\left(\mathcal I^{j\omega}_J\Phi^{j\omega}_{-1-J}\right)(x|z)=\frac{1}{\pi^2}|1+2J||1+\alpha|\int_{\mathbb C}d^2x'd^2z'\left|\left(x-x'\right)^{2J}(z-z')^{\alpha}\right|^2\Phi^{-1-j,\omega}_{-1-J}(x'|z'),$$ where $\alpha$ could depend, in principle, on $j$, $J$, and $\omega$.

In order to explicitly determine the quantities $\alpha$ and $R^{j\omega}_J$ appearing in and , let us apply these to both vertex operators in a two-point function. More precisely, we ask for $$\begin{aligned}
\label{Kyalpha}
&& \left\langle \Phi^{j_1\omega_1}_{J_1}(x_1|z_1) \Phi^{j_2\omega_2}_{J_2}(x_2|z_2)\right\rangle= B^{j_1\omega_1}_{J_1}B^{j_2\omega_2}_{J_2}\int_{\mathbb C}d^2x_1'd^2x_2'd^2z_1'd^2z_2'\left|\left(x_1-x'_1\right)^{2J_1}(z_1-z'_1)^{\alpha_1}\right.\times \nonumber \\
&& ~~~~~ ~~~~~ ~~~~~ ~~~~~ \left.\left(x_2-x'_2\right)^{2J_2}(z_2-z'_2)^{\alpha_2}\right|^2\left\langle \Phi^{-1-j_1,\omega_1}_{-1-J_1}(x'_1|z'_1) \Phi^{-1-j_2,\omega_2}_{-1-J_2}(x'_2|z'_2)\right\rangle \end{aligned}$$ to hold, where, in analogy with , we have introduced $$B^{j\omega}_J=\frac{1}{\pi^2}|1+2J||1+\alpha|R^{j\omega}_J.$$ Under the assumption that $J=J_1\sim J_2$, the correlators on both sides of this expression take the form in Eq. (\[regular\]). We thus get $$\begin{aligned}
\label{Kyalpha2}
&& \left[\delta\left(j_{12}^+\right)+Y^{j_1\omega}_{J}\delta\left(j_{12}\right)\right]\left|x_{12}^{2J}z_{12}^{-2\Delta}\right|^2= B^{j_1\omega}_{J}B^{j_2\omega}_{J}\int_{\mathbb C}d^2x_1'd^2x_2'd^2z_1'd^2z_2'\left|\left(x_1-x'_1\right)^{2J}(z_1-z'_1)^{\alpha_1}\right. \nonumber \\
&& \times \left.\left(x_2-x'_2\right)^{2J}(z_2-z'_2)^{\alpha_2}\right|^2\left[\delta\left(j_{12}^+\right)+Y^{-1-j_1,\omega}_{-1-J}\delta\left(j_{12}\right)\right]\left|x'^{-2-2J}_{12}z'^{-2\Delta-2\omega(1+2J)}_{12}\right|^2,\end{aligned}$$ where $\omega=\omega_1=\omega_2$. If, in addition, we set $j=j_1\sim j_2$, it follows that $$\begin{aligned}
&& Y^{j\omega}_{J}\left|x_{12}^{2J}z_{12}^{-2\Delta}\right|^2= \left(B^{j\omega}_{J}\right)^2Y^{-1-j,\omega}_{-1-J}\int_{\mathbb C}d^2x_1'd^2x_2'd^2z_1'd^2z_2'\left|\left(x_1-x'_1\right)^{2J}(z_1-z'_1)^{\alpha}\right|^2\times \nonumber \\
&& ~~~~~ ~~~~~ ~~~~~ ~~~~~ \left|\left(x_2-x'_2\right)^{2J}(z_2-z'_2)^{\alpha}\right|^2\left|x'^{-2-2J}_{12}z'^{-2\Delta-2\omega(1+2J)}_{12}\right|^2.\nonumber\end{aligned}$$ By using Eq. (\[euler\]) in order to compute all integrals above, together with the following identities $$Y^{j\omega}_J Y^{-1-j,\omega}_J=1, \qquad\qquad Y^{j\omega}_J Y^{j\omega}_{-1-J}= \left[\frac{\pi B_j}{\gamma(-2j)}\right]^2
\frac{\gamma_{\omega}(-1-j-J)\gamma_{\omega}(-j+J)}{\gamma_{\omega}(1+j+J)\gamma_{\omega}(j-J)}\nonumber$$ we obtain $$\alpha = -1+\omega(1+2J),\nonumber$$ and $$\label{K}
B^{j\omega}_{J}=\frac{iB_j|1+2J|}{\pi\gamma(-2j)\gamma(\omega(1+2J))} \sqrt{\frac{\gamma(2\Delta+2\omega(1+2J))\gamma_{\omega}(-1-j-J)\gamma_{\omega}(-j+J)}{\gamma(2\Delta)\gamma_{\omega}(1+j+J)\gamma_{\omega}(j-J)}},$$ so that $$\label{R}
R^{j\omega}_{J}=\frac{i\pi B_j}{\gamma(-2j)}\frac{\gamma(1-\omega(1+2J))}{\omega|1+2J|} \sqrt{\frac{\gamma(2\Delta+2\omega(1+2J))\gamma_{\omega}(-1-j-J)\gamma_{\omega}(-j+J)}{\gamma(2\Delta)\gamma_{\omega}(1+j+J)\gamma_{\omega}(j-J)}}.$$ When $\omega=0$ it follows that $\alpha = -1$, trivializing the worldsheet integration in (\[refle1\]), as expected. Furthermore, by taking $J,\bar J\rightarrow j$ we get $$B^{j\omega}_{J}\rightarrow\frac{B_j}{V_{\mbox{\scriptsize conf}}},\nonumber$$ the 