Asymptotic integration of a function The function $$\int_0^{\frac{\pi }{2}} \exp \left\{-\frac{1}{2}\left [ \sigma ^2 \left(\cos ^2(\theta )+\frac{1}{\cos ^2(\theta )}-2\right)+\frac{x^2 \cos ^2(\theta )}{\sigma ^2}+2 x \left(1-\cos ^2(\theta )\right) \right ] \right\} \, d\theta=\frac{\sqrt{2 \pi } \sigma }{2  R }$$ can be interpreted as the implicit definition of the function $x=f(R,\sigma )$, where $R$ and $\sigma$ are positive parameters. Limiting the attention to $x>0$, is it possible to find an explicit expression for $x$ valid near $\sigma=0$. Thanks in advance.
 A: First of all we note that at $\sigma\ll1 \quad e^{-\frac{\sigma^2}{2}(\cos^2\theta-2)}\approx1-\frac{\sigma^2}{2}(\cos^2\theta-2)\approx 1\,$ - does not contribute into the main asymptotics term. Next, making a substitution $s=\frac{\pi}{2}-\theta$
$$I(\sigma, x)=\int_0^\frac{\pi}{2}\exp\bigg(-\frac{1}{2}\frac{\sigma^2}{\sin^2\theta}-\frac{1}{2}\frac{x^2\sin^2\theta}{\sigma^2}-x(1-\sin^2\theta)\bigg)d\theta$$
Making another substitution $\,t=\sin\theta$
$$ I(\sigma, x)=\int_0^1\exp\bigg(-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}-x(1-t^2)\bigg)\frac{dt}{\sqrt{1-t^2}}$$
Let's consider $f(t)=-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}$. This function has a maximum at $t=\frac{\sigma}{\sqrt x}\,$ and declines rapidly, if we move away from this point (at $\sigma\ll1$).
Indeed, $f'(t)=\frac{\sigma^2}{t^3}-\frac{x^2\,t}{\sigma^2};\quad f''(t)=-3\frac{\sigma^2}{t^4}-\frac{x^2}{\sigma^2};\quad f''\big(\frac{\sigma}{\sqrt x}\big)=-\frac{4x^2}{\sigma^2}\gg1;$ $\,f\big(\frac{\sigma}{\sqrt x}\big)=-x$.
Using the Laplace' method, we decompose $f(t)$ near the point $t=t_0=\frac{\sigma}{\sqrt x}\,$, and our integral takes the form
$$I(\sigma, x)\approx e^{-x}\int_0^1\exp\bigg(-\frac{2x^2}{\sigma^2}(t-t_0)^2-x(1-t^2)\bigg)\frac{dt}{\sqrt{1-t^2}}$$
The exponent declines very sharply; using that $t_0\ll1$, for the main asymptotics term we can simply put $-x(1-t^2)\approx -x(1-t_0^2)\approx -x$ and $\frac{1}{\sqrt{1-t^2}}\approx 1$
Our integral gets the form
$$I(\sigma, x)\approx e^{-2x}\int_0^1\exp\bigg(-\frac{2x^2}{\sigma^2}\big(t-\frac{\sigma}{\sqrt x}\big)^2\bigg)dt$$
With the accuracy up to the exponentially small terms we can expand integration till $\infty$ on the upper bound. Making a change $s=t-\frac{\sigma}{\sqrt x}$
$$\boxed{\,\,I(\sigma, x)\approx e^{-2x}\Big(\,\int_{-\infty}^\infty e^{-\frac{2x^2}{\sigma^2}s^2}ds\,-\,\int_{\frac{\sigma}{\sqrt x}}^\infty e^{-\frac{2x^2}{\sigma^2}s^2}ds\Big)=\sqrt\frac{\pi}{2}\,e^{-2x}\,\frac{\sigma}{x}\Big(1-\frac{1}{2}\operatorname{erfc}(\sqrt{2x}\,)\Big)\,\,}$$
where $\operatorname{erfc}(z)=\frac{2}{\sqrt\pi}\int_z^\infty e^{-t^2}dt$
Please note that this evaluation is valid at $\sigma\ll1$ and $\frac{\sigma}{x}\ll1$, so $x$ cannot be too small. We also kept only linear terms with respect to $\sigma$, dropping its higher powers.

$\mathbf{Numeric \,check}$ at WolframAlpha: at $x=10$ and $\sigma=0.1\quad \displaystyle I=2.639\cdot 10^{-11}$
Approximation: $\,\displaystyle \sqrt\frac{\pi}{2}e^{-2x}\frac{\sigma}{x}=2.583\cdot 10^{-11}$
