$-\frac{1}{2}g_{\mbox{\scriptsize\rm A}}^2$.

Sometimes, the Adler–Weisberger sum rule is also written in the following way $$1-\frac{1}{g_{\mbox{\scriptsize\rm
A}}^2}=\frac{2M_{\mbox{\scriptsize\rm N}}^2} {\pi g_{\pi\mbox{\scriptsize\rm
N}\mbox{\scriptsize\rm N}}^2}\,
\int_{0}^{\infty}\frac{d\nu}{\nu}
\left[\sigma_{\mbox{\scriptsize\rm tot}}^{\pi^{+}
{\mbox{\scriptsize\rm p}}}(\nu)-
\sigma_{\mbox{\scriptsize\rm tot}}^{\pi^{-}{\mbox{\scriptsize\rm p}}}(\nu)
\right].$$ Both forms are equivalent, because of the Goldberger–Treiman relation [@GT58] $$g_{\mbox{\scriptsize\rm A}}M_{\mbox{\scriptsize\rm N}}=
g_{\pi\mbox{\scriptsize\rm N}\mbox{\scriptsize\rm N}}f_{\pi}.$$ The Goldberger–Treiman relation is also a low–energy theorem of QCD.

I would like to come back to the question of an unsubtracted dispersion relation for the isospin–odd amplitude $T^{(-)}(\nu)$. This, of course depends on the high–energy behaviour of the underlying theory. One can consider as a toy model the linear sigma model [@GL60] for constituent quarks. It has been shown [@Per91; @Per92] that in perturbation theory, this model does not require subtractions in the corresponding Adler–Weisberger sum rule and yields, at the one–loop level, the result $$\label{eq:lsm}
g_{\mbox{\scriptsize\rm A}}^2=1-2\frac{M_{Q}^2}{16\pi^2 f_{\pi}^2}\left(
\log\frac{M_{\sigma}^2}{M_{Q}^2}+{{\cal O}}(1)\right),$$ where $M_{Q}$ denotes the mass of the constituent quark in this model. On the other hand, if one considers an effective constituent chiral quark model like the Georgi–Manohar model [@GM84], then it can be shown [@PdeR93] that the corresponding isospin–odd amplitude $T^{(-)}$ does not obey an unsubtracted dispersion relation. Beyond the tree level, ${\mbox{\rm Re}}T^{(-)}(\nu)$ cannot be obtained from a dispersive integral; 