TRIBASIC INTEGRALS AND IDENTITIES

OF ROGERS-RAMANUJAN TYPE

M. E. H. Ismail and D. Stanton

Abstract. Some integrals involving three bases are evaluated as infinite products using complex analysis. Many special cases of these integrals may be evaluated in another way to find infinite sum representations for these infinite products. The resulting identities are identities of Rogers-Ramanujan type. Some integer partition interpretations of these identities are given. Generalizations of the Rogers-Ramanujan type identities involving polynomials are given, again as corollaries of integral evaluations.

1. Introduction. The purpose of this paper is to show that integral evaluations involving infinite products with three independent bases lead to identities of Rogers-Ramanujan type.

In [11] it was shown that the Rogers-Ramanujan identities of modulus 5 follow from evaluating the bibasic integral

\frac{(qt^{2}; q)_{∞}}{2π} ∫_{0}^{π} \frac{(q^{5}, e^{2iθ}, e^{−2iθ}; q^{5})_{∞}}{(te^{iθ}, te^{−iθ}; q)_{∞}} dθ.

We generalize the above integral by replacing the three bases of the infinite products q^{5}, q and q by independent bases s, p and q. The resulting integral can be evaluated as an infinite product for special values of t. By specializing s, p, and q the product sides of identities of Rogers-Ramanujan type appear. The special values of t for which such identities exist can be found by considering the singularities of the integrals as functions of t.

To find the sum sides of these identities of Rogers-Ramanujan type, we use techniques from classical orthogonal polynomials. Once s has been specialized, we realize that evaluating these integrals is equivalent to a certain connection coefficient problem between orthogonal polynomials which are basic hypergeometric series with arbitrary bases p and q. The weight function and generating function for the q-Hermite polynomials motivates our choices of integrals. Judicious choices of the base p will allow for an explicit solution of the connection coefficient problem and thus a different evaluation of the integral as a sum.

The tribasic integrals are given in Theorems 2, 3, 4 and 5 of §2. The details of the orthogonal polynomial method are described in §3. It is applied to the q-Hermite polynomials in §4, and another set of polynomials is similarly considered in §5. The four classes of integrals we evaluate are given in (4.1), (4.2), (4.3), and (5.1). The often lengthy proofs of computational lemmas in §4 and §5 are collected in an Appendix, §9. §6 contains a double series identity of Rogers-Ramanujan type.

Research partially supported by NSF grants DMS 99-70865, DMS 99-70627, and Liu Bie Ju Center for Mathematical Sciences.

It is this master identity that contains all the special cases that lead to the single sum identities of §5. Finally the multisum identities are developed in §7.

In addition we identify polynomials that correspond to the polynomials {a_{n}(q)} and {b_{n}(q)} introduced by Schur [15] in his proof of the Rogers-Ramanujan identities. In particular we extend some of the results in [11] to moduli other than 5. §8 contains remarks about certain aspects of our work. Combinatorial interpretations and positivity results are found throughout the paper.

We will use the standard notation for q-series and infinite products found in [5], [12].

We use the Jacobi triple identity

(1.1)  (x, q/x, q; q)_{∞} = ∑_{n=−∞}^{∞}(−1)^{n}q^{\binom{n}{2}}x^{n},

and the quintuple product identity,

(q, 1/x, qx, q; q)_{∞}(qx^{2}, q/x^{2}; q^{2})_{∞} = ∑_{m=−∞}^{∞} x^{3m}q^{(3m^{2}+m)/2}(1 − q^{−m}/x)
(1.2)  =(q^{3}, −q^{2}x^{3}, −q/x^{3}; q^{3})_{∞} − (q^{3}, −qx^{3}, −q^{2}/x^{3}; q^{3})_{∞}/x.

We will also use methods from complex analysis in §2.

2. Schwarz’s theorem and integral evaluations. In this section we give in Theorems 2 and 3 integral evaluations each involving three independent bases p, q, and s. These evaluations follow from Schwarz’s theorem for the Poisson kernel, the relevant variation is stated in Theorem 1. By specializing one of the bases, the integral of Theorem 3 may be analytic continued as a function of t, the explicit bibasic evaluations are given in Theorems 4 and 5.

Schwarz’s theorem [1, (§6.4)] asserts that if f(θ) is piecewise continuous on [0, 2π] then

lim_{z→re^{iϕ}} ∫_{0}^{2π} \frac{r^{2} − |z|^{2}}{2π|re^{iθ} − z|^{2}} f(θ)dθ = f(ϕ),

provided that f is continuous at ϕ. We shall use a minor modification of Schwarz’s result.

Theorem 1. Let f(θ, z) be continuous in θ for θ ∈ [0, 2π], and for all z so that r ≥ |z| ≥ r − ϵ for some positive ϵ. Assume further that f(θ, z) converges to f(θ, re^{iϕ}) as z → re^{iϕ} uniformly in θ, for θ ∈ [0, 2π]. Then

lim_{z→re^{iϕ}} ∫_{0}^{2π} \frac{(r^{2} − |z|^{2}) f(θ, z)dθ}{2π|re^{iθ} − z|^{2}} = f(ϕ, re^{iϕ}).

Proof. Since ∫_{0}^{2π}[r^{2} − |z|^{2}]dθ/[|re^{iθ} − z|^{2}] = 2π, we obtain

|∫_{0}^{2π} \frac{r^{2} − |z|^{2}}{2π|re^{iθ} − z|^{2}} f(θ, z)dθ − f(ϕ, re^{iϕ})|

= |∫_{0}^{2π} \frac{r^{2} − |z|^{2}}{2π|re^{iθ} − z|^{2}} [f(θ, z) − f(θ, re^{iϕ}) + f(θ, re^{iϕ}) − f(ϕ, re^{iϕ})] dθ|

≤ sup_{θ∈[0,2π]} |f(θ, z) − f(θ, re^{iϕ})| + |∫_{0}^{2π} \frac{r^{2} − |z|^{2}}{2π|re^{iθ} − z|^{2}} f(θ, re^{iϕ})dθ − f(ϕ, re^{iϕ})| .

The first term tends to zero as z → re^{iϕ} by the uniform convergence hypothesis, while the second term tends to zero from Schwarz’s theorem. □

We next apply Theorem 1 to find limiting values of the tribasic integrals

(2.1)  G_{1}(t, p, q, s) = \frac{(−t; q)_{∞}}{2π} ∫_{0}^{π} \frac{(s, e^{2iθ}, e^{−2iθ}; s)_{∞}}{(te^{2iθ}, te^{−2iθ}; p)_{∞}} dθ, |t| < 1.

and Theorem 2. We have

lim_{t→−1} G_{1}(t, p, q, s) = \frac{(q; q)_{∞}}{(p, p; p)_{∞}} (s, −s, −s; s)_{∞}.

Proof. Let

F (z) = (s, z^{2}, 1/z^{2}; s)_{∞}.

Clearly

lim_{t→−1} G_{1}(t, p, q, s) = lim_{t→−1} \frac{(−t; q)_{∞}}{4π} ∫_{0}^{2π} \frac{F (e^{iθ/2}) dθ}{(te^{iθ}, te^{−iθ}; p)_{∞}}
=\frac{(q; q)_{∞}}{8π} lim_{t→−1} ∫_{0}^{2π} \frac{(1 − t^{2})F (e^{iθ/2}) dθ}{|e^{iθ} − t|^{2}(tpe^{iθ}, tpe^{−iθ}; p)_{∞}}
=\frac{1}{4} \frac{(q; q)_{∞}F (e^{iπ/2})}{(−pe^{iπ}, −pe^{−iπ}; p)_{∞}}
=\frac{(q; q)_{∞}}{(p, p; p)_{∞}} (s, −s, −s; s)_{∞},

by applying Theorem 1 with ϕ = π, r = 1, and f(θ, z) = F (e^{iθ/2})/(zpe^{iθ}, zpe^{−iθ}; p)_{∞}. □ Another tribasic integral is defined by

(2.2)  G_{2}(t, p, q, s) = \frac{(pt^{2}; q)_{∞}}{2π} ∫_{0}^{π} \frac{(s, e^{2iθ}, e^{−2iθ}; s)_{∞}}{(te^{iθ}, te^{−iθ}; p)_{∞}} dθ, |t| < 1.

We next consider an analytic continuation and a limiting behavior of G_{2}. Theorem 3. The function G_{2}(t, p, q, s) can be analytically continued to the disc |t| < p^{−1/2}. The analytic continuation to p^{1/2} < |t| < p^{−1/2} is given by

G_{2}(t, p, q, s) = \frac{(pt^{2}; q)_{∞}}{4π} ∫_{0}^{2π} \frac{(s, e^{2iθ}/p, pe^{−2iθ}; s)_{∞}dθ}{(tp^{−1/2}e^{iθ}, tp^{1/2}e^{−iθ}; p)_{∞}} +\frac{(s, 1/t^{2}, st^{2}; s)_{∞}(pt^{2}; q)_{∞}}{2(p, pt^{2}; p)_{∞}} .

Furthermore

lim_{t→p^{−1/2}} G_{2}(t, p, q, s) = \frac{(s, p, s/p; s)_{∞}(q; q)_{∞}}{(p, p; p)_{∞}} .

Proof. We first analytically continue G_{2}(t, p, q, s) to the annulus |t| ∈ (p^{1/2}, p^{−1/2}). Note that for p^{1/2} < t < 1 we have

G_{2}(t, p, q, s) =\frac{(pt^{2}; q)_{∞}}{4π} ∫_{−π}^{π} \frac{(s, e^{2iθ}, e^{−2iθ}; s)_{∞}}{(te^{iθ}, te^{−iθ}; p)_{∞}} dθ
=\frac{(pt^{2}; q)_{∞}}{4π} ∫_{|z|=1} \frac{F (z)}{(tz, t/z; p)_{∞}} \frac{dz}{iz}
=\frac{(pt^{2}; q)_{∞}}{4π} ∫_{|z|=p^{−1/2}} \frac{F (z)}{(tz, t/z; p)_{∞}} \frac{dz}{iz} − R

where R is the sum of residues at the poles in the domain 1 < |z| < p^{−1/2}. The only pole is at z = 1/t, thus

R = −\frac{(s, 1/t^{2}, t^{2}; s)_{∞}(pt^{2}; q)_{∞}}{2(p, t^{2}; p)_{∞}} ,

and

G_{2}(t, p, q, s) = \frac{(pt^{2}; q)_{∞}}{4π} ∫_{0}^{2π} \frac{F (p^{−1/2}e^{iθ})}{(tp^{−1/2}e^{iθ}, tp^{1/2}e^{−iθ}; p)_{∞}} dθ
+\frac{(s, 1/t^{2}, st^{2}; s)_{∞}(pt^{2}; q)_{∞}}{2(p, pt^{2}; p)_{∞}} ,

which is Theorem 3. The right-hand side of G_{2}(t, p, q, s) is clearly analytic in t for p^{1/2} < |t| < p^{−1/2}. The limiting case follows from Theorem 1 with ϕ = 0, r = p^{−1/2}, and

f(θ, z) = F (p^{−1/2}e^{iθ})/(1 − tp^{−1/2}e^{iθ})(zp^{3/2}e^{iθ}, zp^{3/2}e^{−iθ}; p)_{∞}.

□

The next result gives an analytic continuation of G_{2}(t, p, q, s) valid for 1 < |t| < 1/p, but to do so we have to choose q = p.

Theorem 4. An analytic continuation of G_{2}(t, p, p, s) to 1 < |t| < 1/p may be given by

G_{2}(t, p, p, s) =\frac{(pt^{2}; p)_{∞}}{4π} ∫_{0}^{2π} \frac{F (e^{iθ}/p)}{(te^{iθ}/p, tpe^{−iθ}; p)_{∞}} dθ

+ \frac{(s, 1/t^{2}, st^{2}; s)_{∞}}{2(p; p)_{∞}} − \frac{p(s, p^{2}t^{2}, 1/p^{2}t^{2}; s)_{∞}}{2(1 − p)(p; p)_{∞}} .

Furthermore

lim_{t→1/p} G_{2}(t, p, p, s) = \frac{(s, p^{2}, s/p^{2}; s)_{∞}}{(p; p)_{∞}} .

Proof. Theorem 3 gives an analytic continuation of G_{2}(t, p, p, s) to p^{1/2} < |t| < p^{−1/2}. Therefore for 1 < |t| < p^{−1/2}, we have

G_{2}(t, p, p, s)−\frac{(s, 1/t^{2}, st^{2}; s)_{∞}}{2(p; p)_{∞}} = \frac{(pt^{2}; p)_{∞}}{4π} ∫_{|z|=p^{−1/2}} \frac{F (p^{−1/2}z)}{(tp^{−1/2}z, tp^{1/2}/z; p)_{∞}} \frac{dz}{iz} −R,

and R is the contribution from the poles in 1 < |z| < p^{−1/2}. Again the only pole in this annulus is when ztp^{1/2} = 1. Thus

R = \frac{p(s, p^{2}t^{2}, 1/p^{2}t^{2}; s)_{∞}}{2(1 − p)(p; p)_{∞}}

and Theorem 4 follows. The limiting case follows from Theorem 1. □

We now generalize Theorem 4 by analytically continuing G_{2}(t, p, p, s) to annuli of the type p^{1−k/2} < |t| < p^{−k/2}.

