Asymptotic behavior of integral with Laplace's method

Question

I am working on the following integral

$\int_0^1 dx\int_0^1 dT \sqrt{1-(1-\sqrt{x}+\sqrt{xT})^2} e^{-n xT},$

as $n\rightarrow \infty$. The goal is to find the asymptotic behavior of the integral to the leading order of $n$.

Obviously, there is a saddle point at $(x,T)=(0,0)$, which is the main difficulty of this calculation. It seems this is related to the Laplace's method introduced in particular in Chapter VIII of [Wong, R. (2001). Asymptotic approximations of integral]. But when I try to change the variable $y_1=(x+T)/2$, $y_2=(x-T)/2$, $y_1=\sqrt{\xi}\cosh{\mu}$, $y_2=\sqrt{\xi}\cosh{\mu}$. It seems all the higher order terms of $\xi$ in the expansion of $\sqrt{1-(1-\sqrt{x}+\sqrt{xT})^2}$ will contribute to the results.

Thanks for the satisfying solutions. I think the question is already addressed. This question is made up by myself with the goal of addressing a more complex problem posted here Follow up question about Asymptotic behavior of integral with Laplace's method This may need some more efforts and possibly a more general method.

Answer 1

Elaborating on Maxim's comment. If $v=xT$, then \begin{align*} I(n)&=\int_0^1\!\! {\int_0^1 {\mathrm{e}^{ - nxT} \sqrt {1 - (1 - \sqrt x + \sqrt {xT} )^2 } \,\mathrm{d}T}\, \mathrm{d}x} \\& = \int_0^1\!\! {\int_0^x {\mathrm{e}^{ - nv} \frac{{\sqrt {1 - (1 - \sqrt x + \sqrt v )^2 } }}{x}\,\mathrm{d}v} \,\mathrm{d}x} \\ & = \int_0^1 {\mathrm{e}^{ - nv} \int_v^1 {\frac{{\sqrt {1 - (1 - \sqrt x + \sqrt v )^2 } }}{x}\,\mathrm{d}x} \,\mathrm{d}v} . \end{align*} Now, \begin{align*} & \int_v^1 {\frac{{\sqrt {1 - (1 - \sqrt x + \sqrt v )^2 } }}{x}\,\mathrm{d}x} \\ & = 2(\sqrt {1 - v} + (1 + \sqrt v )\arccos (\sqrt v ) - v^{1/4} \sqrt {2 + \sqrt v } \arccos (v + \sqrt v - 1)) \\ & = (\pi + 2) - 2\sqrt 2 \pi v^{1/4} + (\pi + 2)v^{1/2} - \frac{{\sqrt 2 }}{2}\pi v^{3/4} + \frac{1}{3}v + \ldots \end{align*} as $v\to 0^+$. Thus, by Watson's lemma, $$ I(n) \sim \frac{{\pi + 2}}{n} - \frac{{\sqrt 2 \pi \Gamma (1/4)}}{{2n^{5/4} }} + \frac{{(\pi + 2)\sqrt \pi }}{{2n^{3/2} }} - \frac{{3\sqrt 2 \pi \Gamma (3/4)}}{{8n^{7/4} }} + \frac{1}{{3n^2 }} + \ldots $$ as $n\to +\infty$.

Answer 2

As suggested in the comments by Ian, make the change of variables $u=xT$, $v=x$, after which your integral takes the form $$\int_0^1\int_0^v \left(1-(1-\sqrt{v}-\sqrt{u})^2\right)^{1/2}e^{-nu}\frac{1}{v}du\,dv.$$ We Taylor expand the square root term to obtain that the integral equals $$\int_0^1\int_0^v \left(\left(1-(1-\sqrt{v})^2\right)^{1/2}+O(u)\right)e^{-nu}\frac{1}{v}du\,dv=\int_0^1 \frac{\left(1-(1-\sqrt{v})^2\right)^{1/2}}{n}dv+O\left(\frac{1}{n^2}\right).$$ We compute the value of the integral over $v$ to be $\pi/2-2/3$, giving the following final result for the initial expression: $$\frac{\pi/2-2/3}{n}+O\left(\frac{1}{n^2}\right).$$

