In eq. (A.22), the transverse coordinates $x_⊥$, $y_⊥$, $x'_⊥$ and $y'_⊥$ in the NLO,3 CGC correlator depend on $b_⊥$, $r_{xx'}$ and $r_{yy'}$ according to eqs. (5.18). The UV regular term is given by

$$ \left. \frac{d\sigma^{\gamma_T^*+A \to j_f+X}}{d\eta} \right|_{\text{UV-reg}} = \frac{\alpha_{\text{em}} e_f^2 N_c}{(2\pi)^2} \int d^2\mathbf{x}_\perp d^2\mathbf{y}_\perp d^2\mathbf{z}_\perp \frac{\alpha_s}{\pi^2} Q^2 \int_0^1 \frac{dz_g}{z_g} \left\{ \frac{\Theta(1-z-z_g)}{r_{zy}^2} \right. \\ \left. \times \frac{z^2(\bar{z}^2+(1-z)^2)(1-2z(1-z))}{1-z} \left[ K_1^2(QX_R) \frac{z(1-z)\bar{R}_R^2}{X_R^2} 2\mathfrak{Re}\Xi_{\text{NLO},4}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{z}_\perp, \mathbf{x}_\perp) \right. \right. \\ \left. - K_1(\bar{Q}r_{xy})K_1(\bar{Q}|\mathbf{r}_{xy}-\mathbf{r}_{zy}|) \frac{\mathbf{r}_{xy} \cdot (\mathbf{r}_{xy}-\mathbf{r}_{zy})}{r_{xy}|\mathbf{r}_{xy}-\mathbf{r}_{zy}|} 2\mathfrak{Re}C_F\Xi_{\text{LO}}(\mathbf{y}_\perp, \mathbf{x}_\perp, \mathbf{z}_\perp) \right] + \frac{\Theta(1-z-z_g)}{r_{zx}^2} \\ \left. \times \frac{z\bar{z}(z^2+(z+z_g)^2)(1-2\bar{z}(z+z_g))}{z+z_g} \left[ K_1^2(QX_R) \frac{\bar{z}(1-\bar{z})\bar{R}_R^2}{X_R^2} 2\mathfrak{Re}\Xi_{\text{NLO},4}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{z}_\perp, \mathbf{x}_\perp) \right. \right. \\ \left. - K_1(QX_R)K_1(\bar{Q}_{R2}|\mathbf{R}_R|) \frac{\sqrt{\bar{z}(1-\bar{z})}|\mathbf{R}_R|}{X_R} 4\mathfrak{Re}\Xi_{\text{NLO},1}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{z}_\perp, \mathbf{w}_{\perp,r}) \right. \\ \left. + K_1(\bar{Q}_{R2}r_{xy})K_1(\bar{Q}_{R2}|\mathbf{r}_{xy}+\mathbf{r}_{zx}|) \frac{\mathbf{r}_{xy} \cdot (\mathbf{r}_{xy}+\mathbf{r}_{zx})}{r_{xy}|\mathbf{r}_{xy}+\mathbf{r}_{zx}|} 2\mathfrak{Re}C_F\Xi_{\text{LO}}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{z}_\perp) \right] \right\}_{z_f} \quad (A.24) $$