Theorem 5. For k = 1, 2, …, G_{2}(t, p, p, s) can be continued to p^{1−k/2} < |t| < p^{−k/2} through

G_{2}(t, p, p, s) = \frac{(pt^{2}; p)_{∞}}{4π} ∫_{0}^{2π} \frac{F (e^{iθ}p^{−k/2})}{(te^{iθ}p^{−k/2}, tp^{k/2}e^{−iθ}; p)_{∞}} dθ

+\frac{(s; s)_{∞}}{2(1 − t^{2})(p; p)_{∞}} ∑_{j=0}^{k−1} \frac{(t^{2}; p)_{j}(t^{2}p^{2j}, t^{−2}p^{−2j}; s)_{∞}}{(1/p; 1/p)_{j}} .

Furthermore

lim_{t→p^{−k/2}} G_{2}(t, p, p, s) = \frac{1}{2(1 − p^{−k})(p; p)_{∞}} ∑_{j=0}^{k} [ \begin{matrix}k \\ j \end{matrix} ]_{p} (s, p^{2j−k}, p^{k−2j}; s)_{∞}p^{j(j−k)}.

Proof. The proof of the analytic continuation formula is by induction on k and is very similar to the proof of Theorem 4. In going from a value of k to k + 1 we first restrict t to p^{(1−k)/2} < |t| < p^{−k/2}. We then replace the integral on |z| = 1 by an integral on |z| = p^{−1/2}, pick up only one pole in the ring 1 < |z| < p^{−1/2}, namely z = p^{−k/2}/t, compute the residue then replace z by zq^{−1/2} in the integral. To establish the limiting result, let t → p^{−k/2} in G_{2}(t, p, p, s) and apply Theorem 1. The result is that the limit is

∑_{j=0}^{k} \frac{(p^{−k}; p)_{j}(s, p^{2j−k}, p^{k−2j}; s)_{∞}}{2(p; p)_{∞}(1 − p^{−k})(1/p; 1/p)_{j}} ,

and the result follows. □

3. q-Hermite polynomials. The tribasic integrals G_{1}(t, p, q, s) and G_{2}(t, p, q, s) of §2 are closely related to q-Hermite polynomials via an orthogonality relation. In this section we give the basic facts about the q-Hermite polynomials. We then show that evaluating a wide class of integrals, including G_{1}(t, p, q, s) and G_{2}(t, p, q, s) is equivalent to solving a connection coefficient problem for polynomials.

The q-Hermite polynomials H_{n}(x|q) may be defined [12, p. 26] by the generating function

(3.1)  ∑_{n=0}^{∞} \frac{H_{n}(cos θ|q)}{(q; q)_{n}}t^{n} = \frac{1}{(te^{iθ}, te^{−iθ}; q)_{∞}} .

The key fact we need for these polynomials is the orthogonality relation [12, p. 188]

(3.2)  \frac{(q; q)_{∞}}{2π} ∫_{0}^{π} H_{m}(cos θ|q)H_{n}(cos θ|q)(e^{2iθ}, e^{−2iθ}; q)_{∞}dθ = (q; q)_{n}δ_{mn}.

Suppose that r_{n}(x) is another set of polynomials, deg(r_{n}) = n, whose generating function is

R(x, t) = ∑_{n=0}^{∞} r_{n}(x)t^{n}.

To evaluate the integral

(3.3)  I = \frac{(q; q)_{∞}}{2π} ∫_{0}^{π} R(cos θ, t)(e^{2iθ}, e^{−2iθ}; q)_{∞}dθ,

it is sufficient to know the connection coefficients c_{nk}

r_{n}(x) = ∑_{k=0}^{n} c_{n,k}H_{k}(x|q).

The orthogonality relation (3.2) implies

(3.4)  I = ∑_{n=0}^{∞} c_{n,0}t^{n}.

This is our alternate evaluation of an integral, and will be the sum side of a Rogers-Ramanujan identity. What are required are suitable choices of the generating function R(x, t) for which the constants c_{n,0} can be computed.

Two such examples for R(x, t) are given by q-Hermite polynomials themselves,

(3.5a)  ∑_{n=0}^{∞} \frac{H_{2n}(x|q)}{(q^{2}; q^{2})_{n}}t^{n} = \frac{(−t; q)_{∞}}{(te^{2iθ}, te^{−2iθ}; q^{2})_{∞}} ,

(3.5b)  ∑_{n=0}^{∞} \frac{H_{n}(x|q^{2})}{(q; q)_{n}}t^{n} = \frac{(qt^{2}; q^{2})_{∞}}{(te^{iθ}, te^{−iθ}; q)_{∞}} .

Proof of (3.5a) and (3.5b). The Askey-Wilson integral [12, p. 140]

(3.6)  \begin{matrix}\frac{(q; q)_{∞}}{2π} ∫_{0}^{π} & \frac{(e^{2iθ}, e^{−2iθ}; q)_{∞}}{(ae^{iθ}, ae^{−iθ}, be^{iθ}, be^{−iθ}, ce^{iθ}, ce^{−iθ}, de^{iθ}, de^{−iθ}; q)_{∞}} dθ \\ = & \frac{(abcd; q)_{∞}}{(ab, ac, ad, bc, bd, cd; q)_{∞}} , \end{matrix}

implies (3.5a) and (3.5b). To show (3.5a), expand the right-hand side in q-Hermite polynomials as

R = ∑_{m=0}^{∞} b_{m}(t)H_{m}(x|q).

Then multiply R by the q-Hermite generating function 1/(se^{iθ}, se^{−iθ}; q)_{∞}, and integrate against the q-Hermite weight. Then (3.2) and (3.6) with a = \sqrt{t}, b = −\sqrt{t}, c = s, and d = 0 imply

∑_{m=0}^{∞} b_{m}(t)s^{m} = \frac{1}{(ts^{2}; q^{2})_{∞}} .

The q-binomial theorem [12, p. 7] then implies that b_{2m}(t) = t^{m}/(q^{2}; q^{2})_{m} and b_{2m+1}(t) = 0, completing the proof of (3.5a). The proof of (3.5b) is nearly identical. □

Besides the orthogonality relation (3.2) we will use

(3.7) \frac{(q; q)_{∞}}{2π} ∫_{0}^{π} e^{2inθ}(e^{2iθ}, e^{−2iθ}; q)_{∞}dθ = (−1)^{n}q^{\binom{n}{2}}(1 + q^{n})/2, n = 0, 1, · · · .

which is well-known and follows from (1.1).

4. q-Hermite change of base. In this section we consider the integral (3.3) for the generating functions (3.5a), and (3.5b) on base p,

(4.1)  J_{p,q}(t) = G_{1}(t, p^{2}, p, q) = \frac{(q; q)_{∞}(−t; p)_{∞}}{2π} ∫_{0}^{π} \frac{(e^{2iθ}, e^{−2iθ}; q)_{∞}}{(te^{2iθ}, te^{−2iθ}; p^{2})_{∞}} dθ,

(4.2) H_{p,q}(t) = G_{2}(t, p, p^{2}, q^{2}) = \frac{(q^{2}; q^{2})_{∞}(pt^{2}; p^{2})_{∞}}{2π} ∫_{0}^{π} \frac{(e^{2iθ}, e^{−2iθ}; q^{2})_{∞}}{(te^{iθ}, te^{−iθ}; p)_{∞}} dθ,

along with the p-version of (3.1)

(4.3)  I_{p,q}(t) = \frac{G_{2}(t, p, q′, q)}{(pt^{2}; q′)_{∞}} = \frac{(q; q)_{∞}}{2π} ∫_{0}^{π} \frac{(e^{2iθ}, e^{−2iθ}; q)_{∞}}{(te^{iθ}, te^{−iθ}; p)_{∞}} dθ,

where q′ is arbitrary. The bulk of this section is devoted to evaluating these integrals via the method of §3, and giving partition theory consequences of the results.

Theorems 2 and 3 give the infinite product evaluations

(4.1a)  lim_{t→−1} J_{p,q}(t) =\frac{(q, −q, −q; q)_{∞}}{(p, −p, −p; p)_{∞}} ,
(4.2a)  lim_{t→p^{−1/2}} H_{p,q}(t) =\frac{(−p; p)_{∞}(q^{2}, p, q^{2}/p; q^{2})_{∞}}{(p; p)_{∞}} .

while Theorem 3 implies

(4.3a)  lim_{t→p^{−1/2}}(1 − pt^{2})I_{p,q}(t) = \frac{(q, p, q/p; q)_{∞}}{(p, p; p)_{∞}} .

For the sum sides, as in §3 let

H_{n}(x|p) = ∑_{k=0}^{n/2} c_{n,n−2k}(p, q)H_{n−2k}(x|q),

so that

(4.1b)  J_{p,q}(t) = ∑_{n=0}^{∞} \frac{c_{2n,0}(p, q)}{(p^{2}; p^{2})_{n}}t^{n},

(4.2b)  H_{p,q}(t) = ∑_{n=0}^{∞} \frac{c_{2n,0}(p^{2}, q)}{(p; p)_{2n}}t^{2n},

(4.3b)  I_{p,q}(t) = ∑_{n=0}^{∞} \frac{c_{2n,0}(p, q)}{(p; p)_{2n}}t^{2n}.

Thus all that remains is to find c_{n0}(p, q). However c_{nk}(p, q) is known [],[11, (7.2)] to be given as a sum

\begin{aligned}c_{n,n−2k}(p, q) = & ∑_{j=0}^{k}(−1)^{j}p^{k−j}q^{\binom{j+1}{2}} [ \begin{matrix}n − 2k + j \\ j \end{matrix} ]_{q} \\ & × ([ \begin{matrix}n \\ k − j \end{matrix} ]_{p} − p^{n−2k+2j+1} [ \begin{matrix}n \\ k − j − 1 \end{matrix} ]_{p}) \end{aligned}

so that

(4.4)  c_{2n,0}(p, q) = ∑_{j=−n}^{n}(−1)^{j}p^{n−j}q^{\binom{j+1}{2}} [ \begin{matrix}2n \\ n − j \end{matrix} ]_{p} .

We will be concerned with the values of p such that constant term c_{2n,0}(p, q) explicitly factors. For example, it is well-known that

c_{2n,0}(q^{−1}, q) = q^{−n^{2}}(q^{2n}; q^{−1})_{n} = q^{−n^{2}}(q; q)_{2n}/(q; q)_{n}

but other choices of p may also be made (see the Appendix):

(4.5a)  p =q^{2}, c_{2n,0}(q^{2}, q) = (−1)^{n}q^{n^{2}}(q; q^{2})_{n}
(4.5b)  p = − q, c_{2n,0}(−q, q) = (−q)^{n}(−1; q^{2})_{n}
(4.5c)  p =q^{1/2}, c_{2n,0}(q^{1/2}, q) = q^{n/2}(q^{1/2}; q)_{n}
(4.5d)  p =q^{1/3}, c_{2n,0}(q^{1/3}, q) = q^{n/3}(q^{2n/3}; q^{−1/3})_{n}
(4.5e)  p =q^{2/3}, c_{2n,0}(q^{2/3}, q) = q^{2n/3}(q^{1/3}; q^{2/3})_{n}

Proofs of the five evaluations (4.5a)-(4.5e) of the sum (4.4) will be given in the Appendix, §9.

For the J_{p,q}(t) integral (4.1a), the evaluations (4.5) give

(4.6a)  J_{q^{2},q}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{4}; q^{4})_{n}} q^{n^{2}}(−t)^{n},

(4.6b)  J_{−q,q}(t) = ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q^{2}; q^{2})_{n}} (−qt)^{n} = \frac{(qt; q^{2})_{∞}}{(−qt; q^{2})_{∞}} ,

(4.6c)  J_{q,q^{2}}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{n}} (qt)^{n} = \frac{(q^{2}t; q^{2})_{∞}}{(qt; q^{2})_{∞}} ,

(4.6d)  J_{q,q^{3}}(t) = ∑_{n=0}^{∞} \frac{(q; q)_{2n}}{(q; q)_{n}(q^{2}; q^{2})_{n}} (qt)^{n},

(4.6e)  J_{q^{2},q^{3}}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{4}; q^{4})_{n}} (q^{2}t)^{n}.

The product evaluations for (4.5b) and (4.5c) follow from the Askey-Wilson integral (3.6). If t = −1 then (4.5a) factors, by a limiting case of a _{2}ϕ_{1} summation theorem [12, (II.8)], while (4.5d) and (4.5e) give new results for t = −1 via (4.1a).

Theorem 6. We have

∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(−q; −q)_{n}} q^{n} = \frac{(−q^{1}, −q^{5}; q^{6})_{∞}}{(q^{2}, q^{4}; q^{6})_{∞}} = \frac{1}{(q^{1}, q^{4}, q^{5}, q^{7}, q^{8}, q^{11}; q^{12})_{∞}} (A)

∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{4}; q^{4})_{n}} (−q^{2})^{n} = \frac{(q^{2}, q^{6}, q^{6}, q^{10}; q^{12})_{∞}}{(q^{3}, q^{4}, q^{8}, q^{9}; q^{12})_{∞}} (B)

Proof. Use (4.6d), (4.6e), (4.1b) and (4.1a), and replace q by −q to obtain (A). □

Using

\frac{(−q; q^{2})_{n}}{(−q; −q)_{n}} q^{n} = \{ \begin{matrix}\frac{(−q^{1+2N};q^{2})_{N}}{(q^{2};q^{2})_{N}} q^{2N} if n = 2N, \\ \frac{(−q^{3+2N};q^{2})_{N}}{(q^{2};q^{2})_{N}} q^{2N+1} if n = 2N + 1, \end{matrix}

Theorem 6 (A) has the following integer partition interpretation.

