# A NOTE ON CONTOUR INTEGRALS REPRESENTING JACKSON INTEGRALS AND THEIR APPLICATIONS

Changgui Zhang

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Changgui Zhang. A NOTE ON CONTOUR INTEGRALS REPRESENTING JACKSON INTEGRALS AND THEIR APPLICATIONS. 2024. hal-04855040

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A NOTE ON CONTOUR INTEGRALS REPRESENTING
JACKSON INTEGRALS AND THEIR APPLICATIONS

CHANGGUI ZHANG

ABSTRACT. In our paper with M. E. H. Ismail [6], we are interested to study the zeros of a family of entire functions introduced by Ramanujan, using contour integrals and connection formulas for solutions of linear q-difference equations. Many years ago, M. E. H. Ismail provided a simple proof of Ramanujan's $_1\psi_1$ summation formula [5]. In this note, we aim to present an analytic proof of this formula using similar approaches to those in [6]. Furthermore, building on these methods, we derive a decomposition formula for the sum-function of a q-analog of the Euler series, which has the potential to be generalized to a broader class of q-difference equations.

INTRODUCTION

Throughout the whole paper, $q$ denotes a complex number such that $0 < |q| < 1$.
Given a non-zero complex number $\lambda$ and a function defined in a neighborhood of
the $q$-spiral $\lambda q^Z$, the Jackson integral $\int_0^{\lambda\infty} f(x)d_qx$ is defined as follows :

$$
(0.1) \qquad \int_0^{\lambda\infty} f(z)d_qz = (1-q)\lambda \sum_{n\in\mathbb{Z}} f(\lambda q^n)q^n,
$$

provided that the right-hand side converges; see [2, Remark A.5.]. The following
notation will be used:

(1) First of all, the generalized $q$-Pochhammer symbol is defined for positive $n$ or infinite limits as $([3, \S 1.2])$:

$$
(a; q)_n = (1-a)(1-aq)\dots(1-aq^{n-1}), \quad (a; q)_\infty = \prod_{k=0}^{\infty} (1-aq^k).
$$

It follows that

$$
(0.2) \qquad (a; q)_n = \frac{(a; q)_\infty}{(aq^n; q)_\infty}
$$

for any non-negative integer $n$, provided $aq^n \notin q^{\mathbb{Z}_{\le 0}}$. Furthermore, this
expression is extended to $n \in \mathbb{Z}_{<0}$, assuming $aq^n \notin q^{\mathbb{Z}_{\le 0}}$. Consequently, for
$n \in \mathbb{Z}_{\ge 1}$ and $a \notin (q^{n+\mathbb{Z}_{\le 0}} \cup \{0\})$, substituting $n$ with $-n$ into (0.2) yields :

$$
(0.3) \qquad (a; q)_{-n} = \frac{(-1)^n}{a^n (q/a; q)_n} q^{n(n+1)/2}.
$$

<footnote>Date: December 24, 2024.</footnote>

<footnote>We dedicate this note to our dear friend, Professor Mourad E. H. Ismail, in cel-
ebration of his 80th anniversary. This project is supported by the Labex CEMPI (ANR-11-
LABX-0007).</footnote>
Additionally, when $a_1, ..., a_k$ are $k$ complex numbers, we write simply:

$$ (a_1, ..., a_k; q)_n = (a_1; q)_{n-1}...(a_k; q)_n; \quad (a_1, ..., a_k; q)_\infty = (a_1; q)_\infty...(a_k; q)_\infty. $$

(2) Define the Jacobi theta function as:

$$ (0.4) \qquad \theta(z) = \theta(q; z) = \sum_{n \in \mathbb{Z}} q^{n(n-1)/2} z^n, \quad z \in \mathbb{C}^*. $$

The classic Jacobi triple-product formula states:

$$ (0.5) \qquad \theta(z) = \left(q, -z, -\frac{q}{z}; q\right)_\infty. $$

Hence, $\theta(z) = 0$ if and only if $z \in (-q^\mathbb{Z})$. Additionally, the theta function satisfies:

$$ (0.6) \qquad \theta(q^n z) = x^{-n} q^{-n(n-1)/2} \theta(z), \quad n \in \mathbb{Z}. $$

(3) As in [2, (1.9)], the $q$-logarithm $\ell(q; z)$ is defined as follows:

$$ (0.7) \qquad \ell(z) = \ell(q; z) = -z \frac{\theta'(-z)}{\theta(-z)}. $$

Differentiating both sides of Equation (0.6) with respect to $z$, it follows that $\ell(z)$ satisfies the linear $q$-difference equations:

$$ (0.8) \qquad \ell(q^n z) = \ell(z) - n, \quad n \in \mathbb{Z}, $$

and has simple poles at $q^{\mathbb{Z}}$.

