\nu }{A}}\theta (-\frac{l\hat{y}_{2}}{2\pi }+\frac{i}{2}(\mu -\nu ),\frac{Ai%
}{2})  \nonumber \\
&&\times \theta (-\frac{l\hat{y}_{1}}{2\pi }-\frac{1}{2}(\mu +\nu ),\frac{Ai%
}{2})\}^{-1}.\end{aligned}$$the above equation is the generalization of the Boca’s formula Proposition $%
3.1(i)$ [@16].(2) $A$ is an odd number $$\begin{aligned}
P &=&\sum_{s,t=0}^{A-1}e^{-3\pi i\frac{st}{A}}\int d^{2}\mu
d^{2}\nu
f_{1}(\mu )f_{2}(\nu )e^{\pi i\frac{\mu \nu }{A}}  \nonumber \\
&&\times \sum_{a,d=0}^{1}(-1)^{ad}e^{-2\pi i(\frac{sd}{2}+\frac{ta}{2})}\theta %
\left[
\begin{array}{l}
\frac{s}{A} \\
\frac{t}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{2}}{2\pi }+\frac{Ad}{2}+\frac{i}{2}(\mu -\nu ),%
\frac{Ai}{2})\nonumber\\
&&\times\theta \left[
\begin{array}{l}
\frac{t}{A} \\
\frac{s}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{1}}{2\pi }+\frac{Aa}{2}-\frac{1}{2}(\mu +\nu ),%
\frac{Ai}{2})  \nonumber \\
&&\times \{A\int d^{2}\mu d^{2}\nu f_{1}(\mu )f_{2}(\nu )e^{\pi
i\frac{\mu
\nu }{A}}\sum_{a,d=0}^{1}(-1)^{ad}\theta (-\frac{l\hat{y}_{2}}{2\pi }+\frac{Ad}{2}+%
\frac{i}{2}(\mu -\nu ),\frac{Ai}{2})  \nonumber \\
&&\times \theta (-\frac{l\hat{y}_{1}}{2\pi
}+\frac{Aa}{2}-\frac{1}{2}(\mu +\nu ),\frac{Ai}{2})\}^{-1}.\end{aligned}$$Due to $$\begin{aligned}
&&\theta \left[
\begin{array}{l}
\frac{s}{A} \\
\frac{t}{2}%
\end{array}%
\right] (x+\frac{Ad}{2},\tau )  \nonumber \\
&=&\theta \left[
\begin{array}{l}
\frac{s}{2A} \\
0%
\end{array}%
\right] (2x,4\tau )(-1)^{sd}e^{2\pi i\frac{st}{2A}}+\theta \left[
\begin{array}{l}
\frac{s+A}{2A} \\
0%
\end{array}%
\right] (2x,4\tau )(-1)^{(s+A)d}e^{2\pi i\frac{(s+A)t}{2A}}\end{aligned}$$the numerator of $P$ can be written as $$\begin{aligned}
&&2\int d^{2}\mu d^{2}\nu f_{1}(\mu )f_{2}(\nu )e^{\pi i\frac{\mu \nu }{A}%
}\sum_{s,t=0}^{A-1}\{\bar{\theta}_{0}\theta
_{0}++\bar{\theta}_{1}\theta _{0}(-1)^{s}+\bar{\theta}_{0}\theta
_{1}(-1)^{t}+\bar{\theta}_{1}\theta
_{1}(-1)^{s+t-1}\}  \nonumber \\
&=&2\int d^{2}\mu d^{2}\nu f_{1}(\mu )f_{2}(\nu )e^{\pi i\frac{\mu
\nu }{A} }\sum_{s,t=0}^{2A-1}e^{\frac{\pi
i}{A}st}\bar{\theta}_{0}\theta _{0}\end{aligned}$$Where $\bar{\theta}_{\delta }=\theta \left[
\begin{array}{c}
\frac{t+A\delta }{2A} \\
0%
\end{array}%
\right] (2y,4\tau ),\theta _{\delta }=\theta \left[
\begin{array}{c}
\frac{t+A\delta }{2A} \\
0%
\end{array}%
\right] (2x,4\tau ),\delta =0,1$ with $x=-\frac{l\hat{y}_{2}}{2\pi 