Corollary 1. Let A(n) be the number of integer partitions of n into parts congruent to ±1, ±4, or ±5 mod 12. Let B(n) be the number of integer partitions of n

(1) whose odd parts are distinct and lie between the largest even part and twice the largest even part, or

(2) which have a single part of size 1, and whose other odd parts are distinct, greater than two more than the largest even part, and at most one more than twice the largest even part.

Then A(n) = B(n).

In Corollary 1, the partitions for A(10) = 8 are

811, 7111, 55, 541, 511111, 4411, 4111111, 1111111111,

while those for B(10) = 8 are

10, 82, 641, 622, 442, 4222, 22222, 5221.

For the H_{p,q}(t) integral (4.1b), the evaluations (4.5) become

(4.7a)  H_{q^{2},q}(t) = ∑_{n=0}^{∞} \frac{q^{2n^{2}}(−t^{2})^{n}}{(q^{4}; q^{4})_{n}} = (q^{2}t^{2}; q^{4})_{∞},

(4.7b)  H_{iq,q}(t) = ∑_{n=0}^{∞} \frac{(−1; q^{4})_{n}}{(iq; iq)_{2n}} (−qt)^{2n},

(4.7c)  H_{q,q^{2}}(t) = ∑_{n=0}^{∞} \frac{(q^{2}; q^{4})_{n}}{(q; q)_{2n}} (q^{2}t^{2})^{n} = \frac{(−q^{3}t^{2}; q^{2})_{∞}}{(q^{2}t; q^{2})_{∞}} ,

(4.7d)  H_{q,q^{3}}(t) = ∑_{n=0}^{∞} \frac{(−q; q)_{2n}}{(q^{2}; q^{2})_{n}} (qt)^{2n},

(4.7e)  H_{q^{2},q^{3}}(t) = ∑_{n=0}^{∞} \frac{(q^{4}t^{2})^{n}}{(q^{4}; q^{4})_{n}} = \frac{1}{(q^{4}t^{2}; q^{4})_{∞}} .

Again (3.6) establishes the product representations for (4.7a), (4.7c) and (4.7e).

Theorem 7. We have

∑_{n=0}^{∞} \frac{(−q; q)_{2n}}{(q^{2}; q^{2})_{n}} q^{n} = \frac{(q^{6}; q^{6})_{∞}}{(q; q)_{∞}(q^{3}; q^{6})_{∞}} .

Proof. Use (4.2a) and (4.7d). □

Using

\frac{(−q; q)_{2n}}{(q^{2}; q^{2})_{n}} q^{n} = \frac{(−q^{1+n}; q)_{n}}{(q; q)_{n}} q^{n}

we have the following corollary.

Corollary 2. Let A(n) be the number of integer partitions of n into parts not congruent to 0 mod 6, whose parts which are congruent to 3 mod 6 are colored red and blue. Let B(n) be the number of integer partitions of n whose parts are colored red and blue, the red parts are distinct and greater than the largest blue part and at most twice the largest blue part. Then A(n) = B(n).

In Corollary 2, the partitions for A(5) = 9 are

5, 41, 32, \underline{3}2, 311, \underline{3}11, 221, 2111, 11111,

where the red parts are underlined, while those for B(5) = 9 are

11111, \underline{2}111, \underline{3}2, 221, 2111, 32, 311, 41, 5.

Finally we turn to I_{p,q}(t). The evaluations for I_{p,q}(t) which follow from (4.5c) and (4.5d)

I_{q,q^{3}}(t) = \frac{1}{(qt^{2}; q)_{∞}} , I_{q,q^{2}}(t) = \frac{1}{(qt^{2}; q)_{∞}}

are independently established by the The Askey-Wilson integral (2.7). The remaining three evaluations (4.5a), (4.5b) and (4.5e) become

(4.8a)  I_{q^{2},q}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} (−1)^{n}q^{n^{2}}t^{2n}

(4.8b)  I_{−q,q}(t) = ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(−q; −q)_{2n}} (−qt^{2})^{n}

(4.8e)  I_{q^{2},q^{3}}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} (q^{2}t^{2})^{n}

whose (4.3a) evaluations are

(4.9a)  ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} (−1)^{n}q^{(n−1)^{2}−n}(q^{2n−1} + q((1 − q^{4n})(1 + q^{2n−1})) = 0.

(4.9b)  ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n−1}}{(−q; −q)_{2n}} q^{2(n−1)}(1 − q + q^{2} + q^{2n+1}) = \frac{(−1; q^{2})_{∞}}{(−q, q^{2}; q^{2})_{∞}} .

(4.9e)  ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{2n−1}(q^{2n+1} − 1 + q^{4n}) = \frac{(q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} .

One may evaluate the integral I_{−q,q}(t) by other elementary means for special choices of t. Here we take t = 1 and t = q.

Theorem 8. We have

(A)  ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q; q)_{2n}} q^{n} =\frac{(−q; q^{2})_{∞}}{(q; q)_{∞}} (q^{6}, −q^{2}, −q^{4}; q^{6})_{∞},

(B)  ∑_{n=0}^{∞} \frac{(−q^{2}; q^{2})_{n}}{(q; q)_{2n+1}} q^{n} =\frac{(−q; q^{2})_{∞}}{(q; q)_{∞}} (q^{6}, −q^{6}, −q^{6}; q^{6})_{∞}.

Proof. For (A) we evaluate

I_{−q,q}(1) =\frac{(q; q)_{∞}}{2π} ∫_{0}^{π} \frac{(e^{2iθ}, e^{−2iθ}; q)_{∞}}{(e^{iθ}, e^{−iθ}; −q)_{∞}} dθ =
=\frac{(q; q)_{∞}}{4π} ∫_{−π}^{π} \frac{(e^{2iθ}, e^{−2iθ}, q^{2}e^{2iθ}, q^{2}e^{−2iθ}; q^{4})_{∞}(qe^{2iθ}, qe^{−2iθ}; q^{2})_{∞}}{(e^{iθ}, e^{−iθ}, −qe^{iθ}, −qe^{−iθ}; q^{2})_{∞}} dθ
=\frac{(q; q)_{∞}}{4π} ∫_{−π}^{π}(−e^{iθ}, −e^{−iθ}, qe^{iθ}, qe^{−iθ}, qe^{2iθ}, qe^{−2iθ}; q^{2})_{∞}dθ.

Using the Jacobi triple product identity (1.1) we have

I_{−q,q}(1) = \frac{(q; q)_{∞}}{4π(q^{2}; q^{2})_{∞}^{3}} ∑_{m,n,j=−∞}^{∞}(−1)^{n+j}q^{m^{2}+m+n^{2}+j^{2}} ∫_{−π}^{π} e^{i(m+n+2j)θ}(1 + e^{iθ})dθ.

This integral has two terms, the first forces n = −2j − m, the second forces n = −m − 2j − 1. In the first case the power of q is

m^{2} + m + (m + 2j)^{2} + j^{2} = 2(m + j)^{2} + 3j^{2} + m.

So we can shift m down by j and the resulting double sum factors into two Jacobi triple products. For the second term the power of q is

m^{2} + m + (m + 2j + 1)^{2} + j^{2} = 2(m + j)^{2} + 3j^{2} + 3m + 4j + 1,

and again a shift of m down by j allows an evaluation, equal to the first term. The result is, after replacing q by −q,

I_{q,−q}(1) = \frac{(−q; q^{2})_{∞}}{(q; q)_{∞}} (q^{6}, −q^{2}, −q^{4}; q^{6})_{∞}.

For (B) we can similarly show that

I_{q,−q}(q) = ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q; q)_{2n}} q^{3n}

=\frac{(−q; q^{2})_{∞}}{(q; q)_{∞}} (q^{6}, −q^{2}, −q^{4}; q^{6})_{∞} − q(q^{6}, −1, −q^{6}; q^{6})_{∞},

and subtracting this from (A) gives (B). □

For (4.8a) I_{q^{2},q}(1) and I_{q^{2},q}(q) may be evaluated in a similar manner using the quintuple product identity (1.2). The results are equivalent to two given by Slater [16, (117), (118)].

Proposition 1. We have

(A)  ∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{n^{2}} =\frac{(−q^{1}, −q^{5}, −q^{7}, −q^{9}, −q^{13}; q^{14})_{∞}}{(q^{2}, q^{4}, q^{10}, q^{12}; q^{14})_{∞}} ,

(B)  ∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{n^{2}+2n} =\frac{(−q^{3}, −q^{5}, −q^{7}, −q^{9}, −q^{11}; q^{14})_{∞}}{(q^{4}, q^{6}, q^{8}, q^{10}; q^{14})_{∞}} .

It is clear that both sides of both parts Proposition 1 have positive coefficients as power series in q. Here we state the integer partition version of Proposition 1A.

Corollary 3. Let A(n) be the number of integer partitions of n into parts congruent to 2,4, 10 or 12 mod 14 and distinct parts congruent to 1,5,7,9, or 13 mod 14. Let B(n) be the number of integer partitions of n

(1) whose odd parts are consecutive (starting with 1) and have multiplicity one or two,

(2) whose largest even part is at most two more than twice the largest odd part. Then A(n) = B(n).

In Corollary 3, the partitions for A(12) = 12 are

12, 10 2, 921, 75, 741, 7221, 5421, 52221, 444, 4422, 42222, 222222 while those for B(12) = 12 are

44211, 422211, 2222211, 831, 6321, 4431,
43221, 322221, 43311, 332211, 53211, 5331.

The combinatorial version of Proposition 1B changes the parts mod 14 appropriately for A(n) and adds an even part which is one larger than the largest odd part for B(n).

We do not know the analogous specializations of (4.8e).

5. A second set of polynomials. In this section we carry out the program of §3 for the polynomials s_{n}(x|q) defined by the generating function

∑_{n=0}^{∞} s_{n}(x|q)\frac{t^{n}}{(q; q)_{n}} = \frac{(qt^{2}; q)_{∞}}{(te^{iθ}, te^{−iθ}; q)_{∞}} , x = cos θ.

These polynomials were considered in [11], where it was shown that the choice of p = q^{5} immediately leads to the Rogers-Ramanujan identities of modulus 5.

The integral we are evaluating is

(5.1)  S_{p,q}(t) = G_{2}(t, q, q, p) = \frac{(qt^{2}; q)_{∞}(p; p)_{∞}}{2π} ∫_{0}^{π} \frac{(e^{2iθ}, e^{−2iθ}; p)_{∞}}{(te^{iθ}, te^{−iθ}; q)_{∞}} dθ.

Note that Theorems 3 and 4 imply

(5.1a)  lim_{t→q^{−1}} S_{p,q}(t) =\frac{(p, q^{2}, p/q^{2}; p)_{∞}}{(q; q)_{∞}} ,
(5.1b)  lim_{t→q^{−1/2}} S_{p,q}(t) =\frac{(p, q, p/q; p)_{∞}}{(q; q)_{∞}} .

Let

s_{n}(x|q) = ∑_{k=0}^{n} d_{nk}(p, q)H_{k}(x|p).

Only the even degree polynomials s_{2n}(x|q) have a non-zero constant term. Thus

S_{p,q}(t) = ∑_{n=0}^{∞} d_{2n,0}(p, q)\frac{t^{2n}}{(q; q)_{2n}} .

The value of d_{2n,0}(p, q) was found in [11, (7.4)] to be

d_{2n,0}(p, q) = ∑_{j=0}^{n} \frac{(q; q)_{2n}q^{(n−j)(n+j+2)}p^{\binom{j+1}{2}}(−1)^{j}}{(q; q)_{2j}(q; q)_{n−j}(q^{2j+2}; q)_{n−j}} .

We have nine choices of p for which d_{2n,0}(p, q) may be evaluated (see the Appendix for proofs)

(5.2a)  p =q, d_{2n,0}(q, q) = (−1)^{n}q^{\binom{n+1}{2}}(q; q)_{2n}/(q; q)_{n}
(5.2b)  p =q^{2}, d_{2n,0}(q^{2}, q) = (−1)^{n}q^{n(n+1)}(q; q^{2})_{n}
(5.2c)  p =q^{3}, d_{2n,0}(q^{3}, q) = 0, n > 0
(5.2d)  p =q^{4}, d_{2n,0}(q^{4}, q) = q^{n^{2}+2n}(q; q^{2})_{n}
(5.2e)  p =q^{5}, d_{2n,0}(q^{5}, q) = q^{n^{2}+2n}(q; q)_{2n}/(q; q)_{n}
(5.2f)  p =q^{5/2}, d_{2n,0}(q^{5/2}, q) = (−1)^{n}q^{n(3n+2)/2}(q^{1/2}; q)_{n}
(5.2g)  p =q^{7/2}, d_{2n,0}(q^{7/2}, q) = q^{n(n+2)}(q^{1/2}; q)_{n}
(5.2h)  p = − q^{3}, d_{2n,0}(−q^{3}, q) = q^{n^{2}+2n}(−1; q^{2})_{n}

and

(5.2i) p = ωq^{3}, d_{2n,0}(ωq^{3}, q) = (1−ω)q^{n^{2}+2n}(q^{3}; q^{3})_{n−1}/(q; q)_{n−1}, n > 0, ω^{3} = 1.

