the general formulae $$\begin{aligned}
\label{PBorel1}
\mathcal{B}_{M_1^2}\mathcal{B}_{M_2^2}\left \lbrace\frac{\Gamma(\alpha)}{\left(-{\overline{u}}p_1^2-u p_2^2 \right)^{\alpha}} \right \rbrace &= t^{2-\alpha} \delta\left(u-\frac{M_1^2}{M_1^2+M_2^2} \right)\end{aligned}$$ and $$\begin{aligned}
\label{PBorel2}
\mathcal{B}_{M_1^2}\mathcal{B}_{M_2^2}\left \lbrace\frac{1}{\left(m^2_{1}-p_1^2 \right)\left(m^2_{2}-p_2^2 \right)} \right \rbrace &= e^{-m_{1}^2/M_1^2-m_2^2/M_2^2}\mbox{ .}\end{aligned}$$ Here $M_i^2$ is the Borel parameter corresponding to $p_i^2$ and $t=\frac{M_1^2 M_2^2}{M_1^2+M_2^2}$. The continuum subtraction can be accomplished by a simple set of substitution rules [@Belyaev:1994zk]: $$\begin{aligned}
\label{SubstitionOne}
t^3 & \longrightarrow & t^3\left(1-e^{-S_0/t}\left(1+\frac{S_0}{t} +\frac{1}{2} \left(S_0/t \right)^2 \right) \right)\\
\label{SubstitionTwo}
t^2  & \longrightarrow & t^2 \left(1-e^{-S_0/t}\left(1+\frac{S_0}{t} \right) \right)\\
\label{Substitionthree}
t   &\longrightarrow & t\left(1-e^{-S_0/t}\right)\end{aligned}$$ with $S_0$ being the continuum threshold.\
For the process $p \to p \gamma$ the natural choices are $$\begin{aligned}
\label{BorelratioProton}
    &M_1^2=M_2^2  \end{aligned}$$ and the threshold $S_0$ coincides with the normal continuum threshold $s_P$ for the proton.

Putting everything together the final sum rule for $F_2(0)$ takes the following form: $$\begin{aligned}
\label{finalSR}
F_2(0)=& \frac{8 \pi^2  m_p {\left\langle {\overline{q}}q\right\rangle}}{|\lambda_P|^2} e^{m_p^2/t}\bigg[ \frac{e_d}{3} \varphi(1/2) t^2 \left(1-e^{-S_0/t}\left(1+\frac{S_0}{t} \right) \right) 
{\nonumber}\\
&+\left[-\frac{e_d}{4}\left( \mathds{A}(1/2)+\mathds{B}(1/2) \right) \right] t \left(1-e^{-S_0/t} \right)
{\nonumber}\\
&-\frac{e_d}{2}\int_0^{1/2} \!\! d\alpha_q \int_0^{1/2}\!\! d\alpha_{{\overline{q}}} \frac{1}{1-\alpha_q-\alpha_{{\overline{q}}}} 
{\nonumber}\\
&\qquad\quad\times \!\left[\mathcal{S}+\widetilde{\mathcal{S}}+\frac{2e_u}{e_d} \mathcal{S}_{\gamma}\right](\alpha_q,\alpha_{{\overline{q}}},1\!-\alpha_q\!-\alpha_{{\overline{q}}}) t \left(1-e^{-S_0/t} \right)
{\nonumber}\\
&+\frac{e_d}{2}\int_0^{1/2} \!\! d\alpha_q \int_0^{1/2}\!\! d\alpha_{{\overline{q}}} \frac{\alpha_q-\alpha_{{\overline{q}}}}{(1-\!\alpha_q-\!\alpha_{{\overline{q}}})^2}
{\nonumber}\\
&\qquad\quad \times \!\left[\! {T}_2-2{T}_3+{T}_4 -\frac{2e_u}{e_d}{T}^{\gamma}_4\! \right]\!(\alpha_q,\alpha_{{\overline{q}}},1\!-\alpha_q\!-\alpha_{{\overline{q}}}) t \left(1-e^{-S_0/t} \right) \bigg] \mbox{ .}\end{aligned}$$ The first term in Eq. gives the leading twist-2 contribution, which was first obtained in Ref.[@Braun:1988qv]. 