Answer 3

Let's denote $$I=\int_0^1dx\int_0^1dy\sqrt{1-(1-\sqrt x+\sqrt x\sqrt y)^2}e^{-nxy}$$ Making the substitution $s=nx$ $$=\frac{1}{n^{5/4}}\int_0^ns^{1/4}ds\int_0^1dy\sqrt{1-\sqrt y}\sqrt{2-\sqrt\frac{s}{n}+\sqrt\frac{s}{n}\sqrt y }\,\,e^{-sy}$$ Now, let's choose $0<\alpha<1$ and split the interval into two parts: $[0;n^\alpha]$ and $[n^\alpha; n]$. We suppose that $1\ll n^\alpha\ll n$ - a more accurate criterion for $\alpha$ we will define a bit later. Let's denote $$I_1=\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{1/4}ds\int_0^1dy\sqrt{1-\sqrt y}\sqrt{2-\sqrt\frac{s}{n}+\sqrt\frac{s}{n}\sqrt y }\,\,e^{-sy}$$ Due to the fact that $s\in [n^\alpha; n]\,\, s\gg1$, and the integral with respect to $y$ gives the main contribution near $y=0$ - due to the exponent $e^{-sy}$. Noting also that $\sqrt\frac{s}{n}\leqslant 1$, we are allowed to decompose the big square root into the Taylor's series, and evaluate it term by term. $$I_1=\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{1/4}ds\int_0^1dy\,\Big(1-\frac{\sqrt y}{2}\Big)\sqrt{2-\sqrt\frac{s}{n}}\bigg(1+\frac{1}{2}\frac{\sqrt\frac{s}{n}\sqrt y}{2-\sqrt\frac{s}{n}} \bigg)\,e^{-sy}\quad(1)$$ Then we evaluate every term: $I_1=I_1^0+I_1^1+I_1^2$ $$I_1^0=\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{1/4}\sqrt{2-\sqrt\frac{s}{n}}\,ds\int_0^1dy\,e^{-sy}=\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{-3/4}\sqrt{2-\sqrt\frac{s}{n}}(1-e^{-s})\,ds$$ For the term with the exponent we get the following estimation: $$\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{-3/4}\sqrt{2-\sqrt\frac{s}{n}}e^{-s}\,ds<\frac{\sqrt 2}{n^{5/4}}\int_{n^\alpha}^\infty s^{-3/4}e^{-s}\,ds<\frac{\sqrt 2\,n^{-3\alpha/4}}{n^{5/4}}\int_{n^\alpha}^\infty e^{-s}\,ds=\frac{\sqrt 2\,e^{-n^\alpha}}{n^{(5+3\alpha)/4}}$$ The main term in $I_1^0$ gives $$\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{-3/4}\sqrt{2-\sqrt\frac{s}{n}}\,ds=\frac{4}{n}\int_{n^\frac{\alpha-1}{4}}^1ds\sqrt{2-s^2}=\frac{8}{n}\int_{\arcsin\frac{n^\frac{\alpha-1}{4}}{\sqrt2}}^{\pi/4}\cos^2\phi \,d\phi$$ Because we have chosen $1\ll n^\alpha\ll n\,\Rightarrow\, n^\frac{\alpha-1}{4}\ll1$, and, decomposing $\arcsin$, we get the following estimation: $$\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{-3/4}\sqrt{2-\sqrt\frac{s}{n}}\,ds\approx\frac{\pi+2}{n}-\frac{4\sqrt 2}{n\,n^\frac{1-\alpha}{4}}$$ Therefore, the integral $I_1^0$ is evaluated as $$\boxed{\,\,I_1^0=\frac{\pi+2}{n}-\frac{4\sqrt 2}{n\,n^\frac{1-\alpha}{4}}+R_0(n,\alpha);\quad R_0(n,\alpha)< \frac{\sqrt 2\,e^{-n^\alpha}}{n^{(5+3\alpha)/4}}\,\,}\qquad(2)$$ In the same way we get the evaluation of other terms of (1): $$|I_1^1|=\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{1/4}ds\int_0^1dy\,\frac{\sqrt y}{2}\sqrt{2-\sqrt\frac{s}{n}}\,e^{-sy}<\frac{\text{Const}}{n\,n^\frac{1+\alpha}{4}}\quad(3)$$ $$|I_1^2|=\frac{1}{n^{5/4}}\int_{n^\alpha}^ns^{1/4}ds\int_0^1dy\,\frac{\sqrt y}{2}\frac{\sqrt\frac{s}{n}}{\sqrt{2-\sqrt\frac{s}{n}}}\,e^{-sy}<\frac{\text{Const}}{n\sqrt n}\quad(4)$$ In the same fashion we evaluate the integral over the interval $[0;n^\alpha]$ $$I_2=\frac{1}{n^{5/4}}\int_0^{n^\alpha}s^{1/4}ds\int_0^1dy\sqrt{1-\sqrt y}\sqrt{2-\sqrt\frac{s}{n}+\sqrt\frac{s}{n}\sqrt y }\,\,e^{-sy}<\frac{\text{Const}}{n\,n^\frac{1-\alpha}{4}}\quad(5)$$ Gathering the biggest terms together, we get the following evaluation: $$\boxed{\,\,I=I_1+I_2=\frac{\pi+2}{n}+R_1+R_2;\,\,R_1<\frac{C_1}{n\,n^\frac{1-\alpha}{4}}; \,R_2<\frac{C_2}{n\,n^\frac{1+\alpha}{4}}\,\,}\quad(6)$$ We see that the best accuracy we get when we choose $\alpha\to 0$ (though, we have to bear in mind that we still have the requirement $n^\alpha\gg1$). At $\alpha=0$ the estimation of the remainder gives $R_{1,2}\sim\frac{\text{Const}}{n\,n^{1/4}}$, what explains a considerable error even at a big $n$.

