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\begin{document}
\chapter*{EI Exponential Integral}
\label{EI}
\section*{EI.1 Introduction}
\label{EI:intro}


Let $x$ be a complex variable of $\mathbb{C} \setminus \{0,\infty\}$.The function Exponential Integral (noted $\operatorname{Ei}$) is defined by the following second order differential equation
\begin{equation*}
\label{EI:diffeq}
\begin{split}
(1 - x) \frac{\partial y (x)}{\partial x} + x \frac{\partial^{2} y (x)}{\partial x^{2}}& =0.
\end{split}\tag{EI.1.1}
\end{equation*}



The initial conditions of EI.1.1 at $0$ are not simple to state, since $0$ is a (regular) singular point.
\section*{EI.2 Series and asymptotic expansions}
\label{EI:asympt}
\subsection*{EI.2.1 Asymptotic expansion at $0$}
\label{743358320567665023}
\subsubsection*{EI.2.1.1 First terms}
\label{EI:asympt:0:termsec}



\begin{equation*}
\label{EI:asympt:0:terms}
\begin{split}
& \operatorname{Ei} (x)\approx \Biggl(-x - \frac{x^{2}}{4} - \frac{x^{3}}{18} - \frac{x^{4}}{96} - \frac{x^{5}}{600} - \frac{x^{6}}{4320} - \frac{x^{7}}{35280} - \frac{x^{8}}{322560} - \operatorname{ln} (x) + \gamma\ldots\Biggr).
\end{split}\tag{EI.2.1.1.1}
\end{equation*}

\subsubsection*{EI.2.1.2 General form}
\label{743358739634399202}
The general form of is not easy to state and requires to exhibit the basis of formal solutions of ?? (coming soon).\subsection*{EI.2.2 Asymptotic expansion at $\infty$}
\label{743359475924727175}
\subsubsection*{EI.2.2.1 First terms}
\label{743359706704432196}



\begin{equation*}
\begin{split}
& \operatorname{Ei} (x)\approx \operatorname{e} ^{\frac{1}{x}} x y _{0} (x),
\end{split}
\end{equation*}
where
\begin{equation*}
\begin{split}
y _{0} (x)& =1 + x + 2 x^{2} + 6 x^{3} + 2 \ldots
\end{split}
\end{equation*}

\subsubsection*{EI.2.2.2 General form}
\label{743359935502372691}
\paragraph*{EI.2.2.2.1 Auxiliary function $y _{0} (x)$}
\label{743359612975251926}
The coefficients $u (n)$ of $y _{0} (x)$ satisfy the following recurrence
\begin{equation*}
\begin{split}
-u (n) n + u (n - 1) \bigl(-1 + 2 n + (n - 1)^{2}\bigr)& =0
\end{split}
\end{equation*}
whose initial conditions are given by
\begin{equation*}
\begin{split}
u (0)& =1
\end{split}
\end{equation*}
This recurrence has the closed form solution
\begin{equation*}
\begin{split}
u (n)& =\Gamma (n + 1).
\end{split}
\end{equation*}
\end{document}