In this post, Gary gives the following small-$x$ expansion for inverse of standard exponential integral $E_1^{-1}$:
$$ E_1^{ - 1} (x) = - \log x - \log ( - \log x) - \frac{{\log ( - \log x) - 1}}{{\log x}} + \mathcal{O}\!\left( {\frac{{(\log ( - \log x))^2 }}{{\log ^2 x}}} \right), $$
- How do I derive this?
- How can this be extended to to inverse of generalized exponential integral $E_p^{-1}(x)$?
For $y\to+\infty$, the generalised exponential integral possesses the asymptotic expansion $$ x=E_p (y) \sim \frac{{{\rm e}^{ - y} }}{y}\sum\limits_{k = 0}^\infty {( - 1)^k \frac{{(p)_k }}{{y^k }}} = \frac{{{\rm e}^{ - y} }}{{ - y}}\sum\limits_{k = 0}^\infty {\frac{{ - (p)_k }}{{( - y)^k }}} $$ where $(p)_k$ denotes the Pochhammer symbol. Taking reciprocals $$ \frac{1}{x} = \frac{1}{{E_p (y)}} \sim {\rm e}^y y\left( {1 + \frac{p}{y} - \frac{p}{{y^2 }} + \frac{{p(p + 2)}}{{y^3 }} - \ldots } \right) $$ as $y\to+\infty$. Employing Theorem $1$ of this paper with $1/x$ in place of $x$ and $\alpha=-1$, yields $$ E^{-1}_p(x)=y \sim - \log x - \log ( - \log x) - \frac{{\log ( - \log x) - p}}{{\log x}} + \ldots $$ as $x\to 0^+$. The cited paper provides more references on these type of asymptotic inversions.