One way to motivate the choices of p given above is the value

d_{40}(p, q) = (p − q^{3})(p^{2} + pq^{3} − q^{5} − q^{7})

which simplifies for those choices of p.

The first four choices of p evaluate S_{p,q}(t) for all t, which are again equivalent to special cases of the Askey-Wilson integral,

(5.3a)  S_{q,q}(t) = ∑_{n=0}^{∞} \frac{q^{\binom{n}{2}}(−qt^{2})^{n}}{(q; q)_{n}} = (qt^{2}; q)_{∞},

(5.3b)  S_{q^{2},q}(t) = ∑_{n=0}^{∞} \frac{q^{n(n+1)}(−t^{2})^{n}}{(q^{2}; q^{2})_{n}} = (q^{2}t^{2}; q^{2})_{∞},
(5.3c)  S_{q^{3},q}(t) =1,
(5.3d)  S_{q^{4},q}(t) = ∑_{n=0}^{∞} \frac{q^{n(n+2)}(t^{2})^{n}}{(q^{2}; q^{2})_{n}} = (−q^{3}t^{2}; q^{2})_{∞}.

The S series corresponding to the last five choices do not have closed form sums and give new integral evaluations.

(5.3e)  S_{q^{5},q}(t) = ∑_{n=0}^{∞} \frac{q^{n^{2}+2n}t^{2n}}{(q; q)_{n}} ,

(5.3f)  S_{q^{5},q^{2}}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{n(3n+2)}(−t^{2})^{n},

(5.3g)  S_{q^{7},q^{2}}(t) = ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{2n(n+2)}t^{2n},

(5.3h)  S_{−q^{3},q}(t) = ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q; q)_{2n}} q^{n(n+2)}t^{2n}.

(5.3i)  S_{ωq^{3},q}(t) =1 + (1 − ω) ∑_{n=1}^{∞} \frac{(q^{3}; q^{3})_{n−1}}{(q; q)_{2n}(q; q)_{n−1}} q^{n(n+2)}t^{2n}.

The choices of t = 1/q, 1/q^{2} in (5.3e), combined with (5.1a), (5.1b) give the Rogers-Ramanujan identities of modulus 5

(5.4e1)  ∑_{n=0}^{∞} \frac{q^{n^{2}}}{(q; q)_{n}} = \frac{1}{(q^{1}, q^{4}; q^{5})_{∞}}

(5.4e2)  ∑_{n=0}^{∞} \frac{q^{n^{2}+n}}{(q; q)_{n}} = \frac{1}{(q^{2}, q^{3}; q^{5})_{∞}} .

If we use (5.1a) and (5.1b) in (5.3f)-(5.3h), we find six identities which appear in Slater’s [16] list as (19), (15), (32), (33), (48), (47), respectively.

(5.4f1)  ∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{3n^{2}} = \frac{(−q^{3}, −q^{5}, −q^{7}; q^{10})_{∞}}{(q^{4}, q^{6}; q^{10})_{∞}}

(5.4f2)  ∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{3n^{2}−2n} = \frac{(−q^{1}, −q^{5}, −q^{9}; q^{10})_{∞}}{(q^{2}, q^{8}; q^{10})_{∞}}

(5.4g1)  ∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{2n^{2}+2n} = \frac{(−q^{5}, −q^{7}, −q^{9}; q^{14})_{∞}}{(q^{4}, q^{6}, q^{8}, q^{10}; q^{14})_{∞}}

(5.4g2)  ∑_{n=0}^{∞} \frac{(−q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{2n^{2}} = \frac{(−q^{3}, −q^{7}, −q^{11}; q^{14})_{∞}}{(q^{2}, q^{6}, q^{8}, q^{12}; q^{14})_{∞}}

(5.4h1)  ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q; q)_{2n}} q^{n^{2}+n} = \frac{(−q^{2}, −q^{3}, −q^{4}; q^{6})_{∞}}{(q^{2}, q^{3}, q^{4}; q^{6})_{∞}}

(5.4h2)  ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q; q)_{2n}} q^{n^{2}} = \frac{(−q^{1}; q^{2})_{∞}}{(q; q^{2})_{∞}}

Note that (5.4f) and (5.4g) are companion results for Proposition 1. Thus combinatorial interpretations can be stated by changing the multiplicity of the odd parts to change the power of q in the sum.

Proposition 2. Let A(n) be the number of integer partitions of n into parts congruent to 4 or 6 mod 10 and distinct parts congruent to 3,5, or 7 mod 10. Let B(n) be the number of integer partitions of n

(1) whose odd parts are consecutive (starting with 1) and have multiplicity three or four,

(2) whose largest even part is at most two more than twice the largest odd part. Then A(n) = B(n).

Proposition 3. Let A(n) be the number of integer partitions of n into parts congruent to 2,6,8, or 12 mod 14 and distinct parts congruent to 3,7, or 11 mod 14. Let B(n) be the number of integer partitions of n

(1) whose odd parts are consecutive (starting with 1) and have multiplicity two or three,

(2) whose largest even part is at most two more than twice the largest odd part. Then A(n) = B(n).

For (5.4i) we obtain the following results.

Theorem 9. We have

1 + (1 − ω) ∑_{n=1}^{∞} \frac{(q^{3}; q^{3})_{n−1}}{(q; q)_{2n}(q; q)_{n−1}} q^{n(n+1)} = \frac{(ωq^{3}, ωq^{2}, q; ωq^{3})_{∞}}{(q; q)_{∞}}

1 + (1 − ω) ∑_{n=1}^{∞} \frac{(q^{3}; q^{3})_{n−1}}{(q; q)_{2n}(q; q)_{n−1}} q^{n^{2}} = \frac{(ωq^{3}, q^{2}, ωq; ωq^{3})_{∞}}{(q; q)_{∞}} .

Another form of the integral representation (5.3e) is provided by the Nassrallah-Rahman integral [12, (6.4.1)], as _{3}ϕ_{2} on base q^{5}). This gives a quintic transformation which immediately becomes the first Rogers-Ramanujan identity if t = 1/q.

[11, Theorem 7.1]. We have

∑_{n=0}^{∞} \frac{q^{n^{2}+2n}t^{2n}}{(q; q)_{n}} = \frac{(t^{4}q^{9}, t^{2}q^{5}, t^{4}q^{6}; q^{5})_{∞}}{(t^{2}q^{3}; q)_{∞}} _{3}ϕ_{2} ( \begin{matrix}t^{2}q^{2}, & t^{2}q^{3}, & t^{2}q^{5} \\ & t^{4}q^{9}, & t^{4}q^{6} \end{matrix} | q^{5}, t^{2}q^{5}) .

We do not know the analogous extensions of (5.3f), (5.3g) and (5.3h) which include the variable t. These should be 10th degree, 14 th degree, and 12th degree transformations, and may involve multiple sums.

6. m-versions. The Rogers-Ramanujan identities have the natural generalization [10],[11],[14]

(6.1)  ∑_{n=0}^{∞} \frac{q^{n^{2}+mn}}{(q; q)_{n}} = \frac{a_{m}(q)}{(q, q^{4}; q)_{∞}} + \frac{b_{m}(q)}{(q^{2}, q^{3}; q)_{∞}} ,

where a_{m}(q) and b_{m}(q) are Laurent polynomials in q which are explicitly known. We refer to (6.1) as an “m-version” of the Rogers-Ramanujan identities.

In this section we find m-versions for the identities in §5 for m < 0 via Theorem 5. We shall see that Theorem 5 gives a single sum for the appropriate polynomials that replace a_{m}(q) and b_{m}(q). In this section we will also use generating functions to find an alternative representation for these polynomials: in this multisum form the polynomials clearly have a constant sign, and positivity results are immediate. There will also be results given in this form for m > 0.

According to Theorem 5

(6.2) lim_{t→q^{−m/2}} S_{p,q}(t) = \frac{1}{2(1 − q^{−m})} ∑_{j=0}^{m} [ \begin{matrix}m \\ j \end{matrix} ]_{q} \frac{(p, q^{2j−m}, q^{m−2j}; p)_{∞}}{(q; q)_{∞}} q^{j(j−m)},

for m = 1, 2, . . ..

For each choice of p in (5.3) S_{p,q}(t) is an analytic function of t, thus we may replace t by q^{−m/2}. For example in (5.3e)

(6.3)  S_{q^{5},q}(q^{−m/2}) = ∑_{n=0}^{∞} \frac{q^{n^{2}+(2−m)n}}{(q; q)_{n}} .

We next collect several of the cases in §5 into one case. Let s be an odd integer which is relatively prime to a positive integer b. It is clear from (6.2) that S_{q^{s},q^{b}}(q^{−bm/2}) is a linear combination of infinite products

F_{i}(s, b) = \frac{(q^{s}, q^{i}, q^{s−i}; q^{s})_{∞}}{(q^{b}; q^{b})_{∞}} , 1 ≤ i ≤ (s − 1)/2

with rational functions of q as coefficients.

Propositions 4 and 5 give two explicit forms of these rational functions.

Proposition 4. Let s ≥ 3 be an odd positive integer relatively prime to b. Let β be the inverse of b mod s, so that bβ = 1 + T s. Then we have

S_{q^{s},q^{b}}(q^{−bm/2}) = ∑_{i=1}^{(s−1)/2} c_{i}(m, s, b)F_{i}(s, b),

where

c_{i}(m, s, b) = (−1)^{T i+1} ∑_{t} [ \begin{matrix}m \\ (m + βi)/2 + ts \end{matrix} ]_{q^{−b}} \frac{1 − q^{b(2ts+βi)}}{1 − q^{−bm}} q^{e},

e = −bβi − bβT i^{2} + s\binom{T i + 1}{2} − 2b^{2}βit − bst − 2b^{2}st^{2} for m + βi even,

c_{i}(m, s, b) = (−1)^{T i+1+b} ∑_{t} [ \begin{matrix}m \\ (m + βi − s)/2 + ts \end{matrix} ]_{q^{−b}} \frac{1 − q^{b((2t−1)s+βi)}}{1 − q^{−bm}} q^{e},

e = (−bβi + bs − 2tbs)((2t − 1)b + T i + 1) + s\binom{(2t − 1)b + T i + 1}{2}

for m + βi odd.

Proof. Let’s first consider m + βi even so that b(m − 2j) = ±i mod s implies that j = (m ∓ βi)/2 + ts for some integer t. Upon extracting the coefficient of F_{i}(s, b), we see that the choice of +βi gives the same value as −βi, this leaves only the stated single sum and eliminates the denominator factor of 2. The m + βi odd case is done similarly. □

Because of the factor 1 − q^{−bm} in the denominator of Proposition 4, it is not clear that c_{i}(m, s, b) is in fact a polynomial in q^{−b} with non-negative coefficients. However this is the case, if we use the following elementary fact

[ \begin{matrix}m \\ j \end{matrix} ]_{q} \frac{1 − q^{m−2j}}{1 − q^{m}} = [ \begin{matrix}m − 2 \\ j \end{matrix} ]_{q} − q^{m−2j} [ \begin{matrix}m − 2 \\ m − j \end{matrix} ]_{q}

we see that for m + βi even

(−1)^{T i}q^{bβT i^{2}−s\binom{T i+1}{2}}c_{i}(m, s, b) = ∑_{t} [ \begin{matrix}m − 2 \\ (m − βi)/2 − ts \end{matrix} ]_{q^{−b}} q^{−2b^{2}βit+bst−2b^{2}st^{2}}

− ∑_{t} [ \begin{matrix}m − 2 \\ (m + βi)/2 − ts \end{matrix} ]_{q^{−b}} q^{−bβi+2b^{2}βit+bst−2b^{2}st^{2}},

which shows that c_{i}(m, s, b) is a hook difference polynomial (see [6])

Proposition 5. The coefficient of F_{i}(s, b) is a polynomial in 1/q explicitly given by the hook polynomial

c_{i}(m, s, b) = (−1)^{T i}q^{−bβT i^{2}+s\binom{T i+1}{2}}D_{K,I}(N, M; α, β)(q^{−b}), for m + βi even,

K = s, N = (m − βi)/2, M = (m + βi)/2 − 2, I = βi, α = 2b − 1, β = 1. and for m + βi odd,

c_{i}(m, s, b) = (−1)^{T i+b}q^{s\binom{T i+1}{2}−s\binom{b+1}{2}+bβi(b−T i)}D_{K,I}(N, M; α, β)(q^{−b}).

K = s, N = (m−βi+s)/2, M = (m+βi−s)/2−2, I = βi−s, α = 2b−1, β = 1.

We next consider the choices of b and s given in §5. In these cases we shall give recurrences satisfied by the polynomials c_{i}(m, s, b). Solving these recurrences gives manifestly positive expressions for the polynomials, as well as results for positive values of m.

