Moving-cutoff correction to a symmetric exponential integral
= Moving-cutoff correction to a symmetric exponential integral
{title2=$J-I$}
For $I=\int_\varepsilon^\infty e^{-x-\varepsilon/x}dx/x$, the omitted interval becomes $J-I=\int_1^\infty e^{-t-\varepsilon/t}dt/t$. A uniform <Taylor expansion> gives $J-I=E_1(1)-\varepsilon[e^{-1}-E_1(1)]+O(\varepsilon^2)$. Subtracting this correction from the <logarithmic expansion of a symmetric exponential integral> preserves both the constant and first-order coefficient.