\\
\frac{s}{2}%
\end{array}%
\right] (-A\hat{v}+\frac{1}{2}Aa-\frac{1}{2}(\mu +\nu
),\frac{Ai}{2}).\end{aligned}$$Due to $|\omega _{N}\mu |=|\mu |,|\omega _{N}\nu |=|\nu |$, $e^{2\pi i\frac{i}{4A}(|\mu |^{2}+|\nu |^{2})}$ in the above formula can be attributed to $f_{1}(\mu )$ and $f_{2}(\nu )$. Finally, we have $$\begin{aligned}
P &=&\sum_{s,t}u_{1}^{s}u_{2}^{t}\Psi _{st}(u_{1}^{A},u_{2}^{A})
\nonumber
\\
&=&\sum_{s,t}e^{2\pi
i(-\frac{3st}{2A})}\sum_{a,d=0}^{1}(-1)^{ad}\int d^{2}\mu
d^{2}\nu f_{1}(\mu )f_{2}(\nu )e^{\pi i\frac{\mu \nu }{A}}e^{-2\pi i(\frac{sd%
}{2}+\frac{ta}{2})}  \nonumber \\
&&\times \theta \left[
\begin{array}{l}
\frac{s}{A} \\
\frac{t}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{2}}{2\pi }+\frac{1}{2}Ad+\frac{i}{2}(\mu -\nu ),%
\frac{Ai}{2})\theta \left[
\begin{array}{l}
\frac{t}{A} \\
\frac{s}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{1}}{2\pi }+\frac{1}{2}Aa-\frac{1}{2}(\mu +\nu ),%
\frac{Ai}{2})  \nonumber \\
&&\times \{A\sum_{a,d=0}^{1}(-1)^{ad}\int d^{2}\mu d^{2}\nu
f_{1}(\mu )f_{2}(\nu
)e^{\pi i\frac{\mu \nu }{A}}\theta (-\frac{l\hat{y}_{2}}{2\pi }+\frac{1}{2}%
Ad+\frac{i}{2}(\mu -\nu ),\frac{Ai}{2})  \nonumber \\
&&\times \theta (-\frac{l\hat{y}_{1}}{2\pi
}+\frac{1}{2}Aa-\frac{1}{2}(\mu +\nu ),\frac{Ai}{2})\}^{-1}.\end{aligned}$$In the above equation, the two $\theta $ functions can not exchange orders with each other. It holds for any integer number $A$. In the following, we present some discussion.(1) $A$ is an even number, so $\frac{Ad}{2}$ and $\frac{Aa}{2}$ are integers too. Due to (\[eq:34\]) we have$$\begin{aligned}
P &=&\sum_{s,t}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 \theta \left[
\begin{array}{l}
\frac{s}{A} \\
\frac{t}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{2}}{2\pi }+\frac{i}{2}(\mu -\nu ),\frac{Ai}{2}%
)\theta \left[
\begin{array}{l}
\frac{t}{A} \\
\frac{s}{2}%
\end{array}%
\right] (-\frac{l\hat{y}_{1}}{2\pi }-\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