The case (5.3e). In this case we have s = 5 in Propositions 4 and 5

(6.4) S_{q^{5},q}(q^{−m/2}) = Ẽ_{m−2}(q) := ∑_{n=0}^{∞} \frac{q^{n(n+2−m)}}{(q; q)_{n}} = \frac{c_{1}(m, 5, 1)}{(q^{2}, q^{3}; q^{5})_{∞}} + \frac{c_{2}(m, 5, 1)}{(q^{1}, q^{4}; q^{5})_{∞}} .

We now give alternative forms for the polynomials c_{1}(m, 5, 1) and c_{2}(m, 5, 1). The easily verified q-difference equation

(6.5)  Ẽ_{m+1}(q) = Ẽ_{m}(q) + q^{−m}Ẽ_{m−1}(q)

indicates that c_{i}(m + 2, 5, q) also satisfy (6.5). (The reason is that 1/(q, q^{4}; q^{5})_{∞} and 1/(q^{2}, q^{3}; q^{5}) are linear independent over the field of Laurent polynomials, see §8.) The initial conditions are

c_{1}(1, 5, 1) = 1, c_{1}(2, 5, 1) = 0, c_{2}(1, 5, 1) = 0, c_{2}(2, 5, 1) = 1.

(6.5) is a well-studied [3] q-version of the Fibonacci recurrence, the solutions are

(6.6a)  c_{1}(m, 5, 1) = ∑_{j=0}^{(m−3)/2} q^{−j^{2}−j} [ \begin{matrix}m − j − 3 \\ j \end{matrix} ]_{q^{−1}} ,

(6.6b)  c_{2}(m, 5, 1) = ∑_{j=0}^{(m−2)/2} q^{−j^{2}} [ \begin{matrix}m − j − 2 \\ j \end{matrix} ]_{q^{−1}} .

These explicit forms may be found using generating functions, we will carry out the details in the subsequent cases. The equality of the forms (6.6) and those given by Proposition 4 are well-known polynomial identities which imply the Rogers-Ramanujan identities.

The case (5.3f) We set

S_{q^{5},q^{2}}(q^{−m}) =F̃_{m−1}(q) := ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} (−1)^{n}q^{3n^{2}+2n−2mn}

(6.7)  =c_{1}(m, 5, 2)\frac{(q^{5}, q^{1}, q^{4}; q^{5})_{∞}}{(q^{2}, q^{2})_{∞}} + c_{2}(m, 5, 2)\frac{(q^{5}, q^{2}, q^{3}; q^{5})_{∞}}{(q^{2}, q^{2})_{∞}} .

It is easy to verify that F̃ satisfies the recurrence relation

F̃_{m+2} + q^{−1}F̃_{m+1} − F̃_{m} − q^{−1}F̃_{m−1} + q^{−1−2m}F̃_{m−1} = 0,

hence f_{m} = c_{i}(m + 1, 5, 2) satisfy

(6.8)  f_{m+2} + q^{−1}f_{m+1} − f_{m} − q^{−1}f_{m−1} + q^{−1−2m}f_{m−1} = 0.

The initial conditions are:

c_{1}(1, 5, 2) = 0, c_{1}(2, 5, 2) = 1, c_{1}(3, 5, 2) = −q^{−1},
c_{2}(1, 5, 2) = 1, c_{2}(2, 5, 2) = 0, c_{2}(3, 5, 2) = 1.

We next explicitly solve the recurrence relation (6.8) using generating functions. Let f_{m} be a solution to (6.8) and let s = q^{−1} and assume for the time being that |s| < 1. If F (t) = ∑_{n=0}^{∞} f_{n}t^{n} then F satisfies

F (t) = \frac{ϕ(t) − s^{3}t^{3}F (s^{2}t)}{(1 − t^{2})(1 + st)},
ϕ(t) := f_{0}(1 + st) + f_{1}t.

This leads to

F (t) := ∑_{n=0}^{∞} f_{n}t^{n} = ∑_{m=0}^{∞} \frac{(−1)^{m}t^{3m}s^{3m^{2}}ϕ(ts^{2m})}{(−t, t, −st; s^{2})_{m+1}}.

Thus if f_{0} = 0, f_{1} = 1, ϕ(t) = t, we have c_{1}(m + 1, 5, 2) = f_{m},

c_{1}(n + 1, 5, 2) = (−1)^{n−1} ∑_{3m+2j+k+1=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{q^{−4}} [ \begin{matrix}m + k \\ k \end{matrix} ]_{q^{−2}} q^{−3m^{2}−2m−k},

while f_{0} = 1, f_{1} = 0, ϕ(t) = 1 + st yields

c_{2}(n + 1, 5, 2) = (−1)^{n} ∑_{3m+2j+k=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{q^{−4}} [ \begin{matrix}m + k − 1 \\ k \end{matrix} ]_{q^{−2}} q^{−3m^{2}−k}.

We have two forms for the polynomials c_{i}(n, 5, 2), which is a polynomial identity. Somewhat surprisingly, unlike case (5.3e), a new identity, not (5.3f), results from a limit.

From Proposition 4 it is easy to see that if Q = 1/q, |Q| < 1,

lim_{m→∞} c_{i}(m, s, b) =(−1)^{T i+1}Q^{bβT i^{2}−s\binom{T i+1}{2}}/(Q^{b}; Q^{b})_{∞}
× ((Q^{4b^{2}s}, −Q^{2b^{2}(s+βi)−bs}, −Q^{2b^{2}(s−βi)+bs}; Q^{4b^{2}s})_{∞}
− Q^{bβi}(Q^{4b^{2}s}, −Q^{2b^{2}(s+βi)+bs}, −Q^{2b^{2}(s−βi)−bs}; Q^{4b^{2}s})_{∞}).

Evaluating this limit using the positive forms gives

∑_{m=0}^{∞} \frac{q^{3m^{2}+2m}}{(q^{4}; q^{4})_{m}(q; q^{2})_{m+1}} + ∑_{m=0}^{∞} \frac{(−1)^{m+1}q^{3m^{2}+2m}}{(q^{4}; q^{4})_{m}(−q; q^{2})_{m+1}}

= \frac{2q}{(q^{2}; q^{2})_{∞}} ((q^{80}, −q^{54}, −q^{26}; q^{80})_{∞} − q^{6}(q^{80}, −q^{74}, −q^{6}; q^{80})_{∞}).

∑_{m=0}^{∞} \frac{q^{3m^{2}}}{(q^{4}; q^{4})_{m}(q; q^{2})_{m}} + ∑_{m=0}^{∞} \frac{(−1)^{m}q^{3m^{2}}}{(q^{4}; q^{4})_{m}(−q; q^{2})_{m}}

= \frac{2}{(q^{2}; q^{2})_{∞}} ((q^{80}, −q^{38}, −q^{42}; q^{80})_{∞} − q^{2}(q^{80}, −q^{22}, −q^{58}; q^{80})_{∞}).

The case (5.3g). We set

S_{q^{7},q^{2}}(q^{−m}) =G̃_{m−2}(q) := ∑_{n=0}^{∞} \frac{(q; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{2n^{2}+4n−2mn}

=c_{1}(m, 7, 2)\frac{(q^{7}, q^{1}, q^{6}; q^{7})_{∞}}{(q^{2}, q^{2})_{∞}} + c_{2}(m, 7, 2)\frac{(q^{7}, q^{2}, q^{5}; q^{7})_{∞}}{(q^{2}, q^{2})_{∞}}

+ c_{3}(m, 7, 2)\frac{(q^{7}, q^{3}, q^{4}; q^{7})_{∞}}{(q^{2}, q^{2})_{∞}} . (6.9)

Analogous to the case (5.3f) one can verify that

G̃_{m+2} + q^{−1}G̃_{m+1} − (1 + q^{−2−2m})G̃_{m} − q^{−1}G̃_{m−1} = 0.

Again the linear independence of {F_{1}(7, 2), F_{2}(7, 2), F_{3}(7, 2)} over the field of Laurent polynomials (see §8) implies that g_{m} = c_{i}(m + 2, 7, 2) also satisfy (6.10)

(6.10)  g_{m+2} + q^{−1}g_{m+1} − (1 + q^{−2−2m})g_{m} − q^{−1}g_{m−1} = 0.

The initial conditions are

c_{1}(1, 7, 2) = 0, c_{1}(2, 7, 2) = 0, c_{1}(3, 7, 2) = 1,
c_{2}(1, 7, 2) = 1, c_{2}(2, 7, 2) = 0, c_{2}(3, 7, 2) = 1,
c_{3}(1, 7, 2) = 0, c_{3}(2, 7, 2) = 1, c_{3}(3, 7, 2) = 0.

The above initial conditions show that the c_{i}(m + 2, 7, 2), 1 ≤ i ≤ 3 is a basis of solutions to (6.10).

We next solve the recurrence (6.10) using generating functions and find an explicit positive form for the polynomials c_{i}(m + 2, 7, 2).

Let g_{m} be a solution to (6.10) and let s = q^{−1} and assume for the time being that |s| < 1. If G(t) = ∑_{n=0}^{∞} g_{n}t^{n} then G satisfies

G(t) = \frac{ϕ(t) + s^{2}t^{2}G(s^{2}t)}{(1 − t^{2})(1 + st)},
ϕ(t) := g_{0}(1 + st) + g_{1}t + +st^{2}g_{1}.

This leads to

G(t) := ∑_{n=0}^{∞} g_{n}t^{n} = ∑_{m=0}^{∞} \frac{t^{2m}s^{2m^{2}}ϕ(ts^{2m})}{(−t, t, −st; s^{2})_{m+1}} .

TRIBASIC INTEGRALS AND IDENTITIES OF ROGERS-RAMANUJAN TYPE 21

In the case g_{m} = c_{1}(m + 2, 7, 2), ϕ(t) = t and we find

c_{1}(n + 2, 7, 2) = (−1)^{n−1} ∑_{2m+2j+k+1=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{s^{4}} [ \begin{matrix}m + k \\ k \end{matrix} ]_{s^{2}} s^{2m(m+1)+k}.

On the other hand if g_{m} = c_{2}(m + 2, 7, 2), then ϕ(t) = t(1 + st) and we obtain

c_{2}(n + 2, 7, 2) = (−1)^{n−1} ∑_{2m+2j+k+1=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{s^{4}} [ \begin{matrix}m + k − 1 \\ k \end{matrix} ]_{s^{2}} s^{2m(m+1)+k}.

Finally g_{m} = c_{3}(m + 2, 7, 2) makes ϕ(t) = 1 + st and we get

c_{3}(n + 2, 7, 2) = (−1)^{n} ∑_{2m+2j+k=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{s^{4}} [ \begin{matrix}m + k − 1 \\ k \end{matrix} ]_{s^{2}} s^{2m^{2}+k}.

Since the expressions given for c_{i}(n + 2, 7, 2) are polynomials in s we can remove the restriction |s| < 1 and they must hold for all s.

This time when we find lim_{n→∞} c_{i}(n + 2, 7, 2) using the two forms we find

∑_{m=0}^{∞} \frac{q^{2m^{2}+2m}}{(q^{4}; q^{4})_{m}} (\frac{1}{(q; q^{2})_{m+1}} − \frac{1}{(−q; q^{2})_{m+1}} )

= \frac{2q}{(q^{2}; q^{2})_{∞}} ((q^{112}, −q^{74}, −q^{38}; q^{112})_{∞} − q^{8}(q^{112}, −q^{10}, −q^{102}; q^{112})_{∞}).

∑_{m=0}^{∞} \frac{q^{2m^{2}+2m}}{(q^{4}; q^{4})_{m}} (\frac{1}{(q; q^{2})_{m+1}} + \frac{1}{(−q; q^{2})_{m+1}} )

= \frac{2}{(q^{2}; q^{2})_{∞}} ((q^{112}, −q^{46}, −q^{66}; q^{112})_{∞} − q^{6}(q^{112}, −q^{18}, −q^{94}; q^{112})_{∞}).

∑_{m=0}^{∞} \frac{q^{2m^{2}}}{(q^{4}; q^{4})_{m}} (\frac{1}{(q; q^{2})_{m}} + \frac{1}{(−q; q^{2})_{m}} )

= \frac{2}{(q^{2}; q^{2})_{∞}} ((q^{112}, −q^{54}, −q^{58}; q^{112})_{∞} − q^{4}(q^{112}, −q^{26}, −q^{86}; q^{112})_{∞}).

We may also explicitly find the polynomials for c_{i}(n, 7, 2) for n < 0 in the same way. The results are

c_{1}(2 − n, 7, 2) = (−1)^{n−2} ∑_{2m+2j+k+2=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{q^{4}} [ \begin{matrix}m + k \\ k \end{matrix} ]_{q^{2}} q^{m^{2}+2m+1+k},

c_{2}(2 − n, 7, 2) = (−1)^{n−1} ∑_{2m+2j+k+1=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{q^{4}} [ \begin{matrix}m + k − 1 \\ k \end{matrix} ]_{q^{2}} q^{m^{2}+k},

c_{3}(2 − n, 7, 2) =(−1)^{n} ∑_{2m+2j+k=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{q^{4}} [ \begin{matrix}m + k − 1 \\ k \end{matrix} ]_{q^{2}} q^{m^{2}−2m+k}

+ (−1)^{n−1} ∑_{2m+2j+k+1=n} [ \begin{matrix}m + j \\ j \end{matrix} ]_{q^{4}} [ \begin{matrix}m + k \\ k \end{matrix} ]_{q^{2}} q^{m^{2}−1+k}.