(4) Given $\lambda \in \mathbb{C}^*$, let $[\lambda] = [\lambda; q] = \lambda q^{\mathbb{Z}}$ denote the $q$-spiral passing through $\lambda$. Define

$$ \varepsilon(q) = \min \left( \frac{|1 - q^n|}{1 + |q|^n} : n \in \mathbb{Z}_{>0} \right), $$

and let $D(\alpha; r)$ denote the open disc $|z - \alpha| < r$. For any $\epsilon \in (0, \epsilon(q))$ and $\lambda \in \mathbb{C}^*$, notice that:

$$ (q^m \bar{D}(\lambda; \epsilon|\lambda|)) \cap (q^n \bar{D}(\lambda; \epsilon|\lambda|)) = \emptyset, \quad (m, n) \in \mathbb{Z}^2, n \neq m. $$

Define

$$ (0.9) \qquad \bar{D}_{\epsilon}^{[\lambda]} = \bigcup_{n \in \mathbb{Z}} (q^m \bar{D}(\lambda; \epsilon|\lambda|)), \quad C_{\epsilon}^{[\lambda]} = \bigcup_{n \in \mathbb{Z}} (q^n \partial^+ D(\lambda; \epsilon|\lambda|)). $$

From now on, we are ready to represent the Jackson integral (0.1) as follows.

**Lemma 0.1.** Let $\lambda \in \mathbb{C}^*$, $\epsilon \in (0, \epsilon(q))$, and let $U$ be any domain in $\mathbb{C}^*$ with $\bar{D}_{\epsilon}^{[\lambda]} \subset U$. The following equality holds for any analytic function $f$ in $U$:

$$ (0.10) \qquad \int_{0}^{\lambda\infty} f(z) d_q z = \frac{1-q}{2\pi i} \int_{C_{\epsilon}^{[\lambda]}} \ell\left(\frac{x}{\lambda}\right) f(z) dz, $$

provided that the left-hand side converges.

*Proof.* This result follows immediately from Cauchy's residue theorem; see [9, Chap. VI]. □

The remainder of this paper is divided into two parts. In the first part, we use Lemma 0.1 to present an analytic proof of Ramanujan's $_1\psi_1$ summation formula, employing approaches similar to those in [6]. In the second part, building on these methods, we derive a decomposition formula for the sum-function of a $q$-analog of the Euler series.
# 1. A PROOF OF RAMANUJAN'S 1ψ₁ SUM

The following bilateral summation formula, known as Ramanujan's 1ψ₁ sum:

$$ (1.1) \qquad \sum_{n \in \mathbb{Z}} \frac{(a; q)_n}{(b; q)_n} z^n = \frac{(q, b/a; az, q/az; q)_\infty}{(b, q/a, z, b/az; q)_\infty}, \quad |b/a| < |z| < 1, $$

where the left-hand side is by definition 1ψ₁(a; b; q, z), was first given by Ramanujan [4, p. 222, (12.12.2)]. Several mathematicians, including M. E. H. Ismail [5], have published different proofs of (1.1). For further details, see [3, p.126]. This section is to outline a proof based upon the analytic theory of q-difference equations with the help of Lemma 0.1.

Indeed, by applying (0.2), the bilateral basic hypergeometric series 1ψ₁(a; b; q, z) can rewritten as:

$$ (1.2) \qquad 1\psi_1(a; b; q, z) = \frac{(a; q)_\infty}{(b; q)_\infty} \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} z^n. $$

Using (0.2) and (0.3), one obtains that the Laurent series 1ψ₁(a; b; q, z) converges to an analytic function in the annulus $|b/a| < |z| < 1$. Moreover, this function can be continued as a meromorphic solution in $\mathbb{C}^*$ to the q-difference equation:

$$ (1.3) \qquad \left( \frac{b}{q} - az \right) y(qz) = (1-z)y(z) $$

Since $\frac{(az, q/az; q)_\infty}{(z, b/az; q)_\infty}$ is also a solution of Equation (1.3), there exists a meromorphic function $C(a, b; q, z)$ in $\mathbb{C}^*$ such that

$$ (1.4) \qquad 1\psi_1(a; b; q, z) = C(a, b; q, z) \frac{(az, q/az; q)_\infty}{(z, b/az; q)_\infty}, $$

where $C(a, b; q, qz) = C(a, b; q, z)$.

