Asymptotic integration of a function

Question

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.

Answer 1

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.

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$\mathbf{Numeric \,check}$ at WolframAlpha: at $x=10$ and $\sigma=0.1\quad \displaystyle I=2.639\cdot 10^{-11}$

Approximation: $\,\displaystyle \sqrt\frac{\pi}{2}e^{-2x}\frac{\sigma}{x}=2.583\cdot 10^{-11}$

Answer 2

I have to apologise for the negligent approach in the first post. In fact, the result is correct - if we consider only the main term ($\sqrt\frac{\pi}{2}\,e^{-2x}\,\frac{\sigma}{x}$). The point is that the higher derivatives of $f(t)=-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}$ give growing power of $\frac{1}{\sigma}$ at $t=\frac{\sigma}{\sqrt x}$, the decomposition of $f(t)$ near $t=t_0=\frac{\sigma}{\sqrt x}$ is not converging and, therefore, is not valid. I decided to post a new solution and keep the old one - as a learning experience. You don't need to evaluate the new solution - I already got the score for the first post :)

As we are interested in getting the terms $\sim \sigma^3$, we start with the original integral, which looks (after a change of the variable) $$I(\sigma, x)=\int_0^1\exp\bigg(-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}\bigg)e^{-x(1-t^2)-\frac{1}{2}\sigma^2t^2+\sigma^2}\frac{dt}{\sqrt{1-t^2}}$$ The function $\exp\big(-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}\big)$ has a sharp maximum at $t=t_0=\frac{\sigma}{\sqrt x} \ll1$. Therefore, we can decompose the function at $t=0$. As $g(t)$ depends on $t^2$,$\,\,g(t)=g(0)+\frac{1}{2}g''(0)t^2+...$. Keeping only terms which will give $\,\sim \sigma, \,\sigma^3$ $$g(t)=e^{-x(1-t^2)-\frac{1}{2}\sigma^2t^2+\sigma^2}\frac{1}{\sqrt{1-t^2}}=e^{-x+\sigma^2}+ e^{-x}\Big(x+\frac{1}{2}\Big)t^2+...$$ and the integral we have to evaluate $$I(\sigma, x)\approx\int_0^1\exp\bigg(-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}\bigg)\Big(e^{-x+\sigma^2}+ e^{-x}\big(x+\frac{1}{2}\big)t^2\Big)dt$$ $$\boxed{\,\,I(\sigma, x)\approx e^{-x+\sigma^2}J_1(\sigma, x)+e^{-x}\Big(x+\frac{1}{2}\Big)J_2(\sigma, x)\,\,}\qquad (1)$$ where $$J_1(\sigma, x)=\int_0^1\exp\bigg(-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}\bigg)dt=\frac{\sqrt 2\sigma}{x}e^{-x}\int_0^\frac{x}{\sqrt 2\sigma}e^{-\big(s-\frac{x}{2s}\big)^2}ds$$ $$=\frac{\sqrt 2\sigma}{x}e^{-x}\int_0^\infty e^{-\big(s-\frac{x}{2s}\big)^2}ds-\frac{\sqrt 2\sigma}{x}e^{-x}\int_\frac{x}{\sqrt 2\sigma}^\infty e^{-\big(s-\frac{x}{2s}\big)^2}ds$$ The first integral can be evaluated with the help of the Glasser's master theorem, or simply via the substitution $t=\frac{x}{2s}$: $$\frac{\sqrt 2\sigma}{x}e^{-x}\int_0^\infty e^{-\big(s-\frac{x}{2s}\big)^2}ds=\sqrt\frac{\pi}{2}\frac{\sigma}{x}e^{-x}$$ The second integral can be estimated as $$\int_\frac{x}{\sqrt 2\sigma}^\infty e^{-\big(s-\frac{x}{2s}\big)^2}ds<\int_{\frac{x}{\sqrt 2\sigma}-\frac{\sigma}{\sqrt2}}^\infty e^{-t^2}dt<\int_{\frac{x}{\sigma}}^\infty e^{-t^2}dt<\frac{\sigma}{2x}e^{-\frac{x^2}{\sigma^2}}$$ and is exponentially small with regard to our main asymptotics term (we suppose $\frac{\sigma}{x}\ll1$). Therefore, with the accuracy up to exponentially small corrections, we can expand integration t0 $\infty$ in both integrals. Therefore, $$\boxed{\,\,J_1(\sigma, x)\approx\sqrt\frac{\pi}{2}\frac{\sigma}{x}e^{-x}\,\,\,}\qquad (2)$$ As for the second integral, $$J_2(\sigma, x)\approx\int_0^\infty\exp\bigg(-\frac{1}{2}\frac{\sigma^2}{t^2}-\frac{1}{2}\frac{x^2t^2}{\sigma^2}\bigg)t^2dt=\frac{2\sqrt 2\sigma^3}{x^3}\int_0^\infty \exp\Big(-s^2-\frac{x^2}{4s^2}\Big)s^2ds$$ To evaluate the last integral, we notice that $$-4\frac{\partial}{\partial (x^2)}I_0(x)=-4\frac{\partial}{\partial (x^2)}\int_0^\infty e^{-s^2-\frac{x^2}{4s^2}}s^2ds=\int_0^\infty e^{-s^2-\frac{x^2}{4s^2}}ds$$ $$=e^{-x}\int_0^\infty e^{-\big(s-\frac{x}{2s}\big)^2}ds=\frac{\sqrt\pi}{2}e^{-\sqrt{x^2}}$$ $$I_0(x)=-\frac{\sqrt\pi}{8}\int^{x^2}e^{-\sqrt{t^2}}dt+\operatorname{C}=-\frac{\sqrt\pi}{4}\int^xe^{-t}t\,dt+\operatorname{C}=\frac{\sqrt\pi}{4}e^{-x}(x+1)+\operatorname{C}$$ From $\,I_0(x=0)=\int_0^\infty e^{-s^2}s^2ds=\frac{\sqrt\pi}{4}\,$ we find that $\operatorname{C}=0,\,$ and $$\boxed{\,\,J_2(\sigma, x)\approx\frac{2\sqrt 2\sigma^3}{x^3}I_0(x)=\sqrt\frac{\pi}{2}e^{-x}(x+1)\frac{\sigma^3}{x^3}\,\,}\qquad(3)$$ Taking (2) and (3) together, and grouping the terms with corresponding powers of $\sigma$, $$I(\sigma, x)\approx e^{-x+\sigma^2}J_1(\sigma, x)+e^{-x}\Big(x+\frac{1}{2}\Big)J_2(\sigma, x)$$ $$\boxed{\boxed{\,\,I(\sigma, x)\approx \sqrt\frac{\pi}{2}\frac{\sigma}{x}e^{-2x}\Big(1+\sigma^2\frac{4x^2+3x+1}{2x^2}\Big)\,\,}}$$

