k_{\perp}^2 \ll s$ and its contribution, $G^{\gamma}_j$, to the rhs of the IREE is: $$\begin{aligned}
\label{gphot}
G^{(\gamma)}_j = - \frac{1}{8\pi^2} b_j^{(\gamma)} 
&\Big(&\int_{\mu^2}^s \frac{dk_{\perp}^2 }{k_{\perp}^2} 
\ln(s/k_{\perp}^2)A^{(QED)}_j(s, k_{\perp}^2) +
\int_{\mu^2}^{M^2} \frac{dk_{\perp}^2 }{k_{\perp}^2}\ln(s/k_{\perp}^2)
{A'}_j(s, k_{\perp}^2, M^2) + \\ \nonumber 
&&\int_{M^2}^s \frac{dk_{\perp}^2 }{k_{\perp}^2} 
\ln(s/k_{\perp}^2) A'_j(s, k_{\perp}^2, k_{\perp}^2 )\, \Big) ~,\end{aligned}$$ where $$\begin{aligned}
\label{bphot}
b_1^{(\gamma)} = g^2 \sin^2 \theta_W \frac{(Y_2 - Y_1)^2}{4}, 
~~~b_2^{(\gamma)} = g^2 \sin^2 \theta_W \Big[ \frac{1}{6} + 
\frac{(Y_2 - Y_1)^2}{4} \Big] ~,\\ \nonumber
b_3^{(\gamma)} = g^2 \sin^2 \theta_W \frac{(Y_2 + Y_1)^2}{4}, 
~~~b_4^{(\gamma)} = g^2\sin^2 \theta_W \Big[ \frac{1}{6} + 
\frac{(Y_2 + Y_1)^2}{4} \Big] ~. \\ \nonumber\end{aligned}$$

We have used the standard notations in Eq. (\[bphot\]): $g,~ g'$ are the Standard Model couplings, $Y_1 ~(Y_2)$ is the hypercharge of the initial (final) fermions and $\theta_W$ is the Weinberg angle. The logarithmic factors in the integrands of Eq. (\[gphot\]) correspond to the integration in the longitudinal momentum space. The amplitude $A'$ in the last integral of Eq. (\[gphot\]) does not depend on $\mu$ because $k_{\perp}^2 > M^2$. Therefore it can be expressed in terms of $\tilde{A}_j(s,k_{\perp}^2)$ and $\tilde{A}_j^{(QED)}(s,k_{\perp}^2)$ : $${A'}_j(s, k_{\perp}^2, k_{\perp}^2) = \tilde{A}_j(s,
  k_{\perp}^2)-\tilde{A}_j^{(QED)}(s,k_{\perp}^2)~.$$

When the softest boson is either a $Z$ or a $W$, its DL contribution can be factorized in the region 