**Theorem 1.1.** The q-periodic function $C(a, b; q, z)$ defined in the connection formula (1.4) is constant and can be expressed as:

$$ (1.5) \qquad C(a, b; q, z) = \frac{(q, b/a; q)_\infty}{(b, q/a; q)_\infty}. $$

*Proof.* For the rest of this section, we assume that $|b| < |qa|$. Thus, the annulus $|b/a| < |z| < 1$ covers the fundamental domain $|q| \le |z| < 1$ with respect to the action $z \mapsto qz$. It follows that the q-periodic function $C(a, b; q, z)$ defined in (1.4) is constant. Setting $z = q$ in both (1.2) and (1.4), we obtain:

$$ C(a, b; q, q) = \frac{(q, b/qa, a; q)_\infty}{(qa, 1/a, b; q)_\infty} \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n, $$

or, equivalently:

$$ (1.6) \qquad C(a, b; q, q) = -a \frac{(q, b/qa; q)_\infty}{(b, q/a; q)_\infty} \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n. $$

By (0.1), it follows that

$$ \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n = \frac{1}{1-q} \int_0^\infty \frac{(bz; q)_\infty}{(az; q)_\infty} dz. $$
Thus, using (0.10) for $\lambda = 1$ yields:

$$ (1.7) \qquad \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n = \frac{1}{2\pi i} \int_{C_\epsilon^{[1]}} \ell(z) \frac{(bz; q)_\infty}{(az; q)_\infty} dz, $$

where $\epsilon \in (0, \epsilon(q))$. Let

$$ C_{\epsilon, \infty}^{[1/a]} = \cup_{n \in \mathbb{Z}_{\le 0}} (q^n \partial^+ D(1/a; \epsilon/|a|)). $$

Assume that $\epsilon \in (0, \epsilon(q))$ is chosen such that $C_{\epsilon, \infty}^{[1/a]} \cap C_{\epsilon}^{[1]} = \emptyset$. By applying the residue theorem, we have:

$$ \left( \int_{C_{\epsilon}^{[1]}} + \int_{C_{\epsilon, \infty}^{[1/a]}} \right) \ell(z) \frac{(bz; q)_{\infty}}{(az; q)_{\infty}} dz = 0. $$

Finally, we find that

$$ (1.8) \qquad \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n = -\frac{1}{2\pi i} \int_{C_{\epsilon, \infty}^{[1/a]}} \ell(z) \frac{(bz; q)_\infty}{(az; q)_\infty} dz \\ = -\sum_{n=0}^{\infty} \ell\left(\frac{q^{-n}}{a}\right) \mathrm{Res}\left(\frac{(bz; q)_\infty}{(az; q)_\infty} : z = \frac{q^{-n}}{a}\right). $$

By considering (0.8), we have:

$$ (1.9) \qquad \ell\left(\frac{q^{-n}}{a}\right) = \ell\left(\frac{1}{a}\right) + n. $$

A direct computation shows:

$$ \mathrm{Res} \left( \frac{(bz; q)_\infty}{(az; q)_\infty} : z = \frac{q^{-n}}{a} \right) = - \frac{(qa/b; q)_n (b/a; q)_\infty}{a(q; q)_n (q; q)_\infty} \left( \frac{b}{qa} \right)^n $$

Then, it follows from (1.8) that

$$ \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n = \frac{(b/a; q)_\infty}{a(q; q)_\infty} \sum_{n=0}^{\infty} \left( \ell\left(\frac{1}{a}\right) + n \right) \frac{(qa/b; q)_n}{(q; q)_n} \left(\frac{b}{qa}\right)^n \\ = \frac{(b/a; q)_\infty}{a(q; q)_\infty} \left( \ell\left(\frac{1}{a}\right) + \delta \right) {}_1\phi_0\left(\frac{qa}{b}; q, z\right) \Big|_{z=b/qa} $$

where $\delta = z \frac{d}{dz}$ denotes the Euler derivative. Using the q-binomial theorem [3, p.7, (1.3.2)] yields:

$$ {}_1\phi_0\left(\frac{qa}{b}; q, z\right) = \frac{(qa z/b; q)_\infty}{(z; q)_\infty}, $$

which implies:

$$ {}_1\phi_0\left(\frac{qa}{b}; q, z\right)\Big|_{z=b/qa} = 0, \quad \delta {}_1\phi_0\left(\frac{qa}{b}; q, z\right)\Big|_{z=b/qa} = -\frac{(q; q)_\infty}{(b/qa; q)_\infty}. $$

Thus, it follows from the above that

$$ \sum_{n \in \mathbb{Z}} \frac{(bq^n; q)_\infty}{(aq^n; q)_\infty} q^n = - \frac{1}{a(1 - b/qa)}. $$

Using (1.6) gives immediately (1.5).

