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I have the following integral $$I=-\frac{1}{\varepsilon^{4}}\int_{0}^{\infty}dx\left( \sqrt{x^{4}% +2\varepsilon^{4}}-x^{2}\right) e^{-2x}$$ where $\varepsilon\ll1$. This integral is solvable exactly in terms of MeijerG function, and is part of a bigger integral whose integrand has been broken into sums (one of those being the one I'm presenting here).

Because $\varepsilon$ is very small, I'm interested in the first powers of the solution. Because it is summed with other integrals, the final solution, in power series of $\varepsilon$, has its leading term equal to $-\frac{4\sqrt{2}}{9}\varepsilon^2$. This term comes from $I$, and only from the integration interval $\left[0,1\right]$. I know this because I analyzed similar situations and they all come from this "part" of the bigger integral.

So, instead of solving the whole thing (several integrals) exactly and then expanding the solution in powers of $\varepsilon$ to obtain the leading term, I want to expand the integrand of the integral from which the leading term comes (i.e. in this case $I$), and solve only the term proportional to $\varepsilon^2$. Reducing the interval from $\left[0,\infty\right]$ to $\left[0,1\right]$ would make it even easier to find the proper expansion of the integrand and solving it. The thing is, if I expand the integrand around $x=0$ I get $$-\frac{1}{\varepsilon^{4}}\left( \sqrt{x^{4}% +2\varepsilon^{4}}-x^{2}\right)\approx-\frac{\sqrt{2}}{\varepsilon^2}+\frac{x^2}{\varepsilon^4}-...,$$ and increasing powers of $\frac{1}{\varepsilon}$, and thus don't recognize the $\varepsilon^2$ term. If I, however, expand the same integrand around $\varepsilon=0$ I get, $$-\frac{1}{\varepsilon^{4}}\left( \sqrt{x^{4}% +2\varepsilon^{4}}-x^{2}\right)\approx-\frac{1}{x^2}+\frac{\varepsilon^4}{2x^6}-\frac{\varepsilon^8}{2x^{10}}+...,$$ where not only I don't get any $\varepsilon^2$ term but also get divergent integrals at the lower limit $x=0$. I guess that, in the first expansion, I lose the information with which the limit $\varepsilon\rightarrow0$ is well defined, and, in the second expansion, the information in which the limit $x\rightarrow0$ is well defined.

Which aspect of the integrand, and their possible expansion, is causing this behavior? And what would be a better approach for getting the $\varepsilon^2$ term without solving the whole integral?

Thanks

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The analysis here is a bit loose, but time is not on my side today.

Consider the following function $$I\left( a \right)=-\frac{1}{{{a}^{4}}}\int\limits_{0}^{\infty }{\left( \sqrt{{{x}^{4}}+2{{a}^{4}}}-{{x}^{2}} \right){{e}^{-2x}}dx}$$ Which we can write as $$I=-\frac{1}{{{a}^{2}}}\int\limits_{0}^{\infty }{\left( \sqrt{\frac{{{x}^{4}}}{{{a}^{4}}}+2}-\frac{{{x}^{2}}}{{{a}^{2}}} \right){{e}^{-2x}}dx}$$ Now use $u=x/a\Rightarrow du=\frac{1}{a}dx$ $$I=-\frac{1}{a}\int\limits_{0}^{\infty }{\left( \sqrt{{{u}^{4}}+2}-{{u}^{2}} \right){{e}^{-2ua}}du}$$ Observe $$\int\limits_{0}^{\infty }{\left( \sqrt{{{u}^{4}}+2}-{{u}^{2}} \right)du}=\beta \simeq 2.07878...$$ This integral apparently has an expression in terms of elliptic functions (which I suppose isn’t too hard to guess). Mathematica yields the following exact result $$\int\limits_{0}^{\infty }{\left( \sqrt{{{u}^{4}}+2}-{{u}^{2}} \right)du}=-\frac{2}{3}\left( -2 \right){}^{3/4}K\left( 2 \right)+\frac{{{\left( -2 \right)}^{1/4}}}{3\sqrt{\pi }}\Gamma {{\left( \tfrac{1}{4} \right)}^{2}}$$ Where K is the complete elliptic integral of the first kind and $\Gamma $ is the gamma function. Observe that $\left| I\left( a \right) \right|\le \frac{\beta }{a}$, and for very small a, the 1/a term dominates, hence $I\simeq -\frac{\beta }{a}$. It’s actually not quite as simple as that. For note that if we expand the exponential then we obtain $$I=-\int\limits_{0}^{\infty }{\left( \sqrt{{{u}^{4}}+2}-{{u}^{2}} \right)\left( \frac{1}{a}-2u+O\left( a \right) \right)du}$$ The $O\left( a \right)$terms disappear as $a\to 0$ and so we are left with something that behaves like 1/a, and something that becomes infinite. Note however that $$\left( \sqrt{{{u}^{4}}+2}-{{u}^{2}} \right)u=\frac{1}{u}+O\left( {{u}^{-2}} \right)\Rightarrow \int{\left( \sqrt{{{u}^{4}}+2}-{{u}^{2}} \right)u\simeq \ln \left( u \right)+O\left( 1 \right)}$$ for large u. Hence we have two terms $I\simeq -\frac{\beta }{a}+\ln \left( b \right)+O\left( 1 \right)$, where b is large and a is small. We may, however justify all this a tad more by first partitioning the integral into two parts, namely $$I=-\int\limits_{0}^{1/a}{{}}-\int\limits_{1/a}^{\infty }{{}}$$ Doing this then essentially yields for small a $$I\simeq -\frac{\beta }{a}-\ln \left( a \right)$$ plus terms that are $O\left( 1 \right)$.

