Asymptotic expansion for $ F(\beta)=\int_{-a}^{a} dx \, \frac{1}{e^{\beta g(x)}+1}$ as $ \beta \rightarrow +\infty$

Question

I'm trying to find the asymptotic expansion for $$ F(\beta)=\int_{-a}^{a} dx \, \frac{1}{e^{\beta g(x)}+1} \;\;\;\;\;\;\mbox{as}\;\;\;\;\;\; \beta \rightarrow +\infty.$$ where $a>0$ and $g(x)$ is a smooth, bounded and even function.

For example $g(x)=cos(x)$

Answer 1

The expression

$$\frac1{e^{\beta g(x)}+1}$$ corresponds to a sigmoid in the vicinity of the simple roots of $g(x)$, making a transition from $1$ to $0$ or conversely. For large $\beta$, the transition is very sharp and tends to a Heaviside step.

Around a root $r$, using the Taylor development to the first order, $g(x-r)\approx g'_r\,(x-r)$,

$$\int_{-a}^a\frac{dx}{e^{\beta g'_r(x-r)}+1}=2a-\frac1{\beta g'_r}\log\frac{e^{\beta g'_r(a-r)}+1}{e^{\beta g'_r(-a-r)}+1}=a-\frac1{\beta g'_r}\log\frac{e^{\beta g'_ra}e^{-\beta g'_rr}+1}{e^{-\beta g'_rr}+e^{\beta g'_ra}}.$$

When $r=0$, the second term is zero, because the sigmoid is symmetric, and the term $a$ just corresponds to the constant value $1$ in the negatives.

In a more general case of several simple roots, you will get the total length of the intervals where $g$ is negative (a constant), plus a small correction term due to the asymmetric location of the root inside $[-a,a]$.

Answer 2

As it is bounded we have $|g(x)|\leq M$ or equivalently that

$-M\leq g(x)\leq M$

$e^{-\beta M}+1\leq e^{\beta g(x)}+1\leq e^{\beta M}+1$

$\frac{1}{e^{\beta M}+1}\leq \frac{1}{e^{\beta g(x)}+1}\leq \frac{1}{e^{-\beta M}+1}$

$\int_{-a}^{a} dx \,\frac{1}{e^{\beta M}+1}\leq \int_{-a}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}\leq \int_{-a}^{a} dx \,\frac{1}{e^{-\beta M}+1}$

$\frac{2a}{e^{\beta M}+1}\leq \int_{-a}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}\leq \frac{2a}{e^{-\beta M}+1}$

$\lim_{\beta\to\infty}\frac{2a}{e^{\beta M}+1}\leq \lim_{\beta\to\infty}\int_{-a}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}\leq \lim_{\beta\to\infty}\frac{2a}{e^{-\beta M}+1}$

$0\leq \lim_{\beta\to\infty}\int_{-a}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}\leq 2a$

if $g(x)\leq0$ with $x\in[-a,a]$ then $\lim_{\beta\to\infty}\int_{-a}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}=2a$

if $g(x)\geq0$ with $x\in[-a,a]$ then $\lim_{\beta\to\infty}\int_{-a}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}=0$

(As it is even we have also have $0\leq \lim_{\beta\to\infty}\int_{0}^{a} dx \,\frac{1}{e^{\beta g(x)}+1}\leq a)$

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Even though $\cos(x)$ changes of sign in the interval, I think you still can get the limit if you write $a=r+n\frac{\pi}{2}$ with $r\in [0,\frac{\pi}{2}] $:

For $g(x)=\cos(x)$ and $a=\frac{\pi}{2},\ (g(x)>0$ with $x\in[-a,a])$

$$\lim_{\beta\to\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dx \,\frac{1}{e^{\beta g(x)}+1}=2\lim_{\beta\to\infty}\int_{0}^{\frac{\pi}{2}} dx \,\frac{1}{e^{\beta g(x)}+1}=0$$

For $g(x)=-\cos(x)$ and $a=\frac{\pi}{2},\ (g(x)\leq0$ with $x\in[-a,a])$

$$\lim_{\beta\to\infty}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dx \,\frac{1}{e^{\beta g(x)}+1}=2\lim_{\beta\to\infty}\int_{0}^{\frac{\pi}{2}} dx \,\frac{1}{e^{\beta g(x)}+1}=\pi$$

For $g(x)=\cos(x)$, and $a=\pi$

$\lim_{\beta\to\infty}\int_{-\pi}^{\pi} dx \,\frac{1}{e^{\beta \cos(x)}+1}$

$=2\lim_{\beta\to\infty}\int_{0}^{\pi} dx \,\frac{1}{e^{\beta \cos(x)}+1}$

$=2\lim_{\beta\to\infty}\int_{0}^{\frac{\pi}{2}} dx \,\frac{1}{e^{\beta g(x)}+1}+\lim_{\beta\to\infty}\int_{\frac{\pi}{2}}^{\pi} dx \,\frac{1}{e^{\beta \cos(x)}+1}$

$=0+2\lim_{\beta\to\infty}\int_{\frac{\pi}{2}}^{\pi} dx \,\frac{1}{e^{\beta \cos(x)}+1}$

$=2\lim_{\beta\to\infty}\int_{-\frac{\pi}{2}}^{0} d(x+\pi) \,\frac{1}{e^{\beta \cos(x+\pi)}+1}$

$=2\lim_{\beta\to\infty}\int_{-\frac{\pi}{2}}^{0} dx \,\frac{1}{e^{-\beta \cos(x)}+1}$

$=2\lim_{\beta\to\infty}\int_{-\frac{\pi}{2}}^{0} dx \,\frac{1}{e^{-\beta \cos(x)}+1}$

$=2\lim_{\beta\to\infty}\int_{-\frac{\pi}{2}}^{0} dx \,\frac{1}{e^{-\beta \cos(x)}+1}$

$=2\lim_{\beta\to\infty}\int_{0}^{\frac{\pi}{2}} d x \,\frac{1}{e^{-\beta \cos(x)}+1}$

$=\pi$

So we now know how it behaves for $a=n\frac{\pi}{2}\ $ when $n=1$ and $n=2$, it only grows for n even and with n odd it stays the same and from there one should be able to generalize and finally to add a remainder $r$ between $[0,\frac{\pi}{2}]$ that will add $2r$ or $0$

So the limit should be $$\lim_{\beta\to\infty}\int_{-(r+\frac{n\pi}{2})}^{r+\frac{n\pi}{2}} dx \,\frac{1}{e^{\beta \cos(x)}+1}= \begin{cases} n\pi\ \text{, n even}\\ 2r+(n-1)\pi\ \text{, n odd, }r\in[0,\frac{\pi}{2}]\\ \end{cases} $$

It looks like a sigmoid locally in the intervals where there is not sign change as Yves Daoust said, but on the entire interval taking the limit $\beta\to\infty$ and $g(x)=cos(x)$ is more like an staircase.