gets a pole when $D_1=0$ at $q_{\rm on}+i\epsilon$ and two poles when $D_2=0$, at $q_{a_+}+i\epsilon$ and $q_{a_-}-i\epsilon$. Then, when $q_{\rm on}=q_{a_-}$ two poles appear in opposite sides of the real axis at the same energy and the integral path cannot be deformed to avoid the singularity, which thus shows up in the result of the integral. If the width of the $K^*$ is considered, then $q_{\rm on}+i\epsilon\to q_{\rm on}+i\Gamma_{K^*}/2$ and the singularity turns into a peak. If we apply $q_{\rm on}=q_{a_-}$ (see explicit formulae in Ref. [@Bayar:2016ftu]) we find the singularity when the incoming energy in the triangle diagram is 1420 MeV. To show it, we plot in Fig. \[Fig:7\] the result for $|\tilde{t}_T|^2$ of Eq. (\[Eq:17\]). In Fig. \[Fig:7\] we see how an original singularity becomes a broad peak when $\Gamma_{K^*}\neq 0$.

![ Results for the singular diagram, $|\tilde{t}_T|^2$, of Eq. (\[Eq:17\]).[]{data-label="Fig:7"}](fig7.eps "fig:"){width="50.00000%"}\

This softened singularity, together with the propagator of the $f_1(1285)$, is what gives rise to the broad shoulder of the $f_1(1285)\to \pi a_0(980)$ in Fig. \[Fig:6\].

Detailed evaluation with the $I=0$ and $C=+$ parity of the $f_1(1285)$
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So far we did not pay attention to the isospin and $C$-parity structure of the $f_1(1285)$ and $f_1(1420)$, 