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\begin{document}
\chapter*{CHI Hyperbolic Cosine Integral}
\label{CHI}
\section*{CHI.1 Introduction}
\label{CHI:intro}


Let $x$ be a complex variable of $\mathbb{C} \setminus \{0,\infty\}$.The function Hyperbolic Cosine Integral (noted $\operatorname{Chi}$) is defined by the following third order differential equation
\begin{equation*}
\label{CHI:diffeq}
\begin{split}
-x \frac{\partial y (x)}{\partial x} + 2 \frac{\partial^{2} y (x)}{\partial x^{2}} + x \frac{\partial^{3} y (x)}{\partial x^{3}}& =0.
\end{split}\tag{CHI.1.1}
\end{equation*}



The initial conditions of CHI.1.1 at $0$ are not simple to state, since $0$ is a (regular) singular point.


Related function: \href{http://algo.inria.fr/esf/function/SHI/SHI.html#SHI}{Hyperbolic Sine Integral}
\section*{CHI.2 Series and asymptotic expansions}
\label{CHI:asympt}
\subsection*{CHI.2.1 Asymptotic expansion at $0$}
\label{743342370035300293}
\subsubsection*{CHI.2.1.1 First terms}
\label{CHI:asympt:0:termsec}



\begin{equation*}
\label{CHI:asympt:0:terms}
\begin{split}
& \operatorname{Chi} (x)\approx \Biggl(\frac{-x^{2}}{4} - \frac{x^{4}}{96} - \frac{x^{6}}{4320} - \frac{x^{8}}{322560} - \operatorname{ln} (x) + \gamma\ldots\Biggr).
\end{split}\tag{CHI.2.1.1.1}
\end{equation*}

\subsubsection*{CHI.2.1.2 General form}
\label{743342312321897511}
The general form of is not easy to state and requires to exhibit the basis of formal solutions of ?? (coming soon).\subsection*{CHI.2.2 Asymptotic expansion at $\infty$}
\label{743344461159161046}
\subsubsection*{CHI.2.2.1 First terms}
\label{743343262286362344}



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

\subsubsection*{CHI.2.2.2 General form}
\label{743344140288924255}
\paragraph*{CHI.2.2.2.1 Auxiliary function $y _{0} (x)$}
\label{743343116441804875}
The coefficients $u (n)$ of $y _{0} (x)$ satisfy the following recurrence
\begin{equation*}
\begin{split}
& -2u (n) n + u (n - 1) \bigl(-3 + 3 (n - 1)^{2} + 5 n\bigr) +  \\
& u (n - 2) \bigl(8 - 5 n - 4 (n - 2)^{2} - (n - 2)^{3}\bigr)=0
\end{split}
\end{equation*}
whose initial conditions are given by
\begin{equation*}
\begin{split}
u (1)& =\frac{1}{2} \\
u (0)& =\frac{1}{2}
\end{split}
\end{equation*}
This recurrence has the closed form solution
\begin{equation*}
\begin{split}
u (n)& =\frac{\Gamma (n + 1)}{2}.
\end{split}
\end{equation*}
\paragraph*{CHI.2.2.2.2 Auxiliary function $y _{1} (x)$}
\label{743344462714681351}
The coefficients $u (n)$ of $y _{1} (x)$ satisfy the following recurrence
\begin{equation*}
\begin{split}
& -2u (n) n + u (n - 1) \bigl(3 - 3 (n - 1)^{2} - 5 n\bigr) +  \\
& u (n - 2) \bigl(8 - 5 n - 4 (n - 2)^{2} - (n - 2)^{3}\bigr)=0
\end{split}
\end{equation*}
whose initial conditions are given by
\begin{equation*}
\begin{split}
u (1)& =\frac{1}{2} \\
u (0)& =-\frac{1}{2}
\end{split}
\end{equation*}
This recurrence has the closed form solution
\begin{equation*}
\begin{split}
u (n)& =\frac{-(-1)^{n} \Gamma (n + 1)}{2}.
\end{split}
\end{equation*}
\end{document}