Last,

$$ \left. \frac{d\sigma^{\gamma_T^*+A \to j_f+X}}{d\eta} \right|_{\text{NLO-c}} = \frac{\alpha_{\text{em}} e_f^2 N_c}{(2\pi)^2} \int d^2\mathbf{x}_\perp d^2\mathbf{y}_\perp d^2\mathbf{z}_\perp \frac{2\alpha_s}{\pi^2} 2\mathfrak{Re} \int_0^1 \frac{dz_g}{z_g} \left\{ 2z^2(1-z)\Theta(z-z_g) \right. \\ \left. \times \bar{Q}Q \frac{K_1(\bar{Q}|\mathbf{R}_{\text{SE1}}|)}{|\mathbf{R}_{\text{SE1}}|} K_1(QX_V) \left( -\frac{z_g(z-z_g)}{2} \left[ \frac{1}{1-z+z_g} + \frac{(1-z)(z-z_g)}{z^3} \right] \frac{\mathbf{r}_{zx} \cdot \mathbf{R}_{\text{SE1}}}{\mathbf{r}_{zx}^2} \right. \right. \\ \left. - \frac{z_g(z-z_g)(1+z_g-2z)^2}{2z^2(1-z)} \frac{(\mathbf{R}_V \times \mathbf{R}_{\text{SE1}})(\mathbf{r}_{zx} \times \mathbf{r}_{zy})}{\mathbf{r}_{zx}^2 \mathbf{r}_{zy}^2} - [z(z-z_g) + (1-z)(1-z+z_g)] \right. \\ \left. \times \left(1-\frac{z_g}{z}\right) \left(1+\frac{z_g}{1-z}\right) \left(1-\frac{z_g}{2z}-\frac{z_g}{2(1-z+z_g)}\right) \frac{(\mathbf{R}_V \cdot \mathbf{R}_{\text{SE1}})(\mathbf{r}_{zx} \cdot \mathbf{r}_{zy})}{\mathbf{r}_{zx}^2 \mathbf{r}_{zy}^2} \right) \\ \times \Xi_{\text{NLO},1}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{z}_\perp, \mathbf{w}_{\perp,v}) + \Theta(1-z-z_g)Q\bar{Q}_{R2} \frac{K_1(\bar{Q}_{R2}|\mathbf{R}_R|)}{|\mathbf{R}_R|} \frac{K_1(QX_R)}{X_R} \\ \times \left( z(z(1-z)+\bar{z}(1-\bar{z}))(z(1-\bar{z})+\bar{z}(1-z)) \frac{(\bar{\mathbf{R}}_R \cdot \mathbf{R}_R)(\mathbf{r}_{zy} \cdot (\mathbf{r}_{zy}-\mathbf{R}_R))}{\mathbf{r}_{zy}^2 (\mathbf{r}_{zy}-\mathbf{R}_R)^2} \right. \\ \left. +zz_g(\bar{z}-z)^2 \frac{(\bar{\mathbf{R}}_R \times \mathbf{R}_R)(\mathbf{R}_R \times \mathbf{r}_{zy})}{\mathbf{r}_{zy}^2 (\mathbf{r}_{zy}-\mathbf{R}_R)^2} + \frac{zz_g\bar{z}(z(1-z)\bar{z}+(z+z_g)^3)(\mathbf{r}_{zy}-\mathbf{R}_R) \cdot \mathbf{R}_R}{(z+z_g)(1-z)} (\mathbf{r}_{zy}-\mathbf{R}_R)^2 \right) \\ \times \Xi_{\text{NLO},1}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{z}_\perp, \mathbf{w}_{\perp,r}) + \Theta(1-z-z_g)Q^2 \frac{K_1^2(QX_R)}{X_R^2} \left( -\frac{z^3\bar{z}^3z_g}{z+z_g} \frac{\mathbf{r}_{zx} \cdot \mathbf{R}_R}{\mathbf{r}_{zx}^2} + \frac{z^3\bar{z}^2z_g^2}{2(z+z_g)^2} \right. \\ \left. + \frac{\bar{z}^2z^3z_g^2}{2(1-z)^2} - \frac{\bar{z}^2z^4z_g}{1-z} \frac{\mathbf{r}_{zy} \cdot \bar{\mathbf{R}}_R}{\mathbf{r}_{zy}^2} + z^2\bar{z}(z(1-z)+\bar{z}(1-\bar{z}))(z+\bar{z}-2z\bar{z}) \frac{\mathbf{r}_{zx} \cdot \mathbf{r}_{zy}}{\mathbf{r}_{zx}^2 \mathbf{r}_{zy}^2} \mathbf{R}_R \cdot \bar{\mathbf{R}}_R \right) $$
$$
\begin{equation}
\begin{aligned}
& -z^2 \bar{z} z_g (z - \bar{z})^2 \frac{\boldsymbol{r}_{zx}}{\boldsymbol{r}_{zx}^2 \boldsymbol{r}_{zy}^2} \boldsymbol{R}_R \times \bar{\boldsymbol{R}}_R - z^2 \bar{z} z_g \left( \frac{z(1-z)^2}{1-\bar{z}} \frac{\boldsymbol{r}_{zy} \cdot \bar{\boldsymbol{R}}_R}{\boldsymbol{r}_{zy}^2} + \frac{\bar{z}(1-\bar{z})^2}{1-z} \frac{\boldsymbol{r}_{zx} \cdot \boldsymbol{R}_R}{\boldsymbol{r}_{zx}^2} \right) \\
& \times \Xi_{\text{NLO},4}(\boldsymbol{x}_{\perp}, \boldsymbol{y}_{\perp}, \boldsymbol{z}_{\perp}, \boldsymbol{x}_{\perp}) + \frac{\Theta(z-z_g)}{\boldsymbol{r}_{zx}^2} (1-z)(z^2+(1-z)^2) (2z(z-z_g)+z_g^2) \\
& \times \left[ K_1(\bar{Q}|\boldsymbol{R}_{\text{SE1}}|) K_1(QX_V) \frac{\sqrt{z(1-z)}}{X_V} |\boldsymbol{R}_{\text{SE1}}| 4\mathfrak{Re}\Xi_{\text{NLO},1}(\boldsymbol{x}_{\perp}, \boldsymbol{y}_{\perp}, \boldsymbol{z}_{\perp}, \boldsymbol{w}_{\perp,v}) \right. \\
& \qquad \left. - e^{-\frac{\boldsymbol{r}_{zx}^2}{\boldsymbol{r}_{xy}^2 e^{\gamma_E}}} K_0^2(\bar{Q}r_{xy}) 4\mathfrak{Re}C_F \Xi_{\text{LO}}(\boldsymbol{x}_{\perp}, \boldsymbol{y}_{\perp}, \boldsymbol{x}_{\perp}) \right] \Biggr\} \Biggr]_{z_f} .
\end{aligned}
\tag{A.25}
\end{equation}
$$

