Small parameter expansion of the definite integral, divergence of coefficients

Question

I’m considering the following integral with one parameter $\omega$ $$I(\omega):=\int_0^{\infty}(1+x)^2\Bigg(\sqrt{x^2+\frac{\omega^2}{(1+x)^\delta}}-x\Bigg)dx,$$ where $\delta>2$ and $\omega $ is small parameter.

I want to know the expansion of $I(\omega)$ around $\omega=0$.

Naively, it is accomplished by the standard Taylor expansion, but the problem is its coefficient has a divergence, for example,

$$\frac{d^2I(0)}{d\omega^2}= \int_0^{\infty}\frac{1}{(1+x)^{\delta-2}}\frac{1}{x}dx\sim\infty.$$

Does anyone know how to fix this problem?

Answer 1

It turns out that

$$ I(\omega) = \frac{\omega^2\log(1/\omega)}{2} + \mathcal{O}(\omega^2) $$

as $\omega \to 0^+$. The logarithmic correction arises from the asymptotic behavior

$$ \text{[integrand]} \sim \frac{\omega^2}{\sqrt{x^2 + \omega^2} + x} \qquad \text{as } x \to 0^+. $$

Below is a more rigorous argument:

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1. Denote the integrand by $f(x \mid \omega, \delta)$:

$$ f(x \mid \omega, \delta) = (1 + x)^2 \biggl( \sqrt{x^2 + \frac{\omega^2}{(1+x)^{\delta}}} - x \biggr) $$

In order to investigate the asymptotic behavior of $I(\omega)$, we first manipulate the integrand so it is easily analyzed.

\begin{align*} f(x \mid \omega, \delta) &= (1 + x)^2 \frac{\left( x^2 + \frac{\omega^2}{(1+x)^{\delta}} \right) - x^2}{\sqrt{x^2 + \frac{\omega^2}{(1+x)^{\delta}}} + x} \\ &= \omega^2 \frac{1}{(1+x)^{\delta-2}} \frac{1}{\sqrt{x^2 + \frac{\omega^2}{(1+x)^{\delta}}} + x} \end{align*}

This shows that $f(x \mid \omega, \delta) \sim \frac{\omega^2}{(1+x)^{\delta-2}(2x)}$ as $\omega \to 0^+$, suggesting that the primary source of the divergence of $I''(0)$ is the behavior of $f(x \mid \omega, \delta)$ near $x = 0$ as $\omega \to 0^+$. So, we will investigate the integral of $f(x \mid \omega, \delta)$ near $x = 0$ separately from the rest.

2. Define the function $J(\omega)$ by

$$ J(\omega) = \int_{0}^{1} \frac{1}{\sqrt{x^2 + \omega^2} + x} \, \mathrm{d}x. $$

Although $J(\omega)$ admits a closed form in terms of inverse trigonometric functions and hence a detailed asymptotic expansion as $\omega \to 0^+$, let me demonstrate a more elementary approach. Indeed, let us assume $\omega \in (0, 1)$ without losing the generality. Using the inequality

$$ \max\{x, \omega\} \leq \sqrt{x^2 + \omega^2} \leq x + \omega,$$

$J(\omega)$ is bounded as

$$ \int_{0}^{1} \frac{1}{2x+\omega} \, \mathrm{d}x \leq J(\omega) \leq \int_{0}^{1} \frac{1}{x + \max\{x, \omega\}} \, \mathrm{d}x. $$

By evaluating both integrals, we easily find that

$$ J(\omega) = \frac{\log(1/\omega)}{2} + \mathcal{O}(1). $$

3. Finally, decompose $I(\omega)$ into the sum of three integrals $I_1(\omega)$, $I_2(\omega)$, $I_3(\omega)$, where

\begin{align*} I_1(\omega) &= \int_{0}^{1} \frac{\omega^2}{\sqrt{x^2 + \frac{\omega^2}{(1+x)^{\delta}}} + x} \, \mathrm{d}x, \\ I_2(\omega) &= \int_{0}^{1} \Biggl[ f(x \mid \omega, \delta) - \frac{\omega^2}{\sqrt{x^2 + \frac{\omega^2}{(1+x)^{\delta}}} + x} \Biggr] \, \mathrm{d}x, \\ I_3(\omega) &= \int_{1}^{\infty} f(x \mid \omega, \delta) \, \mathrm{d}x \end{align*}

As $\omega \to 0^+$, it is not hard to check that the integrand of each of $I_2(\omega)$ and $I_3(\omega)$ is bounded by $\omega^2$ times an integrable function. Hence, $I_2(\omega) + I_3(\omega) = \mathcal{O}(\omega^2)$. On the other hand,

$$ \omega^2 J(\omega) \leq I_1(\omega) \leq \omega^2 J(\omega\cdot2^{-\delta/2}) $$

shows that

$$ I_1(\omega) = \frac{\omega^2\log(1/\omega)}{2} + \mathcal{O}(\omega^2). $$

4. Combining all the observations together, we conclude

$$ \bbox[padding:5px;border:1px dotted navy;color:navy]{I(\omega) = \frac{\omega^2\log(1/\omega)}{2} + \mathcal{O}(\omega^2)} $$

