# Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?

The formula $$\small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,-q/t;q\right)_\infty}\frac{dt}{t},\quad |p|<|q|,\tag{1}$$ or alternatively in more symmetric form $$\sum_{n=-\infty}^\infty \frac{(bq^n,p/aq^n;p)_\infty}{(-zq^n,-q/zq^n;q)_\infty} =\int_{-\infty}^\infty\frac{\left(bq^x,p/aq^x;p\right)_\infty}{\left(-zq^x,-q/zq^x;q\right)_\infty}dx,\quad |p| < |q|$$ is a q-analogue of $$\small\sum_{n=-\infty}^\infty \frac{e^{i\theta n}}{\Gamma(b+\alpha n)\Gamma(1-a-\alpha n)}=\int\limits_{-\infty}^\infty\frac{e^{i\theta t}dt}{\Gamma(b+\alpha t)\Gamma(1-a-\alpha t)},\quad 0 < \alpha < 1,~-\pi<\theta<\pi.\tag{2}$$ It is known that both the series and the integral in $(2)$ have a closed form $\frac{(1+e^{i\theta})^{b-1-a}e^{i(1-b)\theta}}{\alpha\Gamma(b-a)}$(T. Osler, SIAM J. MATH. ANAL. Vol. 3, No. 1, 1972) and in fact that $(2)$ is a generalized binomial theorem. It is also known that if $p=q,~|b|<|pz/q|<|a|$ then $(1)$ is Ramanujan's psi sum and Ramanujan's beta-integral (see Gasper and Rahman, Basic hypergeometric series, chapters 5-6). Also there is a closed form for the sum when $b=a$ (AMM problem 11742): \begin{align} \small\sum_{n=-\infty}^\infty (aq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}=(q,-z,-{q}/{z};q)_{\infty }(p/q,a/z,pz/aq;p/q)_{\infty }/(p;p)_\infty,\quad |p|<|q|. \end{align}

Q: Is there a closed form for $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$, where $|p|<|q|$ in terms of infinite products for arbitrary $a,b,z$?

Take $p=q^2$,$~z=1$,$~b=q^2a$, then $$\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}=2 \left(q a,q/a;q^2\right){}_{\infty }\sum _{n=-\infty}^\infty \frac{\left(-1/a\right)^n q^{n^2+n}}{1-a q^{2 n+1}}.$$ The sum on the RHS is proportional to Appell-Lerch sum $$m(qa^2,q^2,q^2/a)$$ in the notation of the paper
In general Appell-Lerch sums do not have infinite product representations. For example take $a=q^{-1/2}$, then we get the sum of the type $m(1,q^2,z)$,$~z\neq 1$ which is related to mock theta function of order $2$, formula $(4.2)$ in the paper cited above. Therefore $(1)$ can not have a closed form.