B Calculation of the $k_{\perp}$-dependent single threshold logarithm

In this appendix, we explicitly compute the contribution to the NLO semi-inclusive single-jet production cross-section enhanced by a single logarithm of $1-z$, where $z$ denotes the longitudinal momentum fraction of the jet relative to that of the virtual photon. We detail the calculation for a longitudinally polarized photon but the rationale is identical in the transversely polarized case. Our starting point is eq. (4.8) and more precisely the term proportional to $\ln(k_{\perp}^2 r_{xy} r_{x'y}/c_0^2)$, which we label here as “sl”:

$$
\begin{equation}
\begin{aligned}
\left. \frac{d\sigma \gamma_L^* + A \to j_f + X}{d^2 \boldsymbol{k}_\perp d\eta} \right|_{\text{sl}} &= \left. \frac{\alpha_{\text{em}} e_f^2 N_c}{(2\pi)^4} \int d^2 \boldsymbol{x}_\perp d^2 \boldsymbol{x}'_\perp d^2 \boldsymbol{y}_\perp e^{-i\boldsymbol{k}_\perp \cdot \boldsymbol{r}_{xx'}} 2\mathfrak{Re} [\Xi_{\text{LO}}(\boldsymbol{x}_\perp, \boldsymbol{y}_\perp, \boldsymbol{x}'_\perp)] \right. \\
&\qquad \left. \times 8z^3 (1-z)^2 Q^2 K_0(\bar{Q}r_{xy}) K_0(\bar{Q}r_{x'y}) \frac{\alpha_s C_F}{\pi} \ln \left( \frac{\boldsymbol{k}_\perp^2 r_{xy} r_{x'y}}{c_0^2} \right) \ln \left( \frac{1}{1-z} \right) \right)
\end{aligned}
\tag{B.1}
\end{equation}
$$

and we aim to demonstrate that this term is truly single logarithmic in $1-z$ (it does not contain “hidden” double log in $1-z$). This is a priori not obvious as one could argue that the $K_0$ Bessel functions parametrically enforce $r_{xy} \sim r_{x'y} \sim 1/\bar{Q}$ so that the apparent single log coefficient could behave like $\ln(k_{\perp}^2/\bar{Q}^2)$. Since $\bar{Q}^2 \sim (1-z)Q^2$ for $z$ close to 1, one sees that this naive estimation would make the contribution given by eq. (B.1) double logarithmic in $1-z$.

To do so, we reduce the number of transverse coordinate integrals following appendix A of [67]. Assuming translational invariance in transverse space, and after the change of variables $\boldsymbol{r}_{\perp} = \boldsymbol{r}_{xy}$, $\boldsymbol{r}'_{\perp} = \boldsymbol{r}_{x'y}$, $\boldsymbol{b}_{\perp} = z\boldsymbol{x}_{\perp} + (1-z)\boldsymbol{y}_{\perp}$, we first rewrite the LO semi-inclusive single-jet cross-section as

$$
\begin{equation}
\begin{aligned}
\left. \frac{d\sigma \gamma_L^* + A \to j_f + X}{d^2 \boldsymbol{k}_\perp d\eta} \right|_{\text{LO}} &= \alpha_{\text{em}} e_f^2 N_c 16z(1-z) \mathfrak{Re} \int \frac{d^2 \boldsymbol{b}_\perp}{(2\pi)^2} \int \frac{d^2 \boldsymbol{r}_\perp}{(2\pi)} \frac{d^2 \boldsymbol{r}'_\perp}{(2\pi)} e^{-i\boldsymbol{k}_\perp \cdot (\boldsymbol{r}_\perp - \boldsymbol{r}'_\perp)} \\
&\quad \times \bar{Q}^2 K_0(\bar{Q}r_\perp) K_0(\bar{Q}r') [T_{Y_f}(\boldsymbol{r}_\perp) + T_{Y_f}(-\boldsymbol{r}'_\perp) - T_{Y_f}(\boldsymbol{r}_\perp - \boldsymbol{r}'_\perp)] ,
\end{aligned}
\tag{B.2}
\end{equation}
$$

with $T_{Y_f}(\boldsymbol{r}_\perp) = 1 - \frac{1}{N_c} \langle \text{Tr}[V(\mathbf{0}_\perp)V^\dagger(\boldsymbol{r}_\perp)] \rangle_{Y_f}$. As shown in [67], this expression can be further simplified into

$$
\begin{equation}
\begin{aligned}
\left. \frac{d\sigma \gamma_L^* + A \to j_f + X}{d^2 \boldsymbol{k}_\perp d\eta} \right|_{\text{LO}} &= \alpha_{\text{em}} e_f^2 N_c 16z(1-z) \mathfrak{Re} \int \frac{d^2 \boldsymbol{b}_\perp}{(2\pi)^2} \int \frac{d^2 \boldsymbol{r}_\perp}{(2\pi)} e^{-i\boldsymbol{k}_\perp \cdot \boldsymbol{r}_\perp} T_{Y_f}(\boldsymbol{r}_\perp) \\
&\quad \times \left[ \frac{2\bar{Q}^2}{\boldsymbol{k}_\perp^2 + \bar{Q}^2} K_0(\bar{Q}r_\perp) - \frac{\bar{Q}r_\perp}{2} K_1(\bar{Q}r_\perp) \right] .
\end{aligned}
\tag{B.3}
\end{equation}
$$
We wish to repeat this calculation with the additional $\ln(k_{\perp}^2 r_{xy} r_{x'y}/c_0^2)$ present in eq. (B.1). The terms depending on $T(\mathbf{r}_{\perp})$ and $T(-\mathbf{r}'_{\perp}) = T(\mathbf{r}'_{\perp})^*$ contribute equally and can be computed as follows. Considering the $T(\mathbf{r}_{\perp})$ term alone, one needs the integral