□
Combining (1.4) with (1.5) results in (1.1) under the conditions $|b/a| < |z| < 1$ and $|b/a| < |q|$. It is worth noting that the latter condition, $|b/a| < |q|$, can be removed by employing the standard argument of analytic continuation with respect to the parameter $a$ or $b$.

## 2. q-LAPLACE TRANSFORMS AND APPELL-LERCH SERIES

As a simple example of an irregular singular $q$-difference equation, consider the following $q$-analog of the Euler differential equation:

$$ (2.1) \qquad zy\left(\frac{z}{q}\right) + y(z) = 1. $$

Define the power series $\hat{E}_q(z)$ as:

$$ (2.2) \qquad \hat{E}_q(z) = \sum_{n \ge 0} q^{-n(n-1)/2} (-z)^n $$

It is straightforward to verify that $\hat{E}_q$ is divergent anywhere in $\mathbb{C}^*$, but it formally satisfies Eq. (2.1). To construct an analytic or meromorphic solution from this formal solution, one can employ a $q$-analog of the Borel-Laplace summation method. For more details, see [1], [7] or [8].

In this section, we utilize the $q$-Laplace transform operator $\mathcal{L}_{q;1}^{[\lambda]}$, introduced in [10]. It is defined as:

$$ (2.3) \qquad \mathcal{L}_{q;1}^{[\lambda]} \phi(z) = \sum_{\xi \in [\lambda]} \frac{\phi(\xi)}{\theta(\xi/z)}, $$

where $[\lambda] = \lambda q^Z$ denotes a $q$-spiral, and $\phi$ is a germ of an analytic function at $\xi = 0$ in $\mathbb{C}$ that can be continued into a domain containing $[\lambda]$ and exhibits suitable growth at infinity. Substituting $\xi = \lambda q^n$ into (2.3), the summation can be rewritten as a Jackson integral:

$$ \mathcal{L}_{q;1}^{[\lambda]} \phi(z) = \frac{1}{1-q} \int_{0}^{\lambda\infty} \frac{\phi(\xi)}{\theta(\xi/z)} \frac{d_q \xi}{\xi}. $$

By applying Lemma 0.1, the transform can also be expressed as:

$$ (2.4) \qquad \mathcal{L}_{q;1}^{[\lambda]} \phi(z) = \frac{1}{2\pi i} \int_{C_{\xi}} \ell\left(\frac{\xi}{\lambda}\right) \frac{\phi(\xi)}{\theta(\xi/z)} \frac{d_q \xi}{\xi}, $$

where $\epsilon \in (0, \epsilon(q))$, and $C_{\xi}^{[\lambda]}$ represents a small deformation of the contour near the $q$-spiral $[\lambda]$ (see (0.9)).

Concerning Eq. (2.1) and its formal solution $\hat{E}_q(z)$, introduced in (2.2), the $q$-Borel transform $\hat{B}_{q;1}$ is defined as follows ($|2|$):

$$ \hat{E}_q(z) \mapsto \phi(\xi) := \hat{B}_{q;1} \hat{E}_q(\xi) = \frac{1}{1+\xi}. $$

Letting this into (2.4) yields the sum-function $S_{q;1}^{[\lambda]} \hat{E}_q(z)$ of $\hat{E}_q(z)$ associated with the $q$-spiral $[\lambda]$. It is defined as:

$$ (2.5) \qquad S_{q;1}^{[\lambda]} \hat{E}_q(z) = \frac{1}{2\pi i} \int_{C_{\xi}} \ell\left(\frac{\xi}{\lambda}\right) \frac{1}{(1+\xi)\theta(\xi/z)} \frac{d\xi}{\xi}, $$

where $\lambda \in \mathbb{C}^* \setminus [-1]$ and $z \in \mathbb{C}^* \setminus [-\lambda]$. Notice that $S_{q;1}^{[\lambda]} \hat{E}_q(z)$ is related with Appell-Lerch series and Mordell's integrals (see [11])
**Theorem 2.1.** For $\lambda \in \mathbb{C}^* \setminus [-1]$ and $z \in \mathbb{C}^* \setminus [-\lambda]$, the following identity holds:

$$ (2.6) \qquad S_{q;1}^{[\lambda]} \hat{E}_q(z) = \frac{\ell(-1/\lambda) - \ell(-z/\lambda)}{\theta(-1/z)} - \frac{1}{(q; q)_\infty^3} \sum_{n \in \mathbb{Z}} \frac{(-1)^n n q^{n(n-1)/2}}{1 - q^n z}, $$

where $\ell(z) = \ell(q; z)$ is the q-logarithm as defined in (0.7).

