We are employing the notation ${\mathrm{\Gamma }}_x\equiv \mathrm{\Gamma }(x)$ for the gamma function.

The bulk kernel and contact kernel are given by [25], $$K_B^{(a=0,b=0)}={\left(\frac{z\overline{z}}{w\overline{w}}\right)}^{3/2}\frac{(w-\overline{w})(w^{-1}+{\overline{w}}^{-1}+z^{-1}+{\overline{z}}^{-1}-2)}{{((1-z)(1-\overline{z})(1-w)(1-\overline{w}))}^{3/4}}\frac{x^{\frac{3}{2}}}{64\pi }{}_2{F}_1(\frac{1}{2},\frac{3}{2},2,1-x),$$(30)$$K_C^{(a,b)}=(\frac{(1-w)(1-\overline{w})}{(1-z)(1-\overline{z})}{)}^{\frac{a+b}{2}}\frac{4}{\pi }\frac{1}{{\overline{w}}^2}{\left(\frac{1-{\rho }_z^2{\overline{\rho }}_z^2{\overline{\rho }}_w^2}{(1-{\rho }_z^2)(1-{\overline{\rho }}_z^2)(1-{\overline{\rho }}_w^2)}\right)}^{1/2}\frac{1-{\rho }_z{\overline{\rho }}_z{\overline{\rho }}_w^2}{(1-{\rho }_z{\overline{\rho }}_w)(1-{\overline{\rho }}_z{\overline{\rho }}_w)},$$(31)where ${}_2{F}_1$ is the hypergeometric function, and $x$ and $y$ are defined in (13) and (14).

We define the first-order differential operators ${\mathcal{D}}_2\equiv zw/(z-w)[(1-w){\partial }_w-(1-z){\partial }_z]+(z\leftrightarrow \overline{z})-1$, ${\mathcal{D}}_4\equiv z\overline{z}(w-\overline{w})/[(z-\overline{z})w\overline{w}](zw[(1-w){\partial }_w-(1-z){\partial }_z)/(z-w)+(z\leftrightarrow \overline{z})]$, and the second-order one ${\mathcal{D}}_{{\rho }_z}\equiv {\rho }_z^2{\partial }_{{\rho }_z}^2+2{\rho }_z^2(2a+2b+{\rho }_z){\partial }_{{\rho }_z}/({\rho }_z^2-1)-4ab{\rho }_z/({\rho }_z+1{)}^2$.

This is because we have ${\mathcal{D}}_{(z,\overline{z},w,\overline{w})}^{(24)}[\sqrt{z\overline{z}w\overline{w}}]={\mathcal{D}}_{(\overline{w},\overline{z},w,z)}^{(24)}[\sqrt{z\overline{z}w\overline{w}}]=0$ as well as ${\mathcal{D}}_{(z,\overline{z},w,\overline{w})}^{(24)}[f(x,y)]={\mathcal{D}}_{(\overline{w},\overline{z},w,z)}^{(24)}[f(x,y)]=-f(x,y)$, where $f(x,y)$ is any function of variables $x$ and $y$ in (13) and (14).

Here ${\tilde{S}}^{(a,b)}={\tilde{S}}_a^{(a,b)}+{\tilde{S}}_{-a}^{(a,b)}+{\tilde{S}}_b^{(b,a)}+{\tilde{S}}_{-b}^{(b,a)}$.

In terms of the variables $(w,\overline{w},z,\overline{z})$ or $(U^{\prime },V^{\prime },U,V)$, we have, $$\xi \equiv \frac{1}{2}(1+y-\frac{s}{2})=\frac{1}{2}(1+\sqrt{V{V}^{\prime }}+\frac{U(1-V^{\prime })+U^{\prime }(1-V)}{2\sqrt{U{U}^{\prime }}}),\frac{\eta }{\xi +\eta -\xi \eta }=\frac{2}{1+y+\frac{s}{2}}=\frac{2}{1+\frac{q_1}{2}}$$(32)$$\eta \equiv \frac{1}{2y}(1+y-\frac{s}{2})=\frac{1}{2\sqrt{V{V}^{\prime }}}(1+\sqrt{V{V}^{\prime }}+\frac{U(1-V^{\prime })+U^{\prime }(1-V)}{2\sqrt{U{U}^{\prime }}}),\frac{\xi }{\xi +\eta -\xi \eta }=\frac{2y}{1+y+\frac{s}{2}}=\frac{2}{1+p_2^{-1}}$$(33)where $U^{\prime }\equiv w\overline{w}$, $V^{\prime }\equiv (1-w)(1-\overline{w})$, $U\equiv z\overline{z}$, $V\equiv (1-z)(1-\overline{z})$, and $s\equiv \sqrt{w\overline{w}z\overline{z}}(w^{-1}+{\overline{w}}^{-1}+z^{-1}+{\overline{z}}^{-1}-2)$, and $y=\sqrt{(1-z)(1-\overline{z})(1-w)(1-\overline{w})}$.

