This question has been completely reformulated and a new property for the function $f\_q$ has been added due to a series of helpful comments by fedja. 


Consider the integral from quantum field theory due to F.A. Smirnov (see this [MSE post](https://math.stackexchange.com/questions/2486031/prove-int-0-infty-big-gamma-big-frac13-fraci-alpha-pi-big-big6-alph)):
\begin{align}
&\int\_{-\infty}^{\infty}\prod\_{j=1}^3\Gamma\Bigl(\frac 1 3 -\frac {\alpha-\beta\_j}{2\pi i}\Bigr)
\Gamma\Bigl(\frac 1 3 +\frac {\alpha-\beta\_j}{2\pi i}\Bigr)(3\alpha-\sum\beta\_m)e^{-\frac{\alpha}2}d\alpha\nonumber\\
&=\frac{(2\pi \Gamma(\frac 2 3))^2}{\Gamma(\frac 4 3)}
\prod\_{k\neq j}\Gamma\Bigl(\frac 2 3 -\frac {\beta\_k-\beta\_j}{2\pi i}\Bigr)
e^{-\frac12\sum\beta\_m}\sum e^{\beta\_m},\quad |\text{Im}~\beta\_j|<2\pi/3.
\end{align}
Some analytical and numerical calculations suggest that it has a $q$-analog of the form
\begin{align}
&\int\_{-\infty}^\infty\prod\_{j=1}^3\frac{\Gamma\_q\left(\frac13+\frac{x-\beta\_j}{2\pi i}\right)}{\Gamma\_q\left(\frac23+\frac{x-\beta\_j}{2\pi i}\right)}\frac{\sum q^{-\frac{\beta\_m}{2 \pi i}}-\left(1+q^\frac13+q^{-\frac13}\right) q^{-\frac{x}{2 \pi i}}}{\prod\_{m}\sin\left(\frac{\pi}3-\frac{x-\beta\_m}{2 i}\right)}e^{-\frac{x}2}dx\\
&=\frac{-2\pi i}{(q;q)\_{\infty }^9}\frac{\Gamma\_q^2\left(\frac23\right)}{\Gamma\_q\left(\frac43\right)}\frac{q^{5/9}}{(1-q)^2}e^{-\frac{1}{2} (\beta\_1+\beta\_2+\beta\_3)} q^{-\frac{\beta\_1+\beta\_2+\beta\_3}{6 \pi i}}\prod\_{k\neq j}\Gamma\_q\left(\frac23+\frac{\beta\_j-\beta\_k}{2\pi i}\right)\\
&\times\frac{\theta\_q(q^{\frac{\beta\_1-\beta\_2}{2\pi i}})}{\sin\frac{\beta\_1-\beta\_2}{2 i}}\frac{\theta\_q(q^{\frac{\beta\_2-\beta\_3}{2\pi i}})}{\sin\frac{\beta\_2-\beta\_3}{2 i}} \frac{\theta\_q(q^{\frac{\beta\_3-\beta\_1}{2\pi i}})}{\sin\frac{\beta\_3-\beta\_1}{2 i}} \cdot f\_{q\_1}(\beta\_1,\beta\_2,\beta\_3).\tag{5}
\end{align}
where $q\_1=e^{-\frac{4 \pi ^2}{\ln(1/q)}}$, $\Gamma\_q$ is the [$q$-Gamma function](https://en.wikipedia.org/wiki/Q-gamma_function) and
$$
\theta\_q(z)=(q;q)\_\infty(z;q)\_\infty(q/z;q)\_\infty.
$$
The exact form of the function $f\_q$ is unknown but there is evidence that it has a relatively simple closed form expression in terms of finite combination of theta functions.


Obviously $f\_q(x,y,z)$ is symmetric in all three arguments $x,y,z$. It is known also that it has the following properties
\begin{align}
&f\_q(x+c,y+c,z+c)=e^cf\_q(x,y,z),\tag{1}\\
\\
&f\_q\left(\frac{2\pi i}{3},0,z\right)
\begin{aligned}[t]
 & =\left(e^z-1\right) \frac{\theta\_q\left(e^{\frac{4 i \pi }{3}-z}\right)}{\theta\_q\left(e^{-z}\right)}\\
 & =\left(e^z-1\right)e^{-\frac{\pi i}{3}} q\_1^{-\frac{1}{9}+\frac{z}{6 \pi i}}\frac{\theta\_{q\_1}\left(q\_1^{\frac13+\frac{z}{2\pi i}}\right)}{\theta\_{q\_1}\left(q\_1^{\frac{z}{2\pi i}}\right)}
 \end{aligned} \tag{2}\\
\\
&f\_q(x,y,z)=e^x+e^y+e^z\\
&+\frac32\left(e^x+e^y+e^z\right)\left(1+e^{x-y}+e^{y-x}+e^{y-z}+e^{z-y}+e^{z-x}+e^{x-z}\right)q\\
&+i\frac{\sqrt{3}}{2}\left(e^{-x}+e^{-y}+e^{-z}\right)\left(e^{2x}+e^{2y}+e^{2z}\right)q+O(q^2)
\tag{3}
\\
&f\_{q}(x,y,z)= e^{-\frac{\pi i}{3}} q\_1^{-\frac29} \small{\frac{\left(e^z-e^y\right) q\_1^{\frac{-2 x+y+z}{6 \pi i}}+\left(e^x-e^z\right) q\_1^{\frac{x-2 y+z}{6 \pi i}}+\left(e^y-e^x\right) q\_1^{\frac{x+y-2 z}{6 \pi i}}}{\Big(1-q\_1^{\frac{x-y}{2 \pi i}}\Big) \Big(1-q\_1^{\frac{z-x}{2 \pi i}}\Big) \Big(1-q\_1^{\frac{y-z}{2 \pi i}}\Big)}\left(1+O(q\_1^{\frac13})\right)}\tag{4}
\\
\end{align}


Eq. $(2)$ is a condition that the difference of LHS and RHS of $(5)$ is an entire function. $(3)$ has been extracted from numerical calculations. $(4)$ has been found from exact $q$-series representation of the integral assuming $\beta\_j\in\mathbb{R}$. However numerical calculations show that asymptotics $(4)$ holds if all differences $|\text{Im}~(\beta\_j-\beta\_k)|\leq \frac{2\pi}{3}$, and one can check that $(4)$ is consistent with $(2)$.


**Q:** Can anybody make a good guess from these data $1-4$ what this function $f\_q(x,y,z)$ might be?



$q$-analog of an integral from quantum field theory?
The function
$$
f\_q(x,y,z)=\sum\_{cyc}e^z\frac{\theta\_q\left(e^{\frac{2 \pi i}{3}+x-z}\right) \theta\_q\left(e^{\frac{2 \pi i}{3}+y-z}\right)}{\theta\_q\left(e^{x-z}\right) \theta\_q\left(e^{y-z}\right)}
$$
satisfies all $4$ conditions and also has been confirmed numerically.