$$ \int \frac{d^2 \mathbf{r}'_{\perp}}{(2\pi)} e^{i \mathbf{k}_{\perp} \cdot \mathbf{r}'_{\perp}} K_0(\bar{Q} \mathbf{r}'_{\perp}) \ln \left( \frac{k_{\perp} \bar{Q}}{c_0} \right) = \frac{1}{\mathbf{k}_{\perp}^2 + \bar{Q}^2} \ln \left( \frac{k_{\perp} \bar{Q}}{\mathbf{k}_{\perp}^2 + \bar{Q}^2} \right), \quad (B.4) $$

which has been derived in appendix (A1) of [50]. Therefore, the $T(\mathbf{r}_{\perp}) + T(-\mathbf{r}'_{\perp})$ terms give

$$ \alpha_{\text{em}} e_f^2 N_c 16z (1-z) \mathfrak{R}e \frac{\alpha_s C_F}{\pi} \int \frac{d^2 \mathbf{b}_{\perp}}{(2\pi)^2} \int \frac{d^2 \mathbf{r}_{\perp}}{(2\pi)} e^{-i \mathbf{k}_{\perp} \cdot \mathbf{r}_{\perp}} T_{Y_f}(\mathbf{r}_{\perp}) \times \frac{2 \bar{Q}^2}{\mathbf{k}_{\perp}^2 + \bar{Q}^2} K_0(\bar{Q} \mathbf{r}_{\perp}) \ln \left( \frac{\mathbf{k}_{\perp}^2 \bar{Q} \mathbf{r}_{\perp}}{c_0 (\mathbf{k}_{\perp}^2 + \bar{Q}^2)} \right). \quad (B.5) $$

The term proportional to $T(\mathbf{r}_{\perp} - \mathbf{r}'_{\perp})$ is more complicated, but can be computed in a similar fashion:

$$ \int \frac{d^2 \mathbf{r}_{\perp}}{(2\pi)} \frac{d^2 \mathbf{r}'_{\perp}}{(2\pi)} e^{-i \mathbf{k}_{\perp} \cdot (\mathbf{r}_{\perp} - \mathbf{r}'_{\perp})} T_{Y_f}(\mathbf{r}_{\perp} - \mathbf{r}'_{\perp}) \bar{Q}^2 K_0(\bar{Q} \mathbf{r}_{\perp}) K_0(\bar{Q} \mathbf{r}'_{\perp}) \ln \left( \frac{\mathbf{k}_{\perp}^2 \mathbf{r}_{\perp} \mathbf{r}'_{\perp}}{c_0^2} \right) = \int \frac{d^2 \mathbf{r}_{\perp}}{(2\pi)} e^{-i \mathbf{k}_{\perp} \cdot \mathbf{r}_{\perp}} T_{Y_f}(\mathbf{r}_{\perp}) \int \frac{d^2 \mathbf{l}_{\perp}}{(2\pi)} \frac{e^{-i \mathbf{l}_{\perp} \cdot \mathbf{r}_{\perp}}}{[\mathbf{l}_{\perp}^2 + \bar{Q}^2]^2} \ln \left( \frac{\mathbf{l}_{\perp}^2 \bar{Q}^2}{[\mathbf{l}_{\perp}^2 + \bar{Q}^2]^2} \right). \quad (B.6) $$

We now define the following function

$$ \kappa_L(x) = \frac{2}{x K_1(x)} \int_0^\infty d\ell \frac{\ell J_0(\ell x)}{(\ell^2 + 1)^2} \ln \left( \frac{\ell^2}{(\ell^2 + 1)^2} \right), \quad (B.7) $$

where $x = \bar{Q}r_{\perp}$ in terms of physical variables. Without the logarithm inside the integral over $\ell$, one would simply get $\kappa_L(x) = 1$. When $x$ goes to 0 (meaning $\bar{Q}r_{\perp} \sim \sqrt{1-z}Qr_{\perp} \to 0$ when $z \to 1$), the function $\kappa_L(x)$ has a finite limit

$$ \lim_{x \to 0} \kappa_L(x) = -2. \quad (B.8) $$

This function helps us to write our final exact expression for eq. (B.1) as

$$ \frac{\mathrm{d}\sigma_{\gamma_{\mathrm{L}}^{*}+A \to j_f+X}}{\mathrm{d}^2 \mathbf{k}_{\perp} \mathrm{d}\eta}\bigg|_{\mathrm{sl}}=\alpha_{\mathrm{em}} e_f^2 N_c 16z(1-z) \mathfrak{R}e \ln \left(\frac{1}{1-z}\right) \int \frac{\mathrm{d}^2 \mathbf{b}_{\perp}}{(2\pi)^2} \int \frac{\mathrm{d}^2 \mathbf{r}_{\perp}}{(2\pi)} e^{-i \mathbf{k}_{\perp} \cdot \mathbf{r}_{\perp}} T_{Y_f}(\mathbf{r}_{\perp}) \times \frac{\alpha_s C_F}{\pi} \left[ \frac{2 \bar{Q}^2}{\mathbf{k}_{\perp}^2+\bar{Q}^2} K_0(\bar{Q} \mathbf{r}_{\perp}) \ln \left( \frac{\mathbf{k}_{\perp}^2 \bar{Q} \mathbf{r}_{\perp}}{c_0(\mathbf{k}_{\perp}^2+\bar{Q}^2)} \right) - \frac{\bar{Q} \mathbf{r}_{\perp}}{2} K_1(\bar{Q} \mathbf{r}_{\perp}) \kappa_L(\bar{Q} \mathbf{r}_{\perp}) \right]. \quad (B.9) $$

