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?



Finding Kuramoto Model coupling strength with limits?
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 
$$
 K=\sqrt{\frac{8}{\pi}}\sigma
$$


