SI Sine Integral

SI.1 Introduction

Let x be a complex variable of ℂ \ {∞}.The function Sine Integral (noted Si) is defined by the following third order differential equation

\begin{aligned}& x\frac{∂y(x)}{∂x} + 2\frac{∂^{2}y(x)}{∂x^{2}} + x\frac{∂^{3}y(x)}{∂x^{3}} & = 0. \\ (SI.1.1) \end{aligned}

Although 0 is a singularity of SI.1.1, the initial conditions can be given by

\begin{aligned}& \frac{∂ \frac{Si(x)}{x}}{∂x} & = 1. \\ (SI.1.2) \end{aligned}

Related function: Cosine Integral

SI.2 Series and asymptotic expansions

SI.2.1 Asymptotic expansion at 0. SI.2.1.1 First terms.

\begin{matrix}Si(x) ≈ x(1 − \frac{x^{2}}{18} + \frac{x^{4}}{600} − \frac{x^{6}}{35280} + \frac{x^{8}}{3265920} − \frac{x^{10}}{439084800} + \frac{x^{12}}{80951270400}− \\ \frac{x^{14}}{19615115520000} …). \end{matrix} ; (SI.2.1.1.1)

SI.2.1.2 General form.

\begin{aligned}& Si(x) ≈ x ∑_{n=0}^{∞} u(n)x^{n}. \\ (SI.2.1.2.1) \end{aligned}

The coefficients u(n) satisfy the recurrence

\begin{aligned}& u(n)(n + 1)(−1 − n + (n + 1)^{2}) + u(n − 2)(−1 + n) & = 0. \\ (SI.2.1.2.2) \end{aligned}

Initial conditions of SI.2.1.2.2 are given by

\begin{aligned}u(1) & = 0, \\ u(0) & = 1, \\ u(2) & = −\frac{1}{18}. \end{aligned} ; (SI.2.1.2.3)

The recurrence SI.2.1.2.2 has the closed form solution

\begin{aligned}u(2n + 1) & = 0, \\ u(2n) & = \frac{(−1)^{n}}{(2n + 1)Γ(2n + 2)}. \end{aligned} ; (SI.2.1.2.4)

SI.2.2 Asymptotic expansion at ∞.

SI.2.2.1 First terms.

Si(x) ≈ ser_{[1,1,[[0,[[0, \frac{π}{2} ]]]]]} + e^{(− \frac{RootOf_{ξ,2}(1+ξ^{2})}{x})} xy_{1}(x)+

e^{(− \frac{RootOf_{ξ,1}(1+ξ^{2})}{x})} xy_{2}(x),

where

y_{0}(x) = terms_{[1,1,[[0,[[0, \frac{π}{2} ]]]]]} + …

y_{1}(x) = −\frac{1}{2} − \frac{RootOf_{ξ,2}(1 + ξ^{2})x}{2}−

− (−\frac{1}{4} − \frac{5 RootOf_{ξ,2}(1 + ξ^{2})^{2}}{4} ) x^{2} − −(− RootOf_{ξ,2}(1 + ξ^{2})+

4 RootOf_{ξ,2}(1 + ξ^{2}) (−\frac{1}{4} − \frac{5 RootOf_{ξ,2}(1 + ξ^{2})^{2}}{4} ))x^{3} + 2 …

y_{2}(x) = −\frac{1}{2} − \frac{RootOf_{ξ,1}(1 + ξ^{2})x}{2}−

− (−\frac{1}{4} − \frac{5 RootOf_{ξ,1}(1 + ξ^{2})^{2}}{4} ) x^{2} − −(− RootOf_{ξ,1}(1 + ξ^{2})+

4 RootOf_{ξ,1}(1 + ξ^{2}) (−\frac{1}{4} − \frac{5 RootOf_{ξ,1}(1 + ξ^{2})^{2}}{4} ))x^{3} + 2 …

SI.2.2.2 General form. SI.2.2.2.1 Auxiliary function y_{0}(x). The auxiliary function y_{0}(x) has the exact

form

y_{0}(x) = \frac{π}{2}

SI.2.2.2.2 Auxiliary function y_{1}(x). The coefficients u(n) of y_{1}(x) satisfy the following recurrence

2u(n)n + u(−1 + n)(−2 RootOf_{ξ,2}(1 + ξ^{2})−
5 RootOf_{ξ,2}(1 + ξ^{2})(−1 + n) − 3(−1 + n)^{2} RootOf_{ξ,2}(1 + ξ^{2}))+
u(n − 2)(8 − 5n − 4(n − 2)^{2} − (n − 2)^{3}) = 0

whose initial conditions are given by

\begin{aligned}u(1) & = \frac{− RootOf_{ξ,2}(1 + ξ^{2})}{2} \\ u(0) & = −\frac{1}{2} \end{aligned}

SI.2 SERIES AND ASYMPTOTIC EXPANSIONS 3

SI.2.2.2.3 Auxiliary function y_{2}(x). The coefficients u(n) of y_{2}(x) satisfy the following recurrence

2u(n)n + u(−1 + n)(−2 RootOf_{ξ,1}(1 + ξ^{2})− 5 RootOf_{ξ,1}(1 + ξ^{2})(−1 + n) − 3(−1 + n)^{2} RootOf_{ξ,1}(1 + ξ^{2}))+ u(n − 2)(8 − 5n − 4(n − 2)^{2} − (n − 2)^{3}) = 0

whose initial conditions are given by

\begin{aligned}u(0) & = −\frac{1}{2} \\ u(1) & = \frac{− RootOf_{ξ,1}(1 + ξ^{2})}{2} \end{aligned}