{2} + \frac{a}{2} + c, 1; \\ \frac{3}{4} + \frac{a}{4} + \frac{c}{2}, \frac{3}{2} − \frac{a}{2} + c, \frac{3}{2} + \frac{a}{2} + c, \frac{3}{2} + \frac{a}{2}, 2 + 2c, 2; \end{matrix} −1]

= \frac{(2c + a + 1)}{2 c (2c + a + 3)}Γ [ \begin{matrix}\frac{1}{2}, \frac{3+a}{2}, \frac{3+2c}{2}, \frac{1−a+2c}{2} \\ \frac{2+a}{2}, \frac{2+2c}{2}, \frac{2−a+2c}{2} \end{matrix} ]

− \frac{(1 + a) (1 + 2c) (2c + a + 1)}{2ac (2c + a + 3)}(ψ(\frac{1}{2} + \frac{a}{2} + c) − ψ(\frac{1}{2} − \frac{a}{2} + c)

+ \frac{1}{2}(ψ(\frac{1}{2}) − ψ(\frac{1+a}{2}) − ψ(\frac{1+2c}{2}) + ψ(\frac{1−a+2c}{2}))).  (6.15)

On the other hand, letting c → 1 in (6.13) leads to a q-analogue of [10, (7)]

9Φ_{8}[ \begin{matrix}i(a b)^{1/4}q^{7/4}, −i(a b)^{1/4}q^{7/4}, aq, bq, q, \\ i(a b)^{1/4}q^{3/4}, −i(a b)^{1/4}q^{3/4}, −\sqrt{b}q^{3/2}/\sqrt{a}, −\sqrt{a}q^{3/2}/\sqrt{b}, −\sqrt{a b}q^{3/2}, \end{matrix}

\begin{matrix}−q, −\sqrt{a b q}, −\sqrt{a b q}, q; \\ \sqrt{a b}q^{3/2}, q^{2}, q^{2}; \end{matrix} q, \frac{q^{\frac{3}{2}}}{\sqrt{a b}}] = \frac{(1 − q) (1 − abq) (1 + \begin{matrix}√ & \overline{a q} \\ √ & \overline{b} \end{matrix} ) (1 + \begin{matrix}√ & \overline{b q} \\ √ & \overline{a} \end{matrix} )}{2 (1 − a) (1 − b) (1 + \begin{matrix}√ & \overline{q} \\ √ & \overline{a b} \end{matrix}) q (1 + \sqrt{a b}q^{\frac{3}{2}} )}

(\begin{matrix}(1 − q) & \sqrt{q} \\ √ & \overline{a b} \end{matrix} \frac{(−\sqrt{a b q}, −\sqrt{q}/\sqrt{a b}; q)_{∞}}{(−\sqrt{b q}/\sqrt{a}, −\sqrt{a q}/\sqrt{b}; q)_{∞}} \frac{(a, b, q^{2}/a, q^{2}/b; q^{2})_{∞}}{(q, abq, q, q/ab; q^{2})_{∞}}

ψ̃q(−\begin{matrix}√ & \overline{a q} \\ √ & \overline{b} \end{matrix}) − ψ̃_{q}(−\begin{matrix}√ & \overline{q} \\ √ & \overline{a b} \end{matrix}) + ψ̃q(−\begin{matrix}√ & \overline{b q} \\ √ & \overline{a} \end{matrix}) − ψ̃q(−\sqrt{a b q}) + \frac{2}{1 + q} (ψ̃q^{2}(q) + ψ̃q^{2}(\frac{q}{ab}))

− \frac{2}{1 + q} \frac{(−\sqrt{a b q}, −\sqrt{q}/\sqrt{a b}; q)_{∞}}{(−\sqrt{b q}/\sqrt{a}, −\sqrt{a q}/\sqrt{b}; q)_{∞}} (\frac{(aq, bq, q/a, q/b; q^{2})_{∞}}{(q, abq, q, q/ab; q^{2})_{∞}} (ψ̃q^{2}(\frac{q}{a}) + ψ̃q^{2}(\frac{q}{b}))

− \begin{matrix}√ & \overline{q} \\ √ & \overline{a b} \end{matrix} \frac{(a, b, q^{2}/a, q^{2}/b; q^{2})_{∞}}{(q, abq, q, q/ab; q^{2})_{∞}} (ψ̃q^{2}(\frac{q^{2}}{a}) + ψ̃q^{2}(\frac{q^{2}}{b})))).  (6.16)

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

Acknowledgements. The present investigation was initiated during the firstnamed 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.

References

1. R. Askey, The q-Gamma and q-Beta functions, Appl. Analysis 8 (1978), 125–141.

2. R. Askey, Ramanujan’s extensions of the Gamma and Beta functions, Amer. Math. Monthly 87 (1980), 346–359.

3. W. N. Bailey, Generalized hypergeometric series, Cambridge University Press, Cambridge, 1935.

4. A. Erdélyi (ed.), Higher Transcendental Functions Vol. I, McGraw–Hill, New York, 1953.

5. G. Gasper and M. Rahman, Basic hypergeometric series, Encyclopedia of Mathematics And Its Applications 35, Cambridge University Press, Cambridge, 1990.

6. M.E.H.Ismail, L. Lorch and M. E. Muldoon, Completely monotonic functions associated with the gamma function and its q-analogues, J. Math. Anal. Appl. 116 (1988), 1–9.

7. F. H. Jackson, A generalization of the functions Γ(n) and x^{n}, Proc. Roy. Soc. London 74 (1904), 64–72.

8. V. L. Kocić, A note on q-gamma function, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 1 (1990), 31–34.

9. J. L. Lavoie, Some evaluations for the generalized hypergeometric series, Math. Comp. 46 (1986), 215–218.

10. J. L. Lavoie, Some summation formulas for the series _{3}F_{2}(1), Math. Comp. 49 (1987), 269–274.

11. Y. L. Luke, Mathematical functions and their approximations, Academic Press, New York, 1975.

12. O. I. Marichev and V. K. Tuan, Some properties of the q-gamma function, Dokl. Akad. Nauk BSSR 26 (1982), 488–491, 571.

13. D. S. Moak, The q-gamma function for q > 1, Aequationes Math. 20 (1980), 278–285.

14. D. S. Moak, The q-gamma function for x < 0, Aequationes Math. 21 (1980), 179–191.

15. L. J. Slater, A note on equivalent product theorems, Math. Gazette 38 (1954), 127–128.

16. L. J. Slater, Generalized hypergeometric functions, Cambridge University Press, Cambridge, 1966.

17. H. M. Srivastava, A simple algorithm for the evaluation of a class of generalized hypergeometric series, Stud. Appl. Math. 86 (1992), 79–86.

18. J. Thomae, Beiträge zur Theorie der durch die Heinesche Reihe …, J. Reine Angew. Math. 70 (1869), 258–281.

19. A. Verma and V. K. Jain, Some summation formulae for nonterminating basic hypergeometric series, SIAM J. Math. Anal. 16 (1985), 647–655.

20. G. N. Watson, The integral formula for generalized Legendre functions, Proc. London Math. Soc. (2) 17 (1917), 241–246.

21. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, 4th edition, Cambridge University Press, Cambridge, 1965.

Institut für Mathematik der Universität Wien, Strudlhofgasse 4, A-1090 Wien, Austria. E-mail: KRATT@PAP.UNIVIE.AC.AT.

Department of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W 3P4, Canada. E-mail: HMSRI@UVVM.UVIC.CA or HARIMS@UVVM.UVIC.CA.