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The integral converges as it is easily seen to be upper bounded by $\sqrt{\pi/2}$. However, Laplace's method does not seem to work out as the maxima of the function $S(x) = -a\sqrt{1-e^{-x}}-x^2/2$ is located at the end point $0$. This question enquires about a similar problem, however, with the major difference that the $S(x)$ function there is given by $-a(1-e^{-x})-x^2/2$. The second answer to that problem suggests using a modified form of Laplace's method as given by V. Zorich, Mathematical Analysis II Chap. XIX, Par. 2.4, Theorem 1, to tackle the issue of maxima at an endpoint. However, for the problem at hand, this method breaks down as the function $S(x)$ is not differentiable at $0$. So, Laplace's method cannot be applied here.

I tried using the transformation $1-e^{-x}\to x^2$ to obtain the integral $$\int_0^1 \frac{\exp(-a x-(\ln(1-x^2))^2/2)}{1-x^2}2xdx$$ From this, intuitively, it seems to me that, at least for large $a$, the integrand is concentrated highly around $0$, and there it seems to be approximated ``well'' by $2xe^{-ax}$, which produces a $\sim\frac{1}{a^2}$ trend. However, all this is very intuitive and I don't know how to transform this intuition into rigorous statements. Also, this intuition seems to serve well for getting asymptotics, but my true intention is to obtain tight upper bounds. As Laplace's method seems not to be a suitable choice, I do not have much idea about how to proceed to say anything about an upper bound. Please help.

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  • $\begingroup$ $2(a^{-2}+a^{-3})$ seems pretty tight. $\endgroup$
    – Carlo Beenakker
    Feb 6, 2018 at 12:28
  • $\begingroup$ Does that work as an upper bound? Also, can you kindly suggest how I should begin deriving such an upper bound? Thanks. $\endgroup$
    – Samrat Mukhopadhyay
    Feb 6, 2018 at 13:36

1 Answer 1

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Here again, like in the previous question the Laplace method DLMF applies. With $p(x)=\sqrt{1-\exp(-x)}$ and $q(x)=\exp(-x^2/2)$, we have the expansion $$ p(x)=x^{1/2}-\frac1 4 x^{3/2}+\frac 5{96}x^{5/2}+\cdots$$ which is of the form $$p(x)\sim p(0)+\sum_{s=0}^{\infty}p_{s}t^{s+\mu}$$ with $\mu=\frac1 2$ and $p(0)=0$. One may calculate the coefficients for the expansion using the residue formula in the link above to obtain $$I(a)\sim \frac 2{a^2}+\frac {12}{a^4}+\frac {120}{a^6}+\cdots$$

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  • $\begingroup$ This seems to be exactly what I wanted, thanks! From the link, I see that the expansion of the integral is pretty evident once the expansion of the functions $p(t), q(t)$ are known. But how do you decide the expansion around $0$, is it something that has to be observed and is not fixed, like Taylor expansion? Though I can understand pretty clearly how you obtained the expansion for $\sqrt{1-e^{-x}}$, is there a general rule for this, something similar to the Taylor series? $\endgroup$
    – Samrat Mukhopadhyay
    Feb 6, 2018 at 14:58
  • $\begingroup$ Also, the answer you provided serves fine for the asymptotic expansion, however, is there a method to find tight upper bound, which is true for all $a\ge 0$? $\endgroup$
    – Samrat Mukhopadhyay
    Feb 6, 2018 at 15:00
  • $\begingroup$ Another question: is it guaranteed that every function $p$ has an expansion of the form shown in the link? If some function does not admit such an expansion then what can be done? Please suggest some reference where I can learn these things. $\endgroup$
    – Samrat Mukhopadhyay
    Feb 6, 2018 at 15:31
  • $\begingroup$ The expansion follows from the remark that $p(x)/\sqrt x$ is analytic near the origin. Discussion on the error terms can be useful to analyze the quality of the bound for $a\geq 0$. There are many texts on asymptotic expansions of integrals. You may find what you are looking forin the classical books by Bleistein & Handelsman, Wong or Olver, for example. $\endgroup$
    – Paul Enta
    Feb 6, 2018 at 15:48

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