Note that from (20) one can get all four terms that go into (9), using ${\tilde{S}}_{-a}^{(a,b)}={\tilde{S}}_a^{(-a,-b)}$.

${\mathrm{\Gamma }}_{{\mathrm{\Delta }}_i}^{\mathrm{t},\mathrm{u}}\equiv {\mathrm{\Gamma }}_{\frac{{\mathrm{\Delta }}_1+{\mathrm{\Delta }}_3-\mathrm{t}}{2}}{\mathrm{\Gamma }}_{\frac{{\mathrm{\Delta }}_2+{\mathrm{\Delta }}_4-\mathrm{t}}{2}}{\mathrm{\Gamma }}_{\frac{{\mathrm{\Delta }}_2+{\mathrm{\Delta }}_3-\mathrm{u}}{2}}{\mathrm{\Gamma }}_{\frac{{\mathrm{\Delta }}_1+{\mathrm{\Delta }}_4-\mathrm{u}}{2}}{\mathrm{\Gamma }}_{\frac{\mathrm{t}+\mathrm{u}-{\mathrm{\Delta }}_1-{\mathrm{\Delta }}_2}{2}}{\mathrm{\Gamma }}_{\frac{\mathrm{t}+\mathrm{u}-{\mathrm{\Delta }}_3-{\mathrm{\Delta }}_4}{2}}$.

$$\begin{gathered} H_{a^{\prime }}^{(a,b)}=\sum _{\mathrm{t},\mathrm{u}=0}^{\infty }\frac{(-1{)}^{\mathrm{t}+\mathrm{u}+1}{\mathrm{\Gamma }}_{-\frac{{ba}^{\prime }}{a}+a^{\prime }+1}{\mathrm{\Gamma }}_{\frac{{ba}^{\prime }}{a}-a^{\prime }-\mathrm{t}}{\mathrm{\Gamma }}_{-\frac{{ba}^{\prime }}{a}+u+a^{\prime }}}{2\pi {\mathrm{\Gamma }}_{\mathrm{t}+1}{\mathrm{\Gamma }}_{\mathrm{u}+1}{\mathrm{\Gamma }}_{-u-2a^{\prime }-1}{\mathrm{\Gamma }}_{\frac{{ba}^{\prime }-a(\mathrm{u}+a^{\prime }+1)}{a}}\sin (\frac{\pi (a-b)a^{\prime }}{a})} \\ \times \frac{{}_4{\tilde{F}}_3(1,-\frac{{ba}^{\prime }}{a}+a^{\prime }+1,u+2a^{\prime }+2,u+a^{\prime }-\frac{{ba}^{\prime }}{a}+2;-\frac{{ba}^{\prime }}{a}+a^{\prime }+2,1-\mathrm{t},-\mathrm{t}+a^{\prime }-\frac{{ba}^{\prime }}{a}+1;\frac{1}{v^{\prime }})}{u^{a^{\prime }+\mathrm{t}+\mathrm{u}}{u}^{\prime -a^{\prime }-\mathrm{t}-\mathrm{u}+1}{v}^{\frac{(a+b)(a-a^{\prime })}{2a}-u}{v}^{\prime \frac{1}{2}(-\frac{{ba}^{\prime }}{a}+3a^{\prime }-a-b+2\mathrm{u}+4)}} \end{gathered}$$(34)where ${}_4{\tilde{F}}_3$ is the regularized ${}_4{F}_3$ function.

For the case of identical operators, such results were obtained in [25, 27].