Logarithmic expansion of a symmetric exponential integral ($J(\epsilon )$)

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.

For $I={\int }_{\epsilon }^{\infty }{e}^{-x-\epsilon /x}dx/x$, the omitted interval becomes $J-I={\int }_1^{\infty }{e}^{-t-\epsilon /t}dt/t$. A uniform Taylor expansion gives $J-I=E_1(1)-\epsilon [e^{-1}-E_1(1)]+O({\epsilon }^2)$. Subtracting this correction from the logarithmic expansion of a symmetric exponential integral preserves both the constant and first-order coefficient.