terms that will give rise to contributions of twist 5 or higher. As we will only consider terms up to twist 4, these are not relevant here. Hence, there are only four diagrams that have to be taken into account, see Fig.\[FD\].

It turns out that the diagram in Fig.2a does not give rise to contributions with an even number of $\gamma$-matrices and can be neglected.

After using Fierz identity to decompose ${\left\langle 0 \right|} q_i(x) {\overline{q}}_j(0){\left| 0 \right\rangle}_F$, one can insert the expression for the photon DAs (see Appendix \[A\]) and perform the Fourier transformation.\
We get for diagram b: $$\begin{aligned}
\label{diagramBProton}
    T^{P}_b(p,q) =& \left[ - \frac{ e_d {\left\langle {\overline{q}}q\right\rangle}} { 6 \pi^2  } \int_0^1 \! du \;\varphi(u) \ln \left(\frac{\mu^2}{-{\overline{u}}p_1^2-u p_2^2} \right)\right. 
{\nonumber}\\
    &\left. \phantom{\Big[}+\frac{e_d {\left\langle {\overline{q}}q\right\rangle}}{8 \pi^2} \int_0^1 \! du\; \frac{\mathds{A}(u)+\mathds{B}(u)}{-{\overline{u}}p_1^2-up_2^2} \right] p^{\alpha} p_{\beta} \sigma^{\beta \zeta} f_{\alpha \zeta} + \ldots \mbox{ ,}\end{aligned}$$ here the dots represent terms that do not contribute to the structure $p^{\alpha} p_{\beta} \sigma^{\beta \zeta} f_{\alpha \zeta}$ or that are just polynomials in $p_1^2$ and $p_2^2$. These will vanish after a subsequent double Borel transformation.

The calculation of the remaining two diagrams is analogous and yields $$\begin{aligned}
\label{diagramCProton}
T^{P}_{c}(p,q)=&\left[\frac{e_u {\left\langle {\overline{q}}q\right\rangle}}{2\pi^2} \int_0^1 \!\!du\;\int {\mathcal{D \underline{\alpha}} 