I am trying to understand the saddle point method used in the large N limit of matrix models.
First, for the case of the integral of a single variable I found this notes
There they say that you can approximate the integral
$$I(A)=\int_{-\infty}^\infty e^{Ag(x)} dx$$ by expanding $A g(x)$ in powers of $y=(x-x_0)\sqrt{A}$
$$Ag(x)=Ag(x_0)+\frac{1}{2}g''(x_0)y^2+\frac{g'''(x_0)}{6\sqrt{A}}y^3+\dots$$
where $x_0$ is a maximum of $g(x)$. Then we expand the exponential $e^{Ag(x)}$
$$e^{Ag(x)}=e^{Ag(x_0)}e^{\frac{1}{2}g''(x_0)y^2}(1+\frac{g'''(x_0)}{6\sqrt{A}}y^3+\frac{g''''(x_0)}{72A}y^4+\dots)$$
so now
$$I(A)=e^{Ag(x_0)}\int_{-\infty}^\infty \frac{dy}{\sqrt{A}}e^{\frac{1}{2}g''(x_0)y^2}(1+\frac{g'''(x_0)}{6\sqrt{A}}y^3+\frac{g''''(x_0)}{72A}y^4+\dots) $$
The first integral is a gaussian, the others are known, so this gives an asymptotic series
$$I(A)=e^{Ag(x_0)}\sqrt{\frac{2\pi}{-Ag''(x_0)}}(1+\frac{C_2}{A}+\frac{C_4}{A^2}+\dots) $$
Now, I want to do the same for the large N limit of partition function of a matrix model. The partition function is a multiple integral over a large number of variables
$$Z=\int_{-\infty}^\infty [dm] e^{-N^2 S(m)}$$
$$[dm]=\prod_{j=1}^{N}dm_j$$
Here, the action is a function of $N$ variables. For the gaussian model for example
$$S(m)=\frac{2}{\lambda N}\sum_{j'=1}^N m_{j'}^2-\frac{2}{N^2}\sum_{j'=2}^N\sum_{i=1}^{j'-1}\ln|m_i-m_{j'} |$$
Let's say $m_{(0)}$ is the minimun of $S(m)$. Using Taylor expansion of several variables we have
$$e^{-N^2S(m)}=e^{-N^2S(m_{(0)})} \ e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})} \ e^{-\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})-\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})+\dots}$$
where the vector $u$ is
$$u=N(m-m_{(0)})$$
$$e^{-N^2S(m)}=e^{-N^2S(m_{(0)})} \ e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})} \left(1 -\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})-\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})+\dots\right)$$
so now
$$Z=N^{-N}e^{-N^2S(m_{(0)})}\times\int [du]e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})} \left(1 -\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})-\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})+\dots\right)$$
Now, the first integral is a gaussian. The second integral
$$\int [du]e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})}\frac{1}{3!N}(u\cdot\nabla)^3S(m_{(0)})=0$$
The problem is in the third integral
$$C=\int [du]e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})}\frac{1}{4!N^2}(u\cdot\nabla)^4S(m_{(0)})$$
Since
$$(u\cdot\nabla)^4S(m_{(0)})=\sum u_{i_1}u_{i_2}u_{i_3}u_{i_4}\partial_{i_1}\partial_{i_2}\partial_{i_3}\partial_{i_4}S(m_{(0)})$$
we have
$$C=\frac{1}{4!N^2}\sum \partial_{i_1}\partial_{i_2}\partial_{i_3}\partial_{i_4}S(m_{(0)})\int [du]u_{i_1}u_{i_2}u_{i_3}u_{i_4}e^{-\frac{1}{2}(u\cdot\nabla)^2S(m_{(0)})}$$
The sum over indices will give an order $N^4$ so it seems that $C$ is bigger than the gaussian integral, the opposite of what you expect.
What is the problem here? Have I done the power counting properly? I want to get an asymptotic series where the terms get smaller and smaller.
