Asymptotic expansion of the integral $\int_0^s \sqrt{\frac{1}{E-V(x)}} - \sqrt{\frac{1}{E+x^2}}dx$ as $E\to0$ for $V(x)=-x^2+...$

Question

How can one obtain an asymptotic expansion of an integral of the form $$I(E)=\int_0^s \sqrt{\frac{1}{E-V(x)}} - \sqrt{\frac{1}{E+x^2}}dx$$ as $E\to0^+$ given that $V(x)$ attains its maximum in $[0,s]$ at $x=0$ and that $V(x)$ is analytic around $x=0$ with a finite radius of convergence and that its Taylor expansion begins $V(x)=-x^2+...$.

It is clear that the first term is given by $$I(E)\sim\int_0^s \sqrt{\frac{1}{-V(x)}} - \sqrt{\frac{1}{x^2}}dx$$ I would like to be able to obtain an explicit expression for at least the next term.

Answer 1

It may not be possible to provide a full answer for any potential $V(x)$, but we can outline an expansion that can provide a reasonable approximation to the integral under certain conditions. Consider the potential $V(x)=-x^2-gf(x)$, where $f$ is a continuous function in $(0,s]$ such that $f(x)=\mathcal{O}(x^\rho)$, with $\rho>2$ (at least in some neighborhood around the origin). We also demand that $V(x)$ is concave, with $V'(0)=0$. Then, one can find $I(0^+)$ for this class of potentials explicitly:

$$\lim_{E\to 0+}I(E)=\int_0^s\frac{dx}{x}\left(\frac{1}{\sqrt{1+g\frac{f(x)}{x^2}}}-1\right)$$

For as long as $g f(s)/s^2\ll 1$ the following asymptotic expansion should give reasonable approximations for the requisite integral:

$$I(E)\sim\sum_{n=1}^{\infty}\left(-\frac{g}{4}\right)^n{2n\choose n}\int_0^sdx\frac{(f(x))^n}{(E+x^2)^{n+1/2}}$$

By virtue of the assumptions, for any $n$, the integrals in the summand converge. The asymptotic expansion above may contain terms that are nonanalytic in $E$ as $E\to 0$, so this is probably as good a form as one can get with this general setup. For $f(x)=x^4$, for example, it can be shown that the asymptotic series contains terms of the form

$$I(E)\subset (\alpha_1E+\alpha_2 E^2+...)\log E$$

which cannot be obtained by expanding in powers of the energy. However, these terms are finite as $E\to 0^+$ and hence the leading divergence has been subtracted successfully.

Note: It is interesting to think about how the above series can fail for large $g$. The expansion of the following integral

$$\int_0^s\frac{dx}{\sqrt{E+gx^4}}=\frac{\Gamma^2(1/4)}{(gE)^{1/4}4\sqrt{\pi}}+\frac{1}{g^{1/2}s}\sum_{n=0}^\infty\left(-\frac{E}{4g s^4}\right)^n{2n \choose n}\frac{1}{4n+1}$$

is convergent, but it has a completely different dependence on the energy than the quadratic term and one could reason that for large $g$ the first series must fail at some point, since the expansion in inverse powers of the coupling $g$ has a finite radius of convergence.

Answer 2

Whenever we have a (sum or) difference of square roots, we should always try the trick $\sqrt a-\sqrt b = \frac{a-b}{\sqrt a+\sqrt b}$ and see if it magically helps. Here $$ \sqrt{\frac{1}{E-V(x)}} - \sqrt{\frac{1}{E+x^2}} = \frac{\frac{1}{E-V(x)} -\frac{1}{E+x^2}}{\sqrt{\frac{1}{E-V(x)}} + \sqrt{\frac{1}{E+x^2}}} = \frac{V(x)+x^2}{(E-V(x))(E+x^2)\bigl( \sqrt{\frac{1}{E-V(x)}} + \sqrt{\frac{1}{E+x^2}} \bigr)}. $$ As $E\to0^+$, the denominator approaches $2x^3+\cdots$ and the numerator approaches $cx^3+\cdots$ for some $c\in\Bbb R$, leaving a much more tractable integral to approximate. (How to proceed from here depends on how good an approximation one needs, how much information we have about $V(x)$ and $s$, etc.)