 Moving-cutoff correction to a symmetric exponential integral

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.