Comparing eq. (B.3) with eq. (B.9), it is clear that the integral over $\mathbf{r}_{\perp}$ does not yield an additional $\ln(1-z)$ dependence when $z \to 1$ so that the single log contribution eq. (B.1) is indeed single logarithmic in the threshold limit. In fact, thanks to eq. (B.8), one has

$$ \left. \frac{\mathrm{d}\sigma_{\gamma_{\mathrm{L}}^{*}+A \to j_f+X}}{\mathrm{d}^2 \mathbf{k}_{\perp} \mathrm{d}\eta} \right|_{\mathrm{sl}} \sim \frac{(-2) \alpha_s C_F}{\pi} \ln \left( \frac{1}{1-z} \right) \left. \frac{\mathrm{d}\sigma_{\gamma_{\mathrm{L}}^{*}+A \to j_f+X}}{\mathrm{d}^2 \mathbf{k}_{\perp} \mathrm{d}\eta} \right|_{\mathrm{LO}}, \quad (B.10) $$
which implies that the $k_⊥$-dependent single log coefficient can effectively be replaced by $-2\alpha_s C_F / \pi$ up to power suppressed terms as $z$ goes to 1.

Likewise, the LO SIDIS cross-section for transversely polarized virtual photons can be expressed in terms of $T_{Y_f}(\mathbf{r}_\perp)$ as

$$ \left. \frac{d\sigma_{\gamma_T^*+A \to j_f+X}}{d^2\mathbf{k}_\perp d\eta} \right|_{\text{LO}} = \alpha_{\text{em}} e_f^2 N_c 4z (z^2+(1-z)^2) \mathfrak{Re} \int \frac{d^2\mathbf{b}_\perp}{(2\pi)^2} \int \frac{d^2\mathbf{r}_\perp}{(2\pi)} e^{-i\mathbf{k}_\perp \cdot \mathbf{r}_\perp} T_{Y_f}(\mathbf{r}_\perp) \times \left[ \frac{i\mathbf{k}_\perp \cdot \mathbf{r}_\perp}{\bar{Q}r_\perp} \frac{2\bar{Q}^2}{\mathbf{k}_\perp^2+\bar{Q}^2} K_1(\bar{Q}r_\perp) - K_0(\bar{Q}r_\perp) + \frac{\bar{Q}r_\perp}{2} K_1(\bar{Q}r_\perp) \right], \quad (B.11) $$

while the $k_⊥$-dependent single log coefficient reads

$$ \left. \frac{d\sigma_{\gamma_T^*+A \to j_f+X}}{d^2\mathbf{k}_\perp d\eta} \right|_{\text{sl}} = \alpha_{\text{em}} e_f^2 N_c 4z (z^2 + (1-z)^2) \mathfrak{Re} \frac{\alpha_s C_F}{\pi} \ln\left(\frac{1}{1-z}\right) \int \frac{d^2\mathbf{b}_\perp}{(2\pi)^2} \times \int \frac{d^2\mathbf{r}_\perp}{(2\pi)} e^{-i\mathbf{k}_\perp \cdot \mathbf{r}_\perp} T_{Y_f}(\mathbf{r}_\perp) \left[ \frac{i\mathbf{k}_\perp \cdot \mathbf{r}_\perp}{\bar{Q}r_\perp} \frac{2\bar{Q}^2}{\mathbf{k}_\perp^2 + \bar{Q}^2} K_1(\bar{Q}r_\perp) \left( \ln\left(\frac{\mathbf{k}_\perp \cdot \mathbf{r}_\perp}{c_0}\right) - \frac{1}{2} \ln\left(\frac{\mathbf{k}_\perp^2 + \bar{Q}^2}{\mathbf{k}_\perp^2}\right) \right) + \frac{1}{2} \frac{\bar{Q}^2}{\mathbf{k}_\perp^2} \ln\left(1 + \frac{\mathbf{k}_\perp^2}{\bar{Q}^2}\right) \right] - \left( K_0(\bar{Q}r_\perp) - \frac{\bar{Q}r_\perp}{2} K_1(\bar{Q}r_\perp) \right) \kappa_T(\bar{Q}r_\perp) \right] \quad (B.12) $$

with the function $\kappa_T(x)$ defined by

$$ \kappa_T(x) \equiv \frac{2}{2K_0(x) - xK_1(x)} \int_0^\infty d\ell \frac{\ell^3 J_0(\ell x)}{(\ell^2 + 1)^2} \left[ \frac{1}{\ell} \ln(1 + \ell) - \ln\left(1 + \frac{1}{\ell}\right) \right]. \quad (B.13) $$

The function $\kappa_T(x)$ goes to 0 as $x \to 0$, which implies that the $k_⊥$-dependent coefficient of the single threshold logarithm vanishes in the $z \to 1$ limit.

