I have to calculate the following limit: $$\lim_{N\rightarrow \infty }\frac{1}{N}\log \sqrt{\beta J N} \int_{-\infty}^{\infty} dy e^{-NF(y)} $$ where $\beta, J$ are constants such that $\beta>1/J$ and $F(x)$ has a minimum in $\bar{x}$. Then he wants to use the Laplace Formula so one has to rescale the function in the minimum and rewrite the limite above as $$ \lim_{N\rightarrow \infty }\frac{1}{N}\log \sqrt{\beta J N} e^{-NF(\bar{x})} \int_{-\infty}^{\infty} dy e^{-N(F(y)-F(\bar{x}))}=-F(\bar{x})$$ So my question is: what is the Laplace formula? There's something one can get when there is integrals as $\int_{-\infty}^{\infty} dy e^{-NF(y)}$ and $F(x)\sim x^2$ ? How can have the last equality?
For completeness: the function $F$ is $$F(x)= \frac{\beta J x^2}{2}-\log \cosh [\beta J x]$$
The idea behind the Laplace approach to evaluate the asymptotics of such integral is that most of the integral will be coming from a region near the minimum, here near $\bar{x}$, and so you can Taylor-expand your integral your function $F(y)-F(\bar{x})=\frac{1}{2}F''(\bar{x})(x-\bar{x})^2+\mathcal{O}(x-\bar{x})^3$ around the minimum. If you stop at the quadratic term, you can exactly evaluate the integral in closed-form (widen the integration now back to the entire real line to obtain the asymptotics, since in the limit $N\to\infty$ the region away from $\bar{x}$ will contribute negligibly). You final result for the asymptotics would be
$$\frac{1}{2}\sqrt{\frac{2\pi}{F''(\bar{x})}}\frac{1}{N^{3/2}}\log\left(\beta J N\right)\,e^{-NF(\bar{x})}.$$