In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II. We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$. We now assume that $h > 0$. We set $\gamma_h = G_h''(u_h)$ and we denote $f_h : x \longmapsto \frac{\widehat{G}_h(x)}{x^2}$. Show then that $$\int_{-\infty}^{+\infty} \mathrm{e}^{-n\widehat{G}_h\left(\frac{t}{\sqrt{n}}\right)} \mathrm{d}t \xrightarrow[n \rightarrow +\infty]{} \sqrt{\frac{2\pi}{\gamma_h}}$$ then conclude that $\psi(h) = -G_h(u_h)$.
In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$. We now assume that $h > 0$.
We set $\gamma_h = G_h''(u_h)$ and we denote $f_h : x \longmapsto \frac{\widehat{G}_h(x)}{x^2}$.
Show then that
$$\int_{-\infty}^{+\infty} \mathrm{e}^{-n\widehat{G}_h\left(\frac{t}{\sqrt{n}}\right)} \mathrm{d}t \xrightarrow[n \rightarrow +\infty]{} \sqrt{\frac{2\pi}{\gamma_h}}$$
then conclude that $\psi(h) = -G_h(u_h)$.