## C The Fourier transform of the logarithm as a distribution

The following identities will be useful for the computations in section (5.2):

$$ \int d^2 \mathbf{r}_{\perp} f(\mathbf{r}_{\perp}) \int d^2 \mathbf{k}_{\perp} e^{-i \mathbf{k}_{\perp} \cdot \mathbf{r}_{\perp}} \ln \left( \frac{\mathbf{u}_{\perp}^2 \mathbf{k}_{\perp}^2}{c_0^2} \right) = -(4\pi) \mathcal{P}_{\mathbf{u}_{\perp}} \left( \frac{1}{\mathbf{r}_{\perp}^2} \right) [f], \quad (C.1) $$

$$ \int d^2 \mathbf{r}_{\perp} f(\mathbf{r}_{\perp}) \int d^2 \mathbf{k}_{\perp} e^{-i \mathbf{k}_{\perp} \cdot \mathbf{r}_{\perp}} \ln \left( \frac{\mathbf{r}_{\perp}^2 \mathbf{k}_{\perp}^2}{c_0^2} \right) = -4\pi \int d^2 \mathbf{r}_{\perp} \frac{f(\mathbf{r}_{\perp})}{\mathbf{r}_{\perp}^2}, \quad (C.2) $$

where we defined

$$ \mathcal{P}_{\mathbf{u}_{\perp}} \left( \frac{1}{\mathbf{r}_{\perp}^2} \right) [f] \equiv \int d^2 \mathbf{r}_{\perp} \left[ \frac{f(\mathbf{r}_{\perp}) - f(0) \Theta(\mathbf{u}_{\perp}^2 - \mathbf{r}_{\perp}^2)}{\mathbf{r}_{\perp}^2} \right]. \quad (C.3) $$

We begin with the proof of eq. (C.1) following section (2.9) in [134]:

$$ \mathcal{P}_{\mathbf{u}_{\perp}} \left( \frac{1}{\mathbf{r}_{\perp}^2} \right) [f] = \int d^2 \mathbf{r}_{\perp} \int \frac{d^2 \mathbf{k}_{\perp}}{(2\pi)^2} \tilde{f}(\mathbf{k}_{\perp}) \left[ \frac{e^{i \mathbf{k}_{\perp} \cdot \mathbf{r}_{\perp}} - \Theta(\mathbf{u}_{\perp}^2 - \mathbf{r}_{\perp}^2)}{\mathbf{r}_{\perp}^2} \right], \quad (C.4) $$
where we used the definition of $\mathcal{P}_{u_{\perp}}$ and we traded the function $f(r_{\perp})$ by its Fourier transform $\tilde{f}(k_{\perp})$ defined as

$$ \tilde{f}(k_{\perp}) = \int d^2\mathbf{r}_{\perp} e^{-ik_{\perp} \cdot \mathbf{r}_{\perp}} f(\mathbf{r}_{\perp}). \qquad (C.5) $$

Now one can easily perform the integration over $\mathbf{r}_{\perp}$

$$ \begin{aligned} \int \frac{d^2 \mathbf{r}_{\perp}}{2\pi} \left[ \frac{e^{ik_{\perp} \cdot \mathbf{r}_{\perp}} - \Theta(\mathbf{u}_{\perp}^2 - \mathbf{r}_{\perp}^2)}{\mathbf{r}_{\perp}^2} \right] &= \int \frac{d^2 \mathbf{r}_{\perp}}{2\pi} \left[ \frac{e^{ik_{\perp} \cdot \mathbf{r}_{\perp}} - \Theta(\mathbf{u}_{\perp}^2 - \mathbf{r}_{\perp}^2)}{\mathbf{r}_{\perp}^2} \right] \\ &= \int_0^{\infty} \frac{dr_{\perp}}{r_{\perp}} [J_0(k_{\perp} r_{\perp}) - \Theta(u_{\perp} - r_{\perp})] \\ &= -\frac{1}{2} \ln \left( \frac{\mathbf{u}_{\perp}^2 \mathbf{k}_{\perp}^2}{c_0^2} \right), \end{aligned} \qquad (C.6) $$

where in the last equality we used $\int_0^{\infty} \frac{dz}{z} [J_0(z) - \Theta(1-z)] = \ln(c_0)$ with $c_0 = 2 \exp(-\gamma_E)$. Combining the results in eqs. (C.4) and (C.6), we find

$$ \mathcal{P}_{u_{\perp}} \left( \frac{1}{r_{\perp}^2} \right) [f] = -\frac{1}{4\pi} \int d^2 \mathbf{k}_{\perp} \tilde{f}(\mathbf{k}_{\perp}) \ln \left( \frac{\mathbf{u}_{\perp}^2 \mathbf{k}_{\perp}^2}{c_0^2} \right). \qquad (C.7) $$

Eq. (C.1) is readily obtained by inserting eq. (C.5) into eq. (C.6).

Next we prove the identity in eq. (C.2). We re-express the identity in eq. (C.1) by rewriting the logarithm in the integrand:

$$ \begin{aligned} &-(4\pi)\mathcal{P}_{u_{\perp}}\left(\frac{1}{r_{\perp}^2}\right)[f] \\ &= \int d^2\mathbf{r}_{\perp} f(\mathbf{r}_{\perp}) \int d^2\mathbf{k}_{\perp} e^{-ik_{\perp}\cdot\mathbf{r}_{\perp}} \left[ \ln\left(\frac{(\mathbf{u}_{\perp}^2 + \mathbf{r}_{\perp}^2)\mathbf{k}_{\perp}^2}{c_0^2}\right) + \ln\left(\frac{\mathbf{u}_{\perp}^2}{(\mathbf{u}_{\perp}^2 + \mathbf{r}_{\perp}^2)}\right) \right]. \end{aligned} \qquad (C.8) $$