An unusual theta identity results from evaluating lim_{n→∞} c_{i}(2−n, 7, 2) and using lim_{n→−∞} G̃_{n}(q) = 0. The case (5.3h). We set

(6.11)  H̃_{m}(q) = ∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q; q)_{2n}} q^{n(n−m)},

which satisfies the three term recurrence relation

(6.12)  H̃_{m+2}(q) + [q^{−1} − q^{−1−m}]H̃_{m−2} = [1 + q^{−1} + q^{−1−m}]H̃_{m}(q).

In this case

H̃_{m}(q) = ∑_{j=0}^{m+2} [ \begin{matrix}m + 2 \\ j \end{matrix} ]_{q} \frac{(−q^{3}, q^{2j−m−2}, q^{m+2−2j}; −q^{3})_{∞}}{2(1 − q^{−m−2})(q; q)_{∞}} q^{−j(m+2−j)}.

It is surprising that the right-hand side in the above formula changes dramatically according to whether m is odd or even. It is easy to see that there are Laurent polynomials h_{m}^{(1)}(q), h_{m}^{(2)}(q) and h_{m}^{(3)}(q) such that

(6.13a)  H̃_{2m}(q) = h_{m}^{(1)}(q)\frac{(−q, −q^{3}, −q^{5}; q^{6})_{∞}}{(q, q^{3}, q^{5}; q^{6})_{∞}} ,
(6.13b)  H̃_{2m+1}(q) = h_{m}^{(2)}(q)\frac{(−1, −q^{6}, q^{3}, −q^{3}; q^{6})_{∞}}{(q, q^{2}, q^{4}, q^{5}; q^{6})_{∞}}

+ h_{m}^{(3)}(q)\frac{(−q^{2}, −q^{3}, −q^{4}; q^{6})_{∞}}{(q^{2}, q^{3}, q^{4}; q^{6})_{∞}} .

Therefore (6.12) shows that {h_{m}^{(1)}(q)} satisfies

(6.14a)  h_{m+1}^{(1)}(q) + q^{−1}[1 − q^{−2m}]h_{m−1}^{(1)}(q) = [1 + q^{−1} + q^{−1−2m}]h_{m}^{(1)}(q),

while {h_{m}^{(2)}(q)} and {h_{m}^{(3)}(q)} satisfy

(6.14b)  X_{m+1}(q) + q^{−1}[1 − q^{−2m−1}]X_{m−1}(q) = [1 + q^{−1} + q^{−2−2m}]X_{m}(q).

The initial conditions are

h_{0}^{(1)}(q) = 1, h_{1}^{(1)}(q) = 1 + 2/q,
h_{−1}^{(2)}(q) = 0, h_{0}^{(2)}(q) = 1, h_{−1}^{(3)}(q) = 1, h_{0}^{(3)}(q) = 1.

We can explicitly find h_{n}^{(i)}(q) as in the previous cases, the results are

h_{n}^{(1)}(q) = ∑_{m+j+k=n} [ \begin{matrix}2m + 1 + j \\ j \end{matrix} ]_{q^{−1}} [ \begin{matrix}m \\ k \end{matrix} ]_{q^{−2}} q^{−k^{2}−k−m^{2}},

h_{n}^{(2)}(q) = ∑_{m+j+k=n} [ \begin{matrix}2m + 1 + j \\ j \end{matrix} ]_{q^{−1}} [ \begin{matrix}m \\ k \end{matrix} ]_{q^{−2}} q^{−k^{2}−k−m^{2}−m},

h_{n}^{(3)}(q) = ∑_{m+j+k=n} [ \begin{matrix}2m + j \\ j \end{matrix} ]_{q^{−1}} [ \begin{matrix}m \\ k \end{matrix} ]_{q^{−2}} q^{−k^{2}−k−m^{2}−m}

+ ∑_{m+j+k=n−1} [ \begin{matrix}2m + 1 + j \\ j \end{matrix} ]_{q^{−1}} [ \begin{matrix}m \\ k \end{matrix} ]_{q^{−2}} q^{−k^{2}−k−m^{2}−3m−2}.

One may also appeal to orthogonal polynomials to find explicit formulas for h_{n}^{(i)}(q). The three-term recurrence (6.14a) is a special case of that for the Al-Salam-Chihara polynomials, defined by

p_{n}(cos θ; t_{1}, t_{2}|q) = _{3}ϕ_{2} ( \begin{matrix}q^{−n}, & t_{1}e^{iθ}, & t_{1}e^{−iθ} \\ & t_{1}t_{2}, & 0 \end{matrix} | q, q)

= \frac{(t_{2}e^{−iθ}; q)_{n}t_{1}^{n}e^{inθ}}{(t_{1}t_{2}; q)_{n}} _{2}ϕ_{1} ( \begin{matrix}q^{−n}, t_{1}e^{iθ} \\ q^{1−n}e^{iθ}/t_{2} \end{matrix} | q, \frac{qe^{−iθ}}{t_{2}} ) .

Thus

h_{m}^{(1)}(q) = (−1)^{m}p_{m}(−(q^{1/2} + q^{−1/2})/2; q^{−1/2}, 0|q^{−2}),

and (6.13a) is

∑_{n=0}^{∞} \frac{(−1; q^{2})_{n}}{(q^{2}; q^{2})_{2n}} q^{n(n−2m)} (6.15)

= \frac{(−q, −q^{3}, −q^{5}; q^{6})_{∞}}{(q, q^{3}, q^{5}; q^{6})_{∞}} ∑_{j=0}^{m} [ \begin{matrix}m \\ j \end{matrix} ]_{q^{2}} (−1; q)_{2j} (−1)^{m−j}q^{−j^{2}}.

The recurrence relation (6.14b) corresponds to a special case of the associated Al-Salam-Chihara polynomials [13]. We shall follow the notation in Ismail and Rahman [13]. The polynomials relevant to (6.14b) correspond to b = c = d = 0 in the associated Askey-Wilson recurrence relation [13, (4.14)]

(6.16) [2x − ar^{n+α}] Y_{n}(x; a|r) = a^{−1}Y_{n+1}(x; a|r) + a(1 − r^{n+α})Y_{n−1}(x; a|r)

Ismail and Rahman identified two polynomial solutions to the general recurrence relation of the Askey-Wilson polynomials and in the case considered here they satisfy the initial conditions, see (4.2), (4.14), and (5.2) in [13]

p_{−1}^{α}(x; a|r) = 0, p_{0}^{α}(x; a|r) = 1, q_{−1}^{α}(x; a|r) = 1, q_{0}^{α}(x; a|r) = 1.

Now (4.15) and (8.9) of [13] give

p_{n}^{α}(cos θ; a|r) = ∑_{j=0}^{n} \frac{(r^{−n}, ae^{iθ}, ae^{−iθ}; r)_{j}}{(r; r)_{j}} r^{j}_{3}ϕ_{2} ( \begin{matrix}r^{j−n}, & r^{α}, & r^{j+1} \\ & r^{−n}, & 0 \end{matrix} | r, a^{2}) ,

q_{n}^{α}(cos θ; a|r) = ∑_{j=0}^{n} \frac{(r^{−n}, ae^{iθ}, ae^{−iθ}; r)_{j}}{(r; r)_{j}} r^{j}_{3}ϕ_{2} ( \begin{matrix}r^{j−n}, & r^{α}, & r^{j} \\ & r^{−n}, & 0 \end{matrix} | r, ra^{2}) .

Thus

(6.17a)  h_{m}^{(2)}(q) = (−1)^{m}p_{m}^{1/2}((q^{1/2} + q^{−1/2})/2; −q^{−1/2}|q^{−2})
(6.17b)  h_{m}^{(3)}(q) = (−1)^{m} [2p_{m}^{1/2}((q^{1/2} + q^{−1/2})/2; −q^{−1/2}|q^{−2})
−q_{m}^{1/2}((q^{1/2} + q^{−1/2})/2; −q^{−1/2}|q^{−2})] .

The case (5.3i). Set

Ĩ_{m}(q) = 1 + (1 − ω) ∑_{n=1}^{∞} \frac{(q^{3}; q^{3})_{n−1}}{(q; q)_{2n}(q; q)_{n−1}} q^{n(n−m)}.

It easily follows that Ĩ_{m} satisfies the six term recurrence relation

qĨ_{m+2}(q) + (1 + q^{−1})Ĩ_{m−1}(q) + Ĩ_{m−2}(q)
= Ĩ_{m+1}(q) + (1 + q + q^{−m})Ĩ_{m}(q) + [q^{−1} − q^{−m}]Ĩ_{m−3}(q).

On can also show that Ĩ_{m} satisfies the following inhomogeneous difference equation of order four

Ĩ_{m+2} + q^{−1}[1 − q^{−m}]Ĩ_{m−2} − [1 + q^{−1} + q^{−1−m}]Ĩ_{m} (6.18)

= q^{−1−m}Ĩ_{m−1} − (2 + ω)q^{−1−m}.

By inspection we see that the right-hand side of (6.1) is spanned by {e_{1}, e_{2}, e_{3}, e_{4}},

e_{j} = \frac{(ωq^{3}, q^{j}, q^{3−j}ω; ωq^{3})_{∞}}{(q; q)_{∞}} , 1 ≤ j ≤ 3, e_{4} = \frac{(ωq^{3}, q^{2}ω^{2}, qω^{2}; ωq^{3})_{∞}}{(q; q)_{∞}} .

By extending the proof of Proposition 8, it can shown that {e_{1}, e_{2}, e_{3}, e_{4}} is linearly independent over the field of Laurent polynomials.

The initial conditions are:

Ĩ_{−1}(q) = e_{1}, Ĩ_{0}(q) = e_{2}, Ĩ_{1} = e_{1} + e_{3}, Ĩ_{2} = (1 + q^{−1})e_{2} − ωq^{−1}e_{4}.

Furthermore (6.1) yields

q^{2}Ĩ_{3} = (1 + q^{2})e_{1} + (1 + q + q^{2})e_{3} − ωe_{4}.

Note that the initial conditions and the above value for Ĩ_{3} imply

(6.19)  e_{1} + e_{2} + ωe_{4} = 2 + ω,

which can also be verified using the Jacobi triple product identity (1.1). It is easy to see that Ĩ_{m} − (2 + ω)/3 satisfies the homogeneous recurrence relation

(6.20)  Z_{m+2} = [1 + q^{−1} + q^{−1−m}]Z_{m} + q^{−1−m}Z_{m−1} − q^{−1}[1 − q^{−m}]Z_{m−2}.

In view of (6.19) we set

Ĩ_{m} = \frac{1}{3}[e_{1} + e_{2} + ωe_{4}] + ∑_{j=1}^{4} ϕ_{m}^{(j)}(q)e_{j},

and find that the Laurent polynomials ϕ_{m}^{(j)}(q), 1 ≤ j ≤ 4 satisfy (6.20). As in case g we set

s = 1/q, Z(t) = ∑_{m=0}^{∞} Z_{m}t^{m},

and use (6.20) to derive the functional equation

Z(t) = \frac{ϕ(t) + st^{2}(1 + st + s^{2}t^{2})Z(st)}{(1 − t^{2})(1 − st^{2})},
ϕ(t) = Z_{0}[1 − t^{2}(1 + 2s)] + Z_{1}t + Z_{2}t^{2} − s(1 − s)t^{3}Z_{−1}.

Thus for |s| < 1 we establish

Z(t) = ∑_{m=0}^{∞} \frac{t^{2m}s^{m^{2}}(s^{3}t^{3}; s^{3})_{m}ϕ(ts^{m})}{(t^{2}; s)_{2m+2}(st; s)_{m}}

from which explicit triple sums for the Laurent polynomials ϕ_{m}^{(j)}(q), 1 ≤ j ≤ 4, may be given.

7. Multisum Rogers-Ramanujan identities. It is natural to ask if the kfold sum of the mod 2k + 3 Andrews-Gordon identities appear from the tribasic integrals. For example in Theorems 3 and 4, if q = p, and s = p^{2k+3}, the product sides of these identities do appear. We do have double sums available for any base p, see (4.1b), (4.2b), (4.3b), and (5.1c) along with

b_{n}(p, q) =p^{−n} \frac{c_{2n,0}(p, q)}{(p; p)_{2n}} = ∑_{j=0}^{n} \frac{a_{j}(p, q)}{(p; p)_{n−j}(p; p)_{n+j}}
a_{0}(p, q) =1, a_{j}(p, q) = (−1)^{j}q^{j^{2}/2}(p^{−j}q^{j/2} + p^{j}q^{−j/2}),

β_{n}(p, q) =q^{−n^{2}−2n} \frac{d_{2n,0}(p, q)}{(q; q)_{2n}} = ∑_{j=0}^{n} \frac{α_{j}(p, q)}{(p; p)_{n−j}(p; p)_{n+j}} ,

α_{j}(p, q) =(−1)^{j} \frac{1 − q^{2j+1}}{1 − q} q^{−j^{2}−2j}p^{\binom{j+1}{2}}.

We see that (a_{n}(p, q), b_{n}(p, q)) is a Bailey pair with parameters (1, p), while (α_{n}(p, q), β_{n}(p, q)) is a Bailey pair with parameters (q, q).

