Exponential integral ($E_1,\mathrm{Ei}$)

For real $x>0$, the decaying exponential integral isThe other standard real convention satisfies $\mathrm{Ei}(-x)=-E_1(x)$. The NIST definitions specify the analytic continuation and branch conventions.

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.

As $x\downarrow 0$, the exponential integral haswhere $\gamma$ is the Euler--Mascheroni constant. Integrating the Taylor series of $E_1^{\prime }(x)=-e^{-x}/x$ gives every nonconstant coefficient; the constant follows from the limiting definition of $\gamma$. A logarithm of a stretched variable can create a switchback term in a matched asymptotic expansion.