One then observes that the second term does not contribute as the integral over $\mathbf{k}_{\perp}$ results in a delta function $\delta^{(2)}(\mathbf{r}_{\perp})$ and then this term is proportional to $\ln(\mathbf{u}_{\perp}^2/\mathbf{u}_{\perp}^2) = 0$. Thus, we arrive at an alternative form of eq. (C.1):

$$ -(4\pi)\mathcal{P}_{u_{\perp}}\left(\frac{1}{r_{\perp}^2}\right)[f] = \int d^2\mathbf{r}_{\perp} f(\mathbf{r}_{\perp}) \int d^2\mathbf{k}_{\perp} e^{-ik_{\perp}\cdot\mathbf{r}_{\perp}} \ln\left(\frac{(\mathbf{u}_{\perp}^2 + \mathbf{r}_{\perp}^2)\mathbf{k}_{\perp}^2}{c_0^2}\right). \qquad (C.9) $$

The result in eq. (C.2) is obtained by letting $\mathbf{u}_{\perp} \to 0$ in the equation above.

## D Cancellation of final state interferences in fully inclusive DIS

For longitudinal photon, the total cross-section from these diagrams can be obtained by integrating over $\eta$ the expression given by eq. (5.17) (modulo a factor of 1/2 since eq. (5.17)
accounts for both quark and antiquark tagged jets)

$$
\begin{equation}
\sigma_{\gamma_L^*+A \to X}|_{V3 \times LO^*+R2 \times \overline{R2}^*} + \text{c.c.} =
\end{equation}
\begin{equation}
\begin{split}
& \frac{\alpha_{\text{em}} e_f^2 N_c}{(2\pi)^2} \int_0^1 dz \left[ 8z^2(1-z)Q^2 \right] \int \frac{d^2 \mathbf{b}_\perp}{2\pi} d^2 \mathbf{r}_{xx'} \frac{d^2 \mathbf{r}_{yy'}}{\mathbf{r}_{yy'}^2} \\
& \times \frac{\alpha_s}{\pi} \int_{z-1}^z \frac{dz_g}{z_g} K_0 \left( \bar{Q} \left| (z-z_g)\mathbf{r}_{xx'} + (1-z+z_g)\mathbf{r}_{yy'} \right| \right) K_0 \left( \bar{Q}_{V3} \left| z\mathbf{r}_{xx'} + (1-z)\mathbf{r}_{yy'} \right| \right) \\
& \times (1-z+z_g)(z-z_g)^2 \left[ 1 + \frac{((1-z)\mathbf{r}_{yy'} + (z-z_g)\mathbf{r}_{xx'}) \cdot (z\mathbf{r}_{xx'} + (1-z+z_g)\mathbf{r}_{yy'})}{(1-z)(1-z+z_g)\mathbf{r}_{yy'}^2 - z(z-z_g)\mathbf{r}_{xx'}^2} \right] \\
& \times 2\Re \left[ \Xi_{\text{NLO}}(\mathbf{x}_\perp, \mathbf{y}_\perp, \mathbf{x}'_\perp, \mathbf{y}'_\perp) \right],
\end{split}
\tag{D.1}
\end{equation}
$$

where the transverse coordinates in the color structure are evaluated at the coordinates $\mathbf{x}_{\perp}, \mathbf{y}_{\perp}, \mathbf{x}'_{\perp}, \mathbf{y}'_{\perp}$ are given by eqs. (5.18) in terms of the impact parameter $\mathbf{b}_{\perp}$ and the transverse vectors $\mathbf{r}_{xx'}, \mathbf{r}_{yy'}$. In eq. (D.1), we now perform the change of variable

$$
\zeta = z - z_g, \tag{D.2}
$$

$$
\zeta_g = -z_g. \tag{D.3}
$$

We have then

$$
\int_0^1 dz \int_{-1}^1 dz_g \Theta(z-z_g) \Theta(z_g+1-z) = \int_0^1 d\zeta \int_{-1}^1 d\zeta_g \Theta(\zeta-\zeta_g) \Theta(\zeta_g+1-\zeta), \quad (\text{D.4})
$$

which shows that the longitudinal phase space integrals remains the same. After this change
of variable, eq. (D.1) reads

$$
\sigma_{\gamma_L^*+A \to X}|_{V3 \times LO^*+R2 \times \overline{R2}^*} + \text{c.c.} =
\begin{equation}
\begin{split}
& \frac{\alpha_{\text{em}} e_f^2 N_c}{(2\pi)^2} \int_0^1 d\zeta \left[ 8\zeta^2(1-\zeta)Q^2 \right] \int \frac{d^2\mathbf{b}_\perp}{2\pi} \frac{d^2\mathbf{r}_{yy'}}{\mathbf{r}_{yy'}^2} \\
& \quad \times \frac{\alpha_s}{\pi} \int_{\zeta-1}^{\zeta} \frac{d\zeta_g}{(-\zeta_g)} K_0 \left( \bar{Q} \left| (\zeta-\zeta_g)\mathbf{r}_{xx'} + (1-\zeta+\zeta_g)\mathbf{r}_{yy'} \right| \right) K_0 \left( \bar{Q}_{V3} \left| \zeta\mathbf{r}_{xx'} + (1-\zeta)\mathbf{r}_{yy'} \right| \right) \\
& \quad \times (1-\zeta+\zeta_g)(\zeta-\zeta_g)^2 \left[ 1 + \frac{((1-\zeta)\mathbf{r}_{yy'} + (\zeta-\zeta_g)\mathbf{r}_{xx'}) \cdot (\zeta\mathbf{r}_{xx'} + (1-\zeta+\zeta_g)\mathbf{r}_{yy'})}{(1-\zeta)(1-\zeta+\zeta_g)\mathbf{r}_{yy'}^2 - \zeta(\zeta-\zeta_g)\mathbf{r}_{xx'}^2} \right] \\
& \quad \times 2\Re \left[ \Xi_{\text{NLO},3}(\tilde{\mathbf{x}}_\perp, \tilde{\mathbf{y}}_\perp, \tilde{\mathbf{x}}'_{\perp}, \tilde{\mathbf{y}}'_{\perp}) \right],
\end{split}
\tag{D.5}
\end{equation}
$$