Answer 2

@Sangchul Lee provided an elegant evaluation of the leading asymptotic term. However, for the purpose of practical usage it makes sense finding next asymptotic term. There are two reasons for that:

• logarithm is a very slowly changing function, so the next term also contributes at any reasonable $\omega$
• it can be useful to get dependance on $\delta$

Splitting the interval of integration on $[0;1]$ and $[1;\infty)$ and making the substitution $x=\frac1t$ in the second integral, we get $$I(\omega, \delta)=\int_0^{\infty}(1+x)^2\Big(\sqrt{x^2+\frac{\omega^2}{(1+x)^\delta}}-x\Big)dx$$ $$=\int_0^1(1+x)^2\Big(\sqrt{x^2+\frac{\omega^2}{(1+x)^\delta}}-x\Big)dx+\int_0^1\frac{(1+x)^2}{x^5}\Big(\sqrt{1+\frac{\omega^2x^{\delta+2}}{(1+x)^\delta}}-1\Big)dx$$ $$=I_A+I_B\label{1}\tag{1}$$ where $I_B$ can be presented in the form $$I_B=\omega^2\int_0^1\frac{x^{\delta-3}(1+x)^{2-\delta}}{\sqrt{1+\frac{\omega^2x^{\delta+2}}{(1+x)^\delta}}+1}dx$$ $$=\frac{\omega^2}2\int_0^1x^{\delta-3}(1+x)^{2-\delta}dx+o(\omega^2)\label{2}\tag{2}$$ To evaluate $I_A$ we split the interval of integration on $[0;\sqrt\omega]$ and $[\sqrt\omega;1]$. $$I_A=\int_0^1(1+x)^2\Big(\sqrt{x^2+\frac{\omega^2}{(1+x)^\delta}}-x\Big)dx$$ $$=\omega^2\int_0^1\frac{dx}{x\Big(\sqrt{1+(\frac\omega x)^2\frac1{(1+x)^\delta}}+1\Big)}=\omega^2\left(\int_0^{\sqrt\omega}\,\,+\,\int_{\sqrt\omega}^1\,\,\right)=$$ Then, keeping only the terms $\sim\omega^2\ln\omega\,$ and $\,\sim\omega^2$ $$\sim\omega^2\int_0^{\sqrt\omega}\frac{dx}{x\Big(\sqrt{1+(\frac\omega x)^2}+1\Big)}+\omega^2\int_{\sqrt\omega}^1\frac{(1+x)^{2-\delta}}xdx=\omega^2\left(I_{A1}+I_{A2}\right)\label{3}\tag{3}$$ Making some straightforward transformations, integrating by part and keeping only leading terms, $$I_{A1}=\int_{\sqrt\omega}^\infty\frac{dx}{x(\sqrt{1+x^2}+1)}=\frac12\int_\omega^\infty\frac{dx}{x(\sqrt{1+x}+1)}$$ $$=\frac12\frac{\ln x}{1+\sqrt{1+x}}\bigg|_\omega^1+\frac14\int_\omega^\infty\frac{\ln x}{\sqrt{1+x}(1+\sqrt{1+x}+1)^2}dx$$ $$\sim\frac{\ln\frac1\omega}4+\frac14\int_0^\infty\frac{\ln x}{\sqrt{1+x}(1+\sqrt{1+x}+1)^2}dx$$ $$=\frac{\ln\frac1\omega}4+\frac{1+2\ln2}{4}\label{4}\tag{4}$$ The same approach to $I_{A2}$: $$I_{A2}=\frac{\ln x(1+x)^{2-\delta}}2\bigg|_{\sqrt\omega}^1+\frac{\delta-2}2\int_{\sqrt\omega}^1\ln x(1+x)^{1-\delta}dx$$ $$\sim\frac{\ln\frac1\omega}4+\frac{\delta-2}2\int_0^1\ln x(1+x)^{1-\delta}dx\label{5}\tag{5}$$ Putting \eqref{4} and \eqref{5} into \eqref{3}, and then \eqref{3} and \eqref{2} into \eqref{1}, we finally get: $$I(\omega;\delta)=\frac{\omega^2\ln\frac1\omega}2+\frac{\omega^2}2\left(\frac12+\ln2+(\delta-2)\int_0^1(1+x)^{1-\delta}\ln x\,dx+\int_0^1x^{\delta-3}(1+x)^{2-\delta}dx\right)+o(\omega^2)$$ $$=\frac{\omega^2\ln\frac1\omega}2+\frac{\omega^2}2\left(\frac12+\ln2+(\delta-2)\int_0^1(1+x)^{1-\delta}(1-x^{\delta-3})\ln x\,dx\right)+o(\omega^2)$$ Numeric check confirms the given asymptotics.