To see how the multisums arise from the double sum we use Bailey pairs. If we build a Bailey chain starting with (α_{n}(p, q), β_{n}(p, q)), then the next pair (α′_{n}(p, q), β′_{n}(p, q)) satisfies

α′_{n}(p, q) = q^{n^{2}+n}α_{n}(p, q) = α_{n}(pq^{2}, q).

Thus β′_{n}(q^{7}, q) is a sum of the evaluable β_{n}(q^{5}, q), and (5.1a) and (5.1b) are double sum identities for mod 7. This also works for mod 2k + 3, and for any of the nine evaluations in §5, not just (5.2e).

For the evaluations of §4 we can similarly generate multisum identities using

a′_{n}(p, q) = p^{n^{2}}a_{n}(p, q) = a_{n}(p^{2}, p^{2}q) = a_{n}(q, p^{2}q).

The change of base formulations in [9] are also available.

8. Remarks. The appropriate q → 1 limits of the integrals G_{1} and G_{2} are simple exponential integrals. For example

(8.1)  lim_{q→1} G_{2}(t, q, q, q^{A}) = \frac{e^{−t^{2}}}{\sqrt{2π}} ∫_{−∞}^{∞} e^{−x^{2}/2}e^{\sqrt{A}xt−t^{2}/2}dx = e^{t^{2}(A−3)/2}.

The case A = 5 is the motivating integral of the Introduction.

Using the polynomials

(8.2)  J_{n}(x|q) = ∑_{k=0}^{n} [ \begin{matrix}n \\ k \end{matrix} ]_{q} q^{k(n−k)}e^{i(n−2k)θ}, x = cos θ,

whose generating function is

(8.3)  ∑_{n=0}^{∞} J_{n}(x|q)\frac{t^{n}}{(q; q)_{n}} = \frac{(t^{2}; q)_{∞}}{(te^{iθ}, te^{−iθ}; q)_{∞}} ,

one can evaluate an integral to conclude the following result.

Proposition 6. We have

(A) ∑_{n=0}^{∞} \frac{(−x^{2}p^{1/2−n/2}; p)_{n}}{(p; p)_{n}} p^{3n^{2}/8}x^{−n} = \frac{(p^{3/4}, −p^{3/8}/x, −p^{3/8}x; p^{3/4})_{∞}}{(p; p)_{∞}} .

In fact both the even and odd parts of this sum do sum

(B)  ∑_{n=0}^{∞} \frac{(−p^{1/2}/x^{2}, −p^{1/2}x^{2}; p)_{n}}{(p; p)_{2n}} p^{n^{2}} = \frac{(p^{3}, −p^{3/2}/x^{2}, −p^{3/2}x^{2}; p^{3})_{∞}}{(p; p)_{∞}}

(C) ∑_{n=0}^{∞} \frac{(−p^{1/2}/x^{2}; p)_{n}(−p^{1/2}x^{2}; p)_{n+1}}{(p; p)_{2n+1}} p^{n^{2}+n} = \frac{(p^{3}, −p^{5/2}/x^{2}, −p^{1/2}x^{2}; p^{3})_{∞}}{(p; p)_{∞}}

Propositions 6(B) and 6(C) are in Ramanujan’s Lost Notebook, and are special cases of Lemmas 1 and 2 in [4]. Another combination of these terms may be summed by the quintuple product identity (1.2).

Proposition 7. We have

∑_{n=0}^{∞} \frac{(−1/x^{3}, −p^{2}x^{3}; p^{2})_{n}}{(p^{2}; p^{2})_{2n}} p^{2n^{2}} − \frac{1}{x} ∑_{n=0}^{∞} \frac{(−1/x^{3}; p^{2})_{n}(−p^{2}x^{3}; p^{2})_{n+1}}{(p^{2}; p^{2})_{2n+1}} p^{2n^{2}+2n}
=(1/x, p^{2}x; p^{2})_{∞}(p^{2}x^{2}, p^{2}/x^{2}; p^{4})_{∞}.

We also collect here a proposition which was used in §6.

Proposition 8. Fix a positive integer s, and let G_{i}(q) = (q^{s}, q^{i}, q^{s−i}; q^{s})_{∞} for 1 ≤ i ≤ ⌊s/2⌋. Then there is no non-trivial relation

∑_{i=1}^{⌊s/2⌋} c_{i}(q)G_{i}(q) = 0,

where c_{i}(q) is a polynomial in q.

Proof. By the Jacobi triple product identity we have

G_{i}(q) = 1 + ∑_{j=1}^{∞}(−1)^{j}(q^{sj^{2}/2+(s/2−i)j} + q^{sj^{2}/2−(s/2−i)j}).

For a fixed integer N, G_{i}(q) has exactly two nonzero terms between q^{sN^{2}/2} and q^{s(N+1)^{2}/2}, namely q^{sN^{2}/2+(s/2−i)N} and q^{s(N+1)^{2}/2−(s/2−i)N}. For two different values of i, any pair of these terms have exponents differing by at least N. Thus if N is greater than the degree of any c_{k}(q), the assumed relation forces all c_{k}(q) = 0. □

9. Appendix A. In this Appendix we evaluate the constants of §4 and §5, c_{2n,0}(p, q) and d_{2n,0}(p, q), for the choices of p in (4.5) and (5.2).

Proof of (4.5): For (4.5)

(9.1a)  c_{2n,0}(p, q) = ∑_{j=−n}^{n}(−1)^{j}p^{n−j}q^{\binom{j+1}{2}} [ \begin{matrix}2n \\ n − j \end{matrix} ]_{p}

we note the alternative representation

(9.1b)  c_{2n,0}(p, q) = ∑_{k=0}^{2n}(−1)^{n−k}q^{\binom{n−k}{2}} [ \begin{matrix}2n \\ k \end{matrix} ]_{p} .

One can see that (9.1b) and (9.1a) are equivalent by computing the coefficient of q^{\binom{j+1}{2}} in each term, j ≥ 0.

We need Jackson’s well-poised _{3}ϕ_{2} evaluation [12,(II.15)]

(9.2a)  _{3}ϕ_{2} ( \begin{matrix}r^{−2n}, & b, & c \\ & r^{1−2n}/b, & r^{1−2n}/c \end{matrix} | r, \frac{r^{2−n}}{bc} ) = \frac{(b, c; r)_{n}(r, bc; r)_{2n}}{(r, bc; r)_{n}(b, c; r)_{2n}} .

and the q-Kummer sum [12,(II.9)]

(9.2b)  _{2}ϕ_{1} ( \begin{matrix}r^{−2n}, & b \\ & r^{1−2n}/b, \end{matrix} | r, −\frac{r}{b} ) = \frac{(r^{1−2n}, r^{2−2n}/b^{2}; r^{2})_{n}}{(r^{1−2n}/b; r)_{2n}} .

Rewrite (9.1a) and (9.1b) as

(9.3a)  c_{2n,0}(p, q) =(−1)^{n} ∑_{k=0}^{2n} \frac{(p^{−2n}; p)_{k}}{(p; p)_{k}} p^{k+2nk−\binom{k}{2}}q^{\binom{n−k}{2}},

(9.3b)  c_{2n,0}(p, q) =(−1)^{n} ∑_{k=0}^{2n} \frac{(p^{−2n}; p)_{k}}{(p; p)_{k}} p^{2nk−\binom{k}{2}}q^{\binom{n−k+1}{2}}.

Proof of (4.5ade): These three cases of (4.5) follow from (9.2) and (9.3).

(a) q = p^{1/2}: use (9.3b) with r = p^{1/2}, c = 0, b = −p^{−n} in (9.2a),

(d) q = p^{3}: use (9.3b) with r = p, b, c → ∞ in (9.2a),

(e) q = p^{3/2}: use (9.3a) with r = p^{1/2}, c → ∞. b = −p^{−n} in (9.2a).

Proof of (4.5c): If q = p^{2} (case (4.5c)) we use the r = p, b → ∞ version of (9.2b).

Proof of (4.5b): If q = −p (case (4.5b)) we shall use the q-binomial theorem,

(9.4)  c_{2n,0}(p, −p) = (−1)^{n}(−p)^{\binom{n+1}{2}} ∑_{k=0}^{2n} \frac{(p^{−2n}; p)_{k}}{(p; p)_{k}} p^{k(1−n)}(−1)^{nk+\binom{k}{2}}.

First consider the case when n is even in (9.4). Then (−1)^{nk+\binom{k}{2}} has sign behavior which depends only upon k modulo 4, it is + + −−. So we can evaluate the even terms in the sum by choosing x = i in the q-binomial theorem:

(9.5a) ∑_{k=0, k even}^{2n} \frac{(p^{−2n}; p)_{k}}{(p; p)_{k}} p^{k(1−n)}(−1)^{nk+\binom{k}{2}} = \frac{1}{2}((ip^{1−n}; p)_{2n} + (−ip^{1−n}; p)_{2n}).

The odd terms similarly give

(9.5b) ∑_{k=1, k odd}^{2n} \frac{(p^{−2n}; p)_{k}}{(p; p)_{k}} p^{k(1−n)}(−1)^{nk+\binom{k}{2}} = \frac{1}{2i}((ip^{1−n}; p)_{2n} − (−ip^{1−n}; p)_{2n}),

thus for n even

c_{2n,0}(p, −p) = (−p)^{\binom{n+1}{2}}(\frac{(1 − i)}{2}(ip^{1−n}; p)_{2n} + \frac{(1 + i)}{2}(−ip^{1−n}; p)_{2n}).

Since

(1 − i)(ip^{1−n}; p)_{2n} = (−i)^{n}p^{−\binom{n}{2}}(−1; p^{2})_{n} \frac{1 − ip^{n}}{2} ,

a routine computational completes the proof of (4.5b) for n even.

For n odd the sign behavior in (9.4) is + − −+, again we do the even and odd terms separately to find

c_{2n,0}(p, −p) = −(−p)^{\binom{n+1}{2}}(\frac{(1 + i)}{2}(ip^{1−n}; p)_{2n} + \frac{(1 − i)}{2}(−ip^{1−n}; p)_{2n}).

Again it is routine to verify this case.

Proof of (5.2): For (5.2) we rewrite

d_{2n,0}(p, q) =q^{n^{2}+2n} \frac{(1 − q)(q; q)_{2n}}{(q; q)_{n}(q; q)_{n+1}}
(9.6)  × ∑_{j=0}^{n} \frac{(q^{−n}; q)_{j}}{(q^{n+2}; q)_{j}} \frac{1 − q^{2j+1}}{1 − q} q^{−3\binom{j}{2}+j(n−3)}p^{\binom{j+1}{2}}.

We use the very-well poised _{6}ϕ_{5} evaluation to verify five cases of (5.2),

6ϕ_{5} ( \begin{matrix}a, & r\sqrt{a}, & −r\sqrt{a}, & b, & c, & r^{−n} \\ & \sqrt{a}, & −\sqrt{a}, & ar/b, & ar/c, & ar^{n+1} \end{matrix} | r, \frac{r^{2−n}}{bc} ) =
(9.7)  =\frac{(ar, ar/bc; r)_{n}}{(r, bc; r)_{n}(ar/b, ar/c; r)_{n}} .

Proof of (5.2abcde):

(a) p = q, use (9.7) with r = q, a = q, b = c = 0,

(b) p = q^{2}, use (9.7) with r = q, a = q, b = −q, c = 0,

(c) p = q^{3}, use (9.7) with r = q, a = q, b = c = −q,

(d) p = q^{4}, use (9.7) with r = q, a = q, b = −q, c → ∞,

(e) p = q^{5}, use (9.7) with r = q, a = q, b → ∞, c → ∞. The next two cases use the non-terminating version of Watson’s transformation

[12, (III.36)]

8ϕ_{7} ( \begin{matrix}a, & r\sqrt{a}, & −r\sqrt{a}, & b, & c, & d, & e, & f \\ & \sqrt{a}, & −\sqrt{a}, & ar/b, & ar/c, & ar/d, & ar/e, & ar/f \end{matrix} | r, \frac{a^{2}r^{2}}{bcdef} ) =

\frac{(ar, ar/de, ar/df, ar/ef; r)_{∞}}{(ar/d, ar/e, ar/f, ar/def; r)_{∞}} _{4}ϕ_{3} ( \begin{matrix}ar/bc, & d, & e, & f \\ & ar/b, & ar/c, & def/a \end{matrix} | r, r) +

\frac{(ar, ar/bc, d, e, f, a^{2}r^{2}/bdef, a^{2}r^{2}/cdef; r)_{∞}}{(ar/b, ar/c, ar/d, ar/e, ar/f, a^{2}r^{2}/bcdef, def/ar; r)_{∞}} ×

4ϕ_{3} ( \begin{matrix}ar/de, & ar/df, & ar/ef, & a^{2}r^{2}/bcdef \\ & a^{2}r^{2}/bdef, & a^{2}r^{2}/cdef, & ar^{2}/def \end{matrix} | r, r) .

