 Logarithmic expansion of a symmetric exponential integral

For $J(\epsilon )={\int }_0^{\infty }{e}^{-x-\epsilon /x}dx/x$, reciprocity $x\mapsto \epsilon /x$ and the small-argument expansion of the exponential integral give $J=-\log \epsilon -2\gamma +o(1)$. Differentiation and integration by parts give $\epsilon J^{\prime \prime }+J^{\prime }=J$. Its Frobenius method yieldsThe logarithm is a singular contribution that a naive termwise expansion at $x=0$ misses.