If we numerically evaluate the integral (blue curve) and compare it to the above approximation (yellow curve) for small a, then we have the following graphs enter image description here

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As you wrote, assuming $a>0$, $$J=\int_{0}^{\infty}\left( \sqrt{x^{4}% +2a^{4}}-x^{2}\right) e^{-2x}\,dx$$ has the nice expression (thanks to a CAS) $$J=-\frac{1}{4}-\frac{2\ 2^{3/4}}{\pi ^2 a^3}\,G_{1,5}^{5,1}\left(\frac{a^4}{8}| \begin{array}{c} \frac{9}{4} \\ \frac{3}{4},\frac{3}{2},\frac{7}{4},2,\frac{9}{4} \end{array} \right)$$ Developed as a series (thanks to a CAS), this gives $$J=-\frac{2^{3/4} \pi ^{3/2} }{\Gamma \left(-\frac{1}{4}\right) \Gamma \left(\frac{7}{4}\right)}\,a^3+\frac{1}{2} (4 \log (a)+4 \gamma -5+\log (8))\,a^4-\frac{\sqrt[4]{2} \pi ^{3/2} }{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{9}{4}\right)}\,a^5+O\left(a^6\right)$$ and then $$I=-\frac J {a^4}= \left(\frac{5}{2}-2 \gamma -\frac{3 \log (2)}{2}\right)-2\log(a)+\frac{2^{3/4} \pi ^{3/2}}{ \Gamma \left(-\frac{1}{4}\right) \Gamma \left(\frac{7}{4}\right)}\frac 1a+\frac{\sqrt[4]{2} \pi ^{3/2} }{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{9}{4}\right)}a$$ For a few values of $a$, we get the following values for $I=-a^{-4}\, J$ $$\left( \begin{array}{ccc} a & \text{exact} & \text{approximation}\\ 0.100 & -15.7215 & -15.7156 \\ 0.095 & -16.7205 & -16.7151 \\ 0.090 & -17.8356 & -17.8307 \\ 0.085 & -19.0875 & -19.0832 \\ 0.080 & -20.5023 & -20.4985 \\ 0.075 & -22.1132 & -22.1098 \\ 0.070 & -23.9626 & -23.9596 \\ 0.065 & -26.1064 & -26.1039 \\ 0.060 & -28.6191 & -28.6169 \\ 0.055 & -31.6025 & -31.6006 \\ 0.050 & -35.1992 & -35.1977 \\ 0.045 & -39.6158 & -39.6145 \\ 0.040 & -45.1624 & -45.1614 \\ 0.035 & -52.3274 & -52.3266 \\ 0.030 & -61.9259 & -61.9253 \\ 0.025 & -75.4277 & -75.4273 \\ 0.020 & -95.7771 & -95.7768 \\ 0.015 & -129.856 & -129.856 \\ 0.010 & -198.346 & -198.346 \end{array} \right)$$

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