Finding Kuramoto Model coupling strength with limits? The following is an equation describing the coupled phases of N oscillators according to the Kuramoto model:
$$
1 = K \int_{-\pi/2}^{\pi/2}\cos^{2}\left(\theta\right)\,
{\rm g}\left(KR\sin\left(\theta\right)\right)\,{\rm d}\theta
$$
We assume that g is a symmetric distribution with zero mean, since the equation is written from the point of view of a reference frame rotating together with the oscillator.
I need to take the $\lim_{R \rightarrow0^+}$ to solve for the coupling strength Kc.
This can't be solved with Wolfram Alpha since g is not read as a gaussian distribution but instead as a constant.
How else can you solve this limit?
 A: This problem can be approached by a series in $KR$ and assuming for $g$ a Gaussian distribution. So, we have to manage
$$
   1=K\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\theta\frac{1}{\sqrt{2\pi}\sigma}
e^{-\frac{K^2R^2\sin^2\theta}{2\sigma^2}}d\theta.
$$
The limit $R\rightarrow 0^+$ can be taken under the integrale but we prefer a series in $KR$ that yields
$$
  1=K\frac{1}{\sqrt{2\pi}\sigma}\left(\frac{\pi}{2}-\frac{\pi}{2^4 \sigma ^2} K^2 R^2+\frac{\pi}{2^7 \sigma ^4}K^4 R^4+O(K^6R^6)\right).
$$
The required limit provides 
$$
  K=\sqrt{\frac{8}{\pi}}\sigma
$$
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
1 &= K \int_{-\pi/2}^{\pi/2}\cos^{2}\pars{\theta}
{\rm g}\pars{KR\sin\pars{\theta}}\,\dd\theta\
\\[3mm]&\stackrel{R \sim 0}{\sim}\ K \int_{-\pi/2}^{\pi/2}\cos^{2}\pars{\theta}
\braces{{\rm g}\pars{0} + {\rm g}'\pars{0}KR\sin\pars{\theta}
+
\half\,{\rm g}''\pars{0}\bracks{KR\sin\pars{\theta}}^{2}}\,\dd\theta
\\[3mm]&=K{\rm g}\pars{0}
\underbrace{\int_{-\pi/2}^{\pi/2}\cos^{2}\pars{\theta}\,\dd\theta}
_{\ds{=\ {\pi \over 2}}}
+
{1 \over 8}\,K^{3}{\rm g}''\pars{0}R^{2}
\underbrace{\int_{-\pi/2}^{\pi/2}\sin^{2}\pars{2\theta}\,\dd\theta}_{\ds{=\ {\pi \over 2}}}
\end{align}
$\color{#00f}{\large\ds{K = {2 \over \pi{\rm g}\pars{0}}}}$