Proof of (5.2g): We consider case (5.2g), when p = q^{7/2}. If r = q^{1/2}, a = r, b = ir^{3/2}, c = −ir^{3/2}, d = r^{−n}, e = −r^{−n}, and f → ∞, then we have

c_{2n,0}(q^{7/2}, q) = q^{n^{2}+2n} \frac{(1 − q)(q; q)_{2n}}{(q; q)_{n}(q; q)_{n+1}} ×

lim_{f→∞} _{8}ϕ_{7} ( \begin{matrix}a & r\sqrt{a}, & −r\sqrt{a}, & b, & c, & d, & e, & f \\ & \sqrt{a}, & −\sqrt{a}, & ar/b, & ar/c, & ar/d, & ar/e, & ar/f \end{matrix} | r, \frac{a^{2}r^{2}}{bcdef} ) .

In the second term of Watson’s transformation, all terms have a f → ∞ limit except the pair of infinite products

\frac{(f; r)_{∞}}{(def/ar; r)_{∞}} .

Let f = λr^{−m}, and let m → ∞. Then

\frac{(f; r)_{∞}}{(def/ar; r)_{∞}} = (−1)^{m}r^{(2n+2)m} \frac{(r/λ; r)_{∞}}{(−r^{2n+3}/λ; r)_{∞}} \frac{(λ; r)_{∞}}{(−λr^{−2n−2}; r)_{∞}} .

Since r^{(2n+2)m} → 0 as m → ∞, the second term is zero.

What remains is the first term

\frac{(r^{2}, −r^{2n+2}; r)_{∞}}{(r^{n+2}, −r^{n+2}; r)_{∞}} _{3}ϕ_{2} ( \begin{matrix}r^{−1}, & r^{−n}, & −r^{−n} \\ & ir^{1/2}, & −ir^{1/2} \end{matrix} | r, −r^{2n+2}) .

This is a terminating sum of two terms, the result is (5.2g).

Proof of (5.2f): If p = q^{5/2}, we need the same specializations of the above _{8}ϕ_{7} but with f → 0. However, it is not clear how to evaluate the f → 0 limit of the second term. Thus we take a different choice of parameters which is more delicate. Let

r = q^{1/2}, a = r, e = ir^{3/2}, d = −ϵir^{3/2}, c = −r^{−n}, f = r^{−n},

where ϵ is close to 1. We take the b → 0 limit of Watson’s transformation for this choice. The second _{4}ϕ_{3} is

4ϕ_{3} ( \begin{matrix}1/rϵ, & ir^{n+1/2}/ϵ, & −ir^{n+1/2} & r^{2n+1}/bϵ \\ & r^{n}/ϵ, & −r^{2n+1}/ϵ & r^{n+1}/bϵ \end{matrix} | r, r)

which has a limit as b → 0. The infinite products multiplying this _{4}ϕ_{3} include (f; r)_{∞} = (r^{−n}; r)_{∞} = 0 with no zeros in the denominator, thus the second term is again 0.

The first term is

\frac{(r^{2}, 1/rϵ, −ir^{n+1/2}/ϵ, ir^{n+1/2}; r)_{∞}}{(−ir^{1/2}/ϵ, ir^{1/2}, r^{n+2}, r^{n−1}/ϵ; r)_{∞}} ×

4ϕ_{3} ( \begin{matrix}−r^{n+2}/b, & −ϵir^{3/2}, & ir^{3/2} & r^{−n} \\ & r^{2}/b & −r^{n+2}, & ϵr^{2−n} \end{matrix} | r, r) ,

which terminates and has limit as b → 0. When we let ϵ → 1, two terms of the sum survive, and the result is (5.2f).

Proof of (5.2i): If p = ωq^{3}, where ω = e^{2πi/3}, then we must evaluate

(9.8)  A = ∑_{j=0}^{n} \frac{(q^{−n}; q)_{j}}{(q^{n+2}; q)_{j}} \frac{1 − q^{2j+1}}{1 − q} q^{jn}ω^{\binom{j+1}{2}}.

Since

ω^{\binom{j+1}{2}} = \{ \begin{matrix}1 if j = 0 or 2 mod 3 \\ ω if j = 1 mod 3, \end{matrix}

the coefficient of ω in A can be found by restricting the j-sum in (9.8) to j = 1 mod 3. In this case we can apply (9.7) with

r = q^{3}, a = q^{3}, b = q^{1−n}, c = q^{2−n}, d = q^{3−n},

to evaluate the restricted sum. This gives the coefficient of ω in (5.2i).

For the coefficient of 1, which comes from the terms in (9.8) with j = 0, 2 mod 3, we note by (5.2c) that the case p = q^{3} sums all three terms, to give 0 . Thus the coefficient of 1 is the negative of the coefficient of ω, this gives the factor (1 − ω) in (5.2i).

Proof of (5.2h): If p = −q^{3}, we proceed as in (5.2i). We must evaluate

(9.9)  A = ∑_{j=0}^{n} \frac{(q^{−n}; q)_{j}}{(q^{n+2}; q)_{j}} \frac{1 − q^{2j+1}}{1 − q} q^{jn}(−1)^{\binom{j+1}{2}}.

This time the sign behavior is determined by (−1)^{\binom{j+1}{2}}, which is + − −+ modulo 4. We shall show that the terms arising from j = 0, 3 modulo 4 sum to

(−q^{2}; q^{2})_{n−1} \frac{(q; q)_{n}(q; q)_{n+1}}{(q; q)_{2n}(1 − q)}

as do the terms arising from j = 1, 2 modulo 4. Thus (9.6) shows that adding these two contributions yields (5.2h).

We first do the j = 0, 3 modulo 4 sum in (9.9). It is

8ϕ_{7} ( \begin{matrix}q, & q^{9/2}, & −q^{9/2}, & q, & q^{−n}, & q^{1−n}, & q^{2−n}, & q^{3−n} \\ & q^{1/2}, & −q^{1/2}, & q^{4}, & q^{n+5}, & q^{n+4}, & q^{n+3}, & q^{n+2} \end{matrix} | q^{4}, q^{4n}) +

C _{8}ϕ_{7} ( \begin{matrix}q^{7}, & q^{15/2}, & −q^{15/2}, & q^{4}, & q^{3−n}, & q^{4−n}, & q^{5−n}, & q^{6−n} \\ & q^{7/2}, & −q^{7/2}, & q^{7}, & q^{n+8}, & q^{n+7}, & q^{n+6}, & q^{n+5} \end{matrix} | q^{4}, q^{4n}) , (9.10)

where

C = \frac{(q^{1−n}; q)_{3}}{(q^{n+2}; q)_{3}} q^{3n}.

The above sum of two very-well poised _{8}ϕ_{7}’s can be transformed to an _{8}ψ_{8} using the transformation [12, (III.37)]

\frac{(r, r/bf, r/cf, r/df, r/ef, rf/b, rf/c, rf/d, rf/e; r)_{∞}}{(af, r/af, ar/f, f/a, g/f, fg, rf^{2}; r)_{∞}} ×

8ϕ_{7} ( \begin{matrix}f^{2}, & rf, & −rf, & fb, & fc, & fd, & fe, & fg \\ & f, & −f, & fr/b, & fr/c, & fr/d, & fr/e, & fr/g \end{matrix} | r, \frac{r^{2}}{bcdefg} )

+ a similar term interchanging fand g =

\frac{(ar/b, ar/c, ar/d, ar/e, r/ab, r/ad, r/ae; r)_{∞}}{(af, ag, f/a, g/a, ra^{2}, r/a^{2}; r)_{∞}} ×

8ψ_{8} ( \begin{matrix}qa, & −qa, & ba, & ca, & da, & ea, & fa, & ga \\ a, & −a, & ar/b, & ar/c, & ar/d, & ar/e, & ar/f, & ar/g \end{matrix} | r, \frac{r^{2}}{bcdefg} )

It is remarkable that the choices of r = q^{4}, f = q^{1/2}, b = q^{5/2−n}, c = q^{−1/2−n}, d = q^{1/2−n}, e = q^{3/2−n}, g = q^{7/2}, and a = q make the left-side of the above transformation a multiple of (9.10).

Thus we must evaluate

6ψ_{6} ( \begin{matrix}q^{5}, & −q^{5}, & q^{7/2−n}, & q^{1/2−n}, & q^{3/2−n}, & q^{5/2−n} \\ q, & −q, & q^{5/2+n}, & q^{11/2+n}, & q^{9/2+n}, & q^{7/2+n} \end{matrix} | r, \frac{r^{2}}{bcdefg} ) .

This may be summed using the very-well poised _{6}ψ_{6} evaluation [12, (II.33)]

6ψ_{6} ( \begin{matrix}r\sqrt{a}, & −r\sqrt{a}, & b, & c, & d, & e \\ \sqrt{a}, & −\sqrt{a}, & ar/b, & ar/c, & ar/d, & ar/e \end{matrix} | r, \frac{r^{2}}{bcde}) =

(9.11) \frac{(ar, ar/bc, ar/bd, ar/be, ar/cd, ar/ce, ar/de, r, r/a; r)_{∞}}{(ar/c, ar/d, ar/e, r/b, r/c, r/d, r/e, ra^{2}/bcde; r)_{∞}} .

The appropriate choices in (9.11) are

r = q^{4}, a = q, b = q^{7/2−n}, c = q^{1/2−n}, d = q^{3/2−n}, e = q^{5/2−n}.

The details of simplifying the infinite products are onerous, but the result is that this contributes

(9.12)  q^{n^{2}+2n}(−q^{2}; q^{2})_{n−1}

to (5.2h).

Miraculously, the j = 1, 2 modulo 4 terms contribute an identical amount. The proof is the same, using the_{8}ψ_{8} transformation, this time choosing r = q^{4}, f = q^{3/2}, b = q^{5/2−n}, c = q^{−1/2−n}, d = q^{1/2−n}, e = q^{3/2−n}, g = q^{5/2}, a = q. Again a summable very-well poised _{6}ψ_{6} results, and the infinite products again reduce to (9.12).

10. Appendix B. In this appendix we verify the uniform convergence condition of Theorem 1 for the infinite products in Theorems 2,3,4 and 5.

Lemma. If ∏_{i=1}^{∞} a_{i} and ∏_{i=1}^{∞} b_{i} converge, then

∏_{i=1}^{∞} a_{i} − ∏_{i=1}^{∞} b_{i} = ∑_{j=1}^{∞}(∏_{i=1}^{j} b_{i})(a_{j} − b_{j})( ∏_{i=j+1}^{∞} a_{i}).

Proof. For any integer J a telescoping sum is

∏_{i=1}^{∞} a_{i} − ∏_{i=1}^{∞} b_{i} = ∑_{j=1}^{J}(∏_{i=1}^{j} b_{i})(a_{j} − b_{j})( ∏_{i=j+1}^{∞} a_{i}) + b_{J}( ∏_{i=J+1}^{∞} a_{i} − ∏_{i=J+1}^{∞} b_{i}).

The last term approaches 0 as J → ∞ because the infinite products converge, so their tails approach 1, as do the individual terms. □

For Theorem 2 note that F (e^{iθ/2}) is uniformly bounded for θ ∈ [0, 2π]. We choose

a_{i} = \frac{1}{1 − 2zxp^{i} + z^{2}p^{2i}} , b_{i} = \frac{1}{1 + 2xp^{i} + z^{2}p^{2i}} , x = cos θ,

so

a_{j} − b_{j} = \frac{p^{j}(1 + z)(−2x + p^{j}(1 − z))}{(1 − 2zxp^{i} + z^{2}p^{2i})(1 + 2xp^{i} + z^{2}p^{2i})} ≤ Mp^{j}(1 + z),

for a constant M independent of x. Thus the Lemma implies that

∏_{i=1}^{∞} a_{i} − ∏_{i=1}^{∞} b_{i} ≤ M′(1 + z).

A completely analogous proof works for Theorems 3,4, and 5. Acknowledgments. This research was partially supported by NSF grants DMS

99-70865 and DMS 99-70627. Part of this work was done while the authors were visiting the Liu Bie Ju Center of Mathematical Sciences of City University of Hong Kong and they gratefully acknowledge the generous financial support and the hospitality of the City University of Hong Kong.

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12. G. Gasper and M. Rahman, Basic Hypergeometric Series, Cambridge University Press, Cambridge, 1990.

13. M.E.H. Ismail, and M. Rahman, The associated Askey-Wilson polynomials, Transactions Amer. Math. Soc. 328 (1991), 201-237.

14. M. E.H. Ismail, H. Prodinger, and D. Stanton, Schur’s determinants and partition theorems, Séminaire Lotharingien de Combinatoire B 44a (2000), 10 pages.

15. I. Schur, Ein Beitrag zur additiven Zahlentheorie und zur Theorie der Kettenbrüche, S.-B Preuss. Akad. Wiss. Phys.-Math. Kl. 54 (1917), 302–321.

16. L. Slater, Further identities of the Rogers-Ramanujan type, Proc. Lon. Math. Soc. (2) 54 (1952), 147–167.

17. D. Stanton, Gaussian Integrals and the Rogers-Ramanujan identities, Symbolic Computation, Number Theory, Special Functions, Physics, and Combinatorics (F. Garvan and M.E.H. Ismail, ed.), Kluwer, Dordrecht, 2001, pp. 255-266.

34 M. E. H. ISMAIL AND D. STANTON

Department of Mathematics, University of South Florida, Tampa, Florida 33620-5700, School of Mathematics, University of Minnesota, Minneapolis, Minnesota 55455.