= 1$ contributes also to the forward $e^+e^- \to d \bar{d}$ annihilation, though here $Y_1 = -1, Y_2 = 1/3$ and therefore $b_1^{(\gamma)} \neq 0$. In order to avoid confusion between these cases, we change our notations, denoting $\Phi_1 \equiv F_1,~ \phi_1 \equiv f_1$ and $\phi_1^{(0)} \equiv f_1^{(0)}$ when $Y_1 = Y_2 = -1$. We will also use notations $\Phi_{2,3,4}$ instead of $F_{2,3,4}$ when we discuss the annihilation into leptons. Then we denote $c \equiv c_1 = 1/(8 \pi^2)$. Therefore, the lepton amplitude $\Phi_1 (\omega, \eta)$ for the particular case (\[leptons\]) obeys the Riccati equation $$\label{eqf1}
\frac{\partial \Phi_1}{\partial \eta} = 
c \Big( 2 \phi_1^{(0)}(\omega) \Phi_1 + \Phi_1^2 \Big)~,$$ with the general solution $$\label{f1general}
\Phi_1 = 
\frac{e^{2c \phi_1^{(0)}\eta}}{C \phi_1^{(0)} - 
e^{2c \phi_1^{(0)}\eta}/2 \phi_1^{(0)} } ~.$$ In order to specify $C$, we use the matching (see Eq. (\[masterint\])) $$\label{match}
\Phi_1 = \phi_1(\omega) - \phi_1^{(0)}(\omega)~,$$ when $\eta = 0$, arriving immediately at $$\label{Phi1}
\Phi_1 = 
\frac{2\phi_1^{(0)}(\phi_1 -\phi _1^{(0)})e^{2c\phi_1^{(0)}\eta}}
{\phi_1^{(0)} + \phi_1 - (\phi_1 -\phi_1^{(0)})e^{2c\phi_1^{(0)}\eta}}$$ and therefore to the following expression for the invariant amplitude $L_1 \equiv A_1$ when $Y_1 = Y_2 = -1$: $$\label{aleptf}
L_1 = 
\int_{-\imath \infty}^{\imath \infty} \frac{d \omega}{2\pi\imath}\,
\Big( \frac {s}{\mu^2}\Big)^{\omega} \phi_1^{(0)}(\omega) + 
\int_{-\imath \infty}^{\imath \infty} \frac{d \omega}{2\pi\imath}\,
\Big( \frac 