In this appendix we are going to see how the saddle point approximation works in general. Let us define the class of integrals:

${\displaystyle I_{N}=\int e^{Nf(x)}dx\quad \quad N\gg 1}$
Suppose that ${\displaystyle f(x)}$ has a unique maximum at ${\displaystyle x=x^{*}}$ and that ${\textstyle \lim _{|x|\to \infty }f(x)=-\infty }$. Then, expanding ${\displaystyle f}$ around ${\displaystyle x^{*}}$ we have:
${\displaystyle f(x)=f(x^{*})+{\frac {(x-x^{*})^{2}}{2}}f''(x^{*})+{\frac {(x-x^{*})^{3}}{6}}f'''(x^{*})+\cdots }$
(and obviously ${\displaystyle f'(x^{*})=0}$ because ${\displaystyle x^{*}}$ is a maximum). Setting ${\displaystyle z=x-x^{*}}$ we can write, stopping the expansion at the third order:
${\displaystyle I_{N}\approx e^{Nf(x^{*})}\int e^{N{\frac {z^{2}}{2}}f''(x^{*})+N{\frac {z^{3}}{6}}f'''(x^{*})}dz}$
Calling ${\displaystyle y=z{\sqrt {N}}}$ and remembering that ${\textstyle f''(x^{*})<0}$ since ${\displaystyle x^{*}}$ is a maximum, we have:
${\displaystyle I_{N}=e^{Nf(x^{*})}{\frac {1}{\sqrt {N}}}\int e^{-{\frac {y^{2}}{2}}|f''(x^{*})|+{\frac {y^{2}}{6{\sqrt {N}}}}f'''(x^{*})}dy}$
Therefore for very large ${\displaystyle N}$ the term proportional to ${\displaystyle f'''(x^{*})}$ in the exponential (like all the following terms of the complete expansion) is negligible, so:
${\displaystyle I_{N}\approx e^{Nf(x^{*})}{\frac {1}{\sqrt {N}}}\int e^{-{\frac {y^{2}}{2}}|f''(x^{*})|}dy}$
and computing the Gaussian integral:
${\displaystyle I_{N}\approx {\sqrt {\frac {2\pi }{N|f''(x^{*})|}}}\cdot e^{Nf(x^{*})}}$
Therefore we see that the saddle point approximation essentially states that an integral of the form ${\displaystyle I_{N}}$ can be approximated, provided that ${\displaystyle N}$ is large, with the value of the integrand calculated at its maximum (up to a multiplicative factor).