*Proof.* Let $\lambda \in \mathbb{C}^* \setminus [-1]$, and let $K$ be any non-empty compact of $\mathbb{C}^* \setminus [-\lambda]$. We only need to prove (2.6) for $z \in K$. To do this, fix $\epsilon \in (0, \epsilon(q))$ such that the three integral contours $C_{\epsilon}^{[-z]}$, $C_{\epsilon}^{[\lambda]}$ and $C(-1; \epsilon) = \partial^{+}D(-1; \epsilon)$ are mutually disjoint for all $z \in K$. Differentiating both sides of Equation (0.6) with respect to $z$ and letting $z = -1$ yields:

$$ (2.7) \qquad \theta'(-q^n) = (-1)^n q^{-n(n+1)/2} \theta'(-1), $$

from what we deduce the expression:

$$ (2.8) \qquad \frac{1}{\theta(-1/z)} = -\frac{1}{\theta'(-1)} \sum_{n \in \mathbb{Z}} \frac{(-1)^n q^{n(n-1)/2}}{1 - q^n z}. $$

Using the residue theorem and using equation (2.7), it follows from (2.5) that:

$$ \begin{align} S_{q;1}^{[\lambda]} \hat{E}_q(z) &= -\frac{1}{2\pi i} \left( \int_{C_{\epsilon}^{[-z]}} + \int_{C(-1;\epsilon)} \right) \ell\left(\frac{\xi}{\lambda}\right) \frac{1}{(1+\xi)\theta(\xi/z)} \frac{d\xi}{\xi} \tag{2.9} \\ &= \frac{\ell(-1/\lambda)}{\theta(-1/z)} + \frac{1}{\theta'(-1)} \sum_{n \in \mathbb{Z}} \ell\left(-\frac{q^n z}{\lambda}\right) \frac{(-1)^n q^{n(n-1)/2}}{1-q^n z}. \nonumber \end{align} $$

Substituting $n$ with $-n$ into (1.9) gives:

$$ \ell(-\frac{q^n z}{\lambda}) = \ell(-\frac{z}{\lambda}) - n. $$

With the help of (2.8), we deduce:

$$ (2.10) \qquad \sum_{n \in \mathbb{Z}} \ell(-\frac{q^n z}{\lambda}) \frac{(-1)^n q^{n(n-1)/2}}{1-q^n z} = -\theta'(-1) \frac{\ell(-z/\lambda)}{\theta(-1/z)} - \sum_{n \in \mathbb{Z}} \frac{(-1)^n n q^{n(n-1)/2}}{1-q^n z}. $$

Gathering (2.9) with (2.10) and using $\theta'(-1) = (q; q)_\infty^3$, we arrive at the desired identity (2.6), completing the proof. $\square$

Using (2.6), we derive the Stokes multiplier as follows:

$$ (2.11) \qquad S_{q;1}^{[\lambda]} \hat{E}_q(z) - S_{q;1}^{[\mu]} \hat{E}_q(z) = \frac{1}{\theta(-1/z)} \left( \ell(-\frac{1}{\lambda}) - \ell(-\frac{z}{\lambda}) - \ell(-\frac{1}{\mu}) + \ell(-\frac{z}{\mu}) \right), $$

where both $\lambda$ and $\mu$ belong to $\mathbb{C}^* \setminus [-1]$. It is worth noting that Eq. (2.11) provides a new expression of the q-periodic factor $K$ as defined in [2, Theorem 3.13].

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CHANGGUI ZHANG, LABORATOIRE P. PAINLEVÉ (UMR - CNRS 8524) & CEMPI, FST, UNIVERSITÉ DE LILLE, CITÉ SCIENTIFIQUE, 59655 VILLENEUVE D'ASCQ CEDEX, FRANCE
*Email address: changgui.zhang@univ-lille.fr*