Logarithmic asymptotics of an integral: order unity correction

Question

For a physics problem I'm working on I am struggling with the following integral: $$I(\varepsilon)=\int_0^\infty \frac{x}{ 1+e^{\varepsilon x}\bigl(1+2x^2+2x\sqrt{1+x^2}\bigr) }\,dx,\;\;\varepsilon\geq 0.$$ I have not found a closed-form expression (I don't think it exists), but my main interest is in the small-$\varepsilon$ asymptotics. Since for $\epsilon=0$ the integrand decays for large $x$ as $1/4x$, the leading small-$\varepsilon$ behaviour is logarithmic, $$I(\varepsilon)=-\tfrac{1}{4}\ln\varepsilon+{\cal O}(1).$$
I would want to know the order unity correction, what is the limit $$\lim_{\varepsilon\rightarrow 0}\left[I(\varepsilon)+\tfrac{1}{4}\ln\varepsilon\right]=c_0.$$ A numerical integration gives $c_0\approx -0.096$. Is there an exact value for this coefficient?

Answer 1

Somehow posting the question forced me to think this through more clearly, and I see that I can now answer it myself:

$$c_0=\tfrac{1}{4} (\ln 2 -\gamma_{\rm{Euler}} -1/2)=-0.0960171\cdots.$$

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Derivation: use that $$J(\varepsilon)=\int_0^\infty \frac{e^{-\varepsilon x}}{1+4x}\,dx=-\tfrac{1}{4} e^{\varepsilon/4} \text{Ei}\left(-\varepsilon/4\right)$$ has the same logarithmic asymptotics as $I(\varepsilon)$, $$J(\varepsilon)=\tfrac{1}{4} (-\ln \varepsilon-\gamma_{\rm{Euler}} +2\ln 2)+{\cal O}(\varepsilon).$$ By subtracting $J$ from $I$ we remove the logarithmic term and can then simply take the $\varepsilon\rightarrow 0$ limit, $$I(\varepsilon)-J(\varepsilon)=\tfrac{1}{8} (-1-2\ln 2)+{\cal O}(\varepsilon).$$ Upon combination of these two expansions we arrive at the desired result, $$\lim_{\varepsilon\rightarrow 0}[I(\varepsilon)+\tfrac{1}{4} \ln \varepsilon]=\tfrac{1}{8} (-1-2\ln 2)+\tfrac{1}{4} (-\gamma_{\rm{Euler}} +2\ln 2)\equiv c_0.$$