= Logarithmic expansion of a symmetric exponential integral
{title2=$J(\varepsilon)$}
For $J(\varepsilon)=\int_0^\infty e^{-x-\varepsilon/x}dx/x$, reciprocity $x\mapsto\varepsilon/x$ and the <small-argument expansion of the exponential integral> give $J=-\log\varepsilon-2\gamma+o(1)$. Differentiation and <integration by parts> give $\varepsilon J''+J'=J$. Its <Frobenius method> yields
$$
J=-\log\varepsilon-2\gamma+\varepsilon[-\log\varepsilon+2-2\gamma]+O(\varepsilon^2|\log\varepsilon|).
$$
The logarithm is a singular contribution that a naive termwise expansion at $x=0$ misses.
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