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](https://www.researchgate.net/publication/238877348_A_Further_Extension_of_the_Leibniz_Rule_to_Fractional_Derivatives_and_Its_Relation_to_Parseval's_Formula)) 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 $a,b,z$?
> 
> 
> 



Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?
It turns out that the answer is quite easy.


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 


E. Mortenson, D. Hickerson, *Hecke-type double sums, Appell-Lerch sums, and mock theta functions (I)*,
Proceedings of the London Mathematical Society, (3) 109 (2014), no. 2, 382-422.


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. 


