summation over colour index is assumed. The couplings of the ground state meson $M=(\pi, a_1, f_1)$ with quarks, $A_M$, and the corresponding couplings of its radially excited state $M'=(\pi', a_1', f_1')$, $B_M$, can be written in a form $$\begin{aligned}
\label{coefficients}
&&\!\!\!\! A_M = A_M^0\left[g_M \sin\theta_{M}^+ +g'_M f(k_\perp^2)\sin\theta_M^ -\right]\! , \nonumber \\
&&\!\!\!\! B_M =-A_M^0\left[g_M \cos\theta_{M}^+ +g'_{M} f (k_\perp^2)\cos\theta_M^ -\right].\end{aligned}$$ These expressions are obtained as a result of several procedures. The renormalization factors $g_M$ and $g_M'$ eliminate the divergent parts of the amplitudes describing the one-quark-loop self-energy transitions $M\to M$ and $M'\to M'$ correspondingly. In the NJL model they are fully determined by the requirement that such transitions generate a free Lagrangian of meson fields. Their specific values are expressed through the divergent quark-loop integrals $$I_{2, n} =
-i\frac{N_{c}}{(2\pi)^{4}}\int\frac{f^{n}(k_\perp^2)}{(m^{2} - k^2)^{2}}\theta(\Lambda_{3}^2 - k_\perp^2)
\mathrm{d}^{4}k,$$ which are regularized by imposing the three-dimensional cutoff on $|k_\perp|\leq \Lambda_3 = 1.03$ GeV (in accord with eq.(\[ff2\])), after carrying out the $k_0$ integration in the self-energy diagrams [@Volkov97; @Volkov17]. In particular, we have $$\begin{aligned}
&&g_{f_1} = g_{a_{1}}=\sqrt{\frac{3}{2I_{2,0}}}, \quad g_{\pi}=\sqrt{\frac{Z_\pi}{4I_{2,0}}}, \nonumber\\
&&g'_{f_1} = g'_{a_1} = \sqrt{\frac{3}{2I_{2,2}}}, \quad g'_{\pi}=\sqrt{\frac{1}{4I_{2,2}}}.\end{aligned}$$ Note that $Z_{\pi}$ is the factor induced by a diagonalization of the free Lagrangian for ground state mesons (the 