with obviously, $\bar{Q}^2 = \zeta(1-\zeta)Q^2$ and $\bar{Q}_{V3}^2 = (\zeta - \zeta_g)(1 - \zeta + \zeta_g)Q^2$. Note the important minus sign in the denominator $d\zeta_g/(-\zeta_g)$. The CGC correlator is evaluated at the transverse coordinate $\tilde{\mathbf{x}}_{\perp}, \tilde{\mathbf{y}}_{\perp}, \tilde{\mathbf{x}}'_{\perp}, \tilde{\mathbf{y}}'_{\perp}$, which are functions of the new variables $\zeta, \zeta_g$:

$$
\begin{align}
\tilde{\mathbf{x}}_{\perp} &= \mathbf{b}_{\perp} + (1-\zeta)[(\zeta-\zeta_g)\mathbf{r}_{xx'} + (1-\zeta+\zeta_g)\mathbf{r}_{yy'}], \nonumber \\
\tilde{\mathbf{y}}_{\perp} &= \mathbf{b}_{\perp} - \zeta[(\zeta-\zeta_g)\mathbf{r}_{xx'} + (1-\zeta+\zeta_g)\mathbf{r}_{yy'}], \nonumber \\
\tilde{\mathbf{x}}'_{\perp} &= \mathbf{b}_{\perp} + (1-\zeta+\zeta_g)[\zeta\mathbf{r}_{xx'} + (1-\zeta)\mathbf{r}_{yy'}], \nonumber \\
\tilde{\mathbf{y}}'_{\perp} &= \mathbf{b}_{\perp} - (\zeta-\zeta_g)[\zeta\mathbf{r}_{xx'} + (1-\zeta)\mathbf{r}_{yy'}]. \tag{D.6}
\end{align}
$$
It is clear that the transverse coordinates $\tilde{x}_{\perp}$, $\tilde{y}_{\perp}$, $\tilde{x}_{\perp}'$ and $\tilde{y}_{\perp}'$ are nothing but $x'_{\perp}$, $y'_{\perp}$, $x_{\perp}$ and $y_{\perp}$ expressed in terms of $\zeta$ and $\zeta_g$. Further, the color structure satisfies the identity

$$2\Re[\Xi_{\text{NLO},3}(x_{\perp}, y_{\perp}, x'_{\perp}, y'_{\perp})] = 2\Re[\Xi_{\text{NLO},3}(x'_{\perp}, y'_{\perp}, x_{\perp}, y_{\perp})], \quad (\text{D.7})$$

which follows from $\Xi_{\text{NLO},3}(x'_{\perp}, y'_{\perp}, x_{\perp}, y_{\perp}) = \Xi^*_{\text{NLO},3}(x_{\perp}, y_{\perp}, x'_{\perp}, y'_{\perp})$. Hence, one gets

$$\sigma_{\gamma_L^*+A \to X}|_{V3 \times LO^*+R2 \times \overline{R2}^*} + \text{c.c.} = - \left( \sigma_{\gamma_L^*+A \to X}|_{V3 \times LO^*+R2 \times \overline{R2}^*} + \text{c.c.} \right), \quad (\text{D.8})$$

which implies that the sum of the total cross-section associated with diagrams $V3 \times LO^*$, $R2 \times \overline{R2}^*$ and their complex conjugate is exactly zero.

We emphasize that it is crucial to use the proper definition of the impact parameter $b_{\perp} = (z-z_g)x_{\perp} + (1-z+z_g)y_{\perp}$ for diagrams $V3 \times LO^*$ and $R2 \times \overline{R2}^*$ to have the cancellation of these diagrams also differentially in impact parameter space, namely

$$\left. \frac{d\sigma_{\gamma_L^*+A \to X}}{d^2b_{\perp}} \right|_{V3 \times LO^*+R2 \times \overline{R2}^*} + \left. \frac{d\sigma_{\gamma_L^*+A \to X}}{d^2b_{\perp}} \right|_{LO \times V3^*+R2 \times \overline{R2}^*} = 0. \quad (\text{D.9})$$

This cancellation is a requirement for the optical theorem to work differentially with the impact parameter $b_{\perp}$ (as it should), since a contribution coming from $V3 \times LO^*$ and $R2 \times \overline{R2}^*$ would contradict the results obtained in [18–20] (indeed, these results do not depend on the quadrupole correlator for instance)

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited.

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