For example, at $n=1000\,\, I=0.00384236... $, and $\frac{\pi+2}{n}=0.00514159... $

The rough evaluation of $R$ for $n=1000$ gives $\frac{R}{I}\sim\frac{\text{Const}}{n^{1/4}}=0.18\cdot\text{Const}$, what is in agreement with $\frac{\frac{\pi+2}{1000}-I(1000)}{I(1000)}=0.34$

$\text{Addendum}$

$\frac{s}{n}\ll1$ for $s\in[0;n^\alpha]$, so we can drop all such terms in eq. (5) $$I\sim\frac{\sqrt 2}{n^{5/4}}\int_0^{n^\alpha}s^{1/4}ds\int_0^1\sqrt{1-\sqrt y}\,e^{-sy}dy$$ It would be great to evaluate this integral (so far, I don't know how to approach it). But we can also try the following: use $\beta<\alpha$, and split $[0;n^\alpha]$ into $[0; n^\beta]$ and $[n^\beta;n^\alpha].\, I=I_1+I_2$; $$I_1=\frac{\sqrt 2}{n^{5/4}}\int_0^{n^\beta}s^{1/4}ds\int_0^1\sqrt{1-\sqrt y}\,e^{-sy}dy<\frac{4\sqrt 2\,n^{\beta/4}}{n^{5/4}}+O(n^{-5/4})$$ $$I_2=\frac{\sqrt 2}{n^{5/4}}\int_{n^\beta}^{n^\alpha}s^{1/4}ds\int_0^1\sqrt{1-\sqrt y}\,e^{-sy}dy\sim\frac{4\sqrt 2\,n^{\alpha/4}}{n^{5/4}}+O(\frac{n^{\beta/4}}{n^{5/4}})$$