Saddle point method and end-to-end distribution of polymer models

Introduction
Saddle point methods are widely used in estimation of integrals with form
$$I=\int \exp (-f(x))\mathrm{d}x$$
where function $f$ can be approximated by first 2 terms of its Taylor series around some $x_0$, i.e.
$$f(x)\approx f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{1}{2}{f}^{\prime \prime }(x_0)(x-x_0{)}^2$$
The integral is thus approximated by its saddle point, where $f^{\prime }(x_0)=0$ and $f^{\prime \prime }(x_0)>0$:
$$\begin{array}{rl}I& \approx \int \exp (-f(x_0)-\frac{1}{2}{f}^{\prime \prime }(x_0)(x-x_0{)}^2)\mathrm{d}x\\ & =\exp (-f(x_0))\sqrt{\frac{2\pi }{f^{\prime \prime }(x_0)}}\end{array}$$

Examples

• Stirling's formula:

With the knowledge of $\mathrm{\Gamma }$ function we know that
$$N!={\int }_0^{\mathrm{\infty }}\exp (-x)x^N\mathrm{d}x$$
let $f(x):=x-N\ln (x)$, with large $N$, the negative part is negligible, solving $f^{\prime }(x)=0$, we have:
$$N!\approx \exp (-N+N\ln (N))\sqrt{2\pi N}=\sqrt{2\pi N}{\left(\frac{N}{e}\right)}^N$$

• Partition function:

$$Z=\int \exp (-\beta U(\mathbf{x}))\mathrm{d}\mathbf{x}$$
with
$$U(\mathbf{x})\approx U({\mathbf{x}}_0)+\frac{1}{2}(\mathbf{x}-{\mathbf{x}}_0{)}^{T}H[U](\mathbf{x}-{\mathbf{x}}_0)$$
where $H$ represents hessian matrix.

End-to-end distribution function of random walk model of polymer chains
For an $N$-step random walk model, the exact end-to-end vector distribution is
$$\begin{array}{rl}P(\mathbf{Y})& =\frac{1}{(2\pi {)}^3}\int \mathrm{d}\mathbf{k}\exp (-i\mathbf{k}\cdot \mathbf{Y}){\tilde{\varphi }}^N\\ & ={\int }_0^{\mathrm{\infty }}k\sin (kY){\left(\frac{\sin (kb)}{kb}\right)}^N\mathrm{d}k\end{array}$$
with $\varphi (\mathbf{x})=\frac{1}{4\pi b^2}\delta (|\mathbf{x}|-b)$ is the distribution of one step vector (length=$b$) and $\tilde{\varphi }$ is the characteristic function of $\varphi$; $\mathbf{Y}:=\sum _{i=1}^N{\mathbf{x}}_i$ is the end-to-end vector. Let $s=kb$ and $f(s):=i\frac{Y}{Nb}s-\ln \frac{\sin (s)}{s}$ then we have:
$$\begin{array}{rl}P& =\frac{i}{4{\pi }^2{b}^{2}Y}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}s\exp (-is\frac{Y}{b}){\left(\frac{\sin (s)}{s}\right)}^N\mathrm{d}s\\ & =\frac{i}{4{\pi }^2{b}^{2}Y}\int s\exp (-Nf(s))\mathrm{d}s\end{array}$$
in this step, the integral is extened to $(-\mathrm{\infty },+\mathrm{\infty })$ due to the symmetry of sin/cos function: the first sin function is replaced with form of $\exp (-ix)$ by Eular's equation: $\exp (ix)=\cos (x)+i\sin (x)$.

Solving for $f^{\prime }(s)=0$, one could find that $is$ satisfies
$$\coth (is)-\frac{1}{is}=\frac{Y}{Nb}$$
i.e. the Langevin function, $i{s}_0=L^{-1}(\frac{Y}{Nb})$. We therefore have:
$$\begin{array}{rl}P& \approx \frac{s_0}{4{\pi }^2{b}^{2}Y}\sqrt{\frac{2\pi }{N{f}^{\prime \prime }(s_0)}}{e}^{-Nf(s_0)}\\ & =\frac{1}{(2\pi N{b}^2{)}^{3/2}}\frac{L^{-1}(x{)}^2}{x(1-(L^{-1}(x)\csc (L^{-1}(x)){)}^2{)}^{1/2}}\\ & \times {\left(\frac{\sinh (L^{-1}(x))}{L^{-1}(x)\exp (x{L}^{-1}(x))}\right)}^N\end{array}$$
with $x:=\frac{Y}{Nb}$.