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$\mathbf{Numerical \,\,verification}$ at WolframAlpha: at $\sigma=0.1$ and $x=2\quad\displaystyle I=0.0011813...$

In turn, the approximation gives $\quad\displaystyle \sqrt\frac{\pi}{2}\frac{\sigma}{x}e^{-2x}\Big(1+\sigma^2\frac{4x^2+3x+1}{2x^2}\Big) =0.00118082... $

what gives the accuracy of 0.05% - as it should be. It seems the approximation works appropriately now :)

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$\mathbf{Addendum}$

1. $\big(s-\frac{x}{2s}\big)^2>\big(s-\frac{x}{2a}\big)^2$ for $s\in[a;\infty)\,\,\Rightarrow\,\,$ $$\int_a^\infty e^{-\big(s-\frac{x}{2s}\big)^2}ds<\int_a^\infty e^{-\big(s-\frac{x}{2a}\big)^2}ds=\int_{a-\frac{x}{2a}}^\infty e^{-s^2}ds$$

2. integrating by part $$\int_{a}^\infty e^{-s^2}ds=-\frac{1}{2}\int_{a}^\infty \frac{1}{s}d(e^{-s^2})=\frac{e^{-a^2}}{2a}-\int_{a}^\infty \frac{e^{-s^2}}{s^2}ds<\frac{e^{-a^2}}{2a}$$ For $a\gg1$ we have an exponentially small term $e^{-a^2}$.