As a first step, the log-sine integral package needs to be loaded in some way.
Ultimately, this step is hoped to become unnecessary if the functionality is integrated into the core of SAGE.
{{{id=1| attach "/home/armin/docs/math/sage/logsine.sage" /// }}}Log-sine integrals can now be entered as in the examples below.
If the option typeset is selected for this notebook then the resulting output matches the notation used in mathematical papers.
{{{id=6| ls(5,1,pi/3) /// \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{5}^{(1)}(\frac{1}{3} \, \pi) }}} {{{id=8| ls(5,0,pi/3) /// \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{5}(\frac{1}{3} \, \pi) }}}The main functionality of this package is the conversion of log-sine integrals to polylogarithms. This is achieved with the command lstoli as illustrated in the next few examples:
{{{id=37| lstoli(-ls(4,2,pi)) /// \newcommand{\Bold}[1]{\mathbf{#1}}\frac{3}{2} \, \pi \zeta(3) }}} {{{id=38| lstoli(ls(4,1,pi)) /// \newcommand{\Bold}[1]{\mathbf{#1}}-\frac{11}{720} \, \pi^{4} - 2 \, \textrm{Gl}_{3,1}(\pi) }}} {{{id=39| lstoli(-ls(5,1,pi)) /// \newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{4} \, \pi^{2} \zeta(3) + \frac{105}{32} \, \zeta(5) - 6 \, \textrm{Cl}_{3,1,1}(\pi) }}} {{{id=40| lstoli(-ls(6,1,pi)) /// \newcommand{\Bold}[1]{\mathbf{#1}}-\frac{3}{1120} \, \pi^{6} + 3 \, \zeta(3)^{2} + 24 \, \textrm{Gl}_{3,1,1,1}(\pi) - 18 \, \textrm{Gl}_{5,1}(\pi) }}} {{{id=43| lstoli(ls(5,1,pi/3)) /// \newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{12} \, \pi^{2} \zeta(3) + \frac{3}{2} \, \pi \textrm{Cl}_{4}(\frac{1}{3} \, \pi) - \frac{19}{4} \, \zeta(5) }}} {{{id=44| lstoli(ls(5,0,pi/3)) /// \newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1543}{19440} \, \pi^{5} + 6 \, \textrm{Gl}_{4,1}(\frac{1}{3} \, \pi) }}} {{{id=5| for n in [2..8]: view(ls(n,0,pi) == lstoli(ls(n,0,pi))) /// \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{2}(\pi) = 0 \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{3}(\pi) = -\frac{1}{12} \, \pi^{3} \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{4}(\pi) = \frac{3}{2} \, \pi \zeta(3) \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{5}(\pi) = -\frac{19}{240} \, \pi^{5} \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{6}(\pi) = \frac{5}{4} \, \pi^{3} \zeta(3) + \frac{45}{2} \, \pi \zeta(5) \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{7}(\pi) = -\frac{275}{1344} \, \pi^{7} - \frac{45}{2} \, \pi \zeta(3)^{2} \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{8}(\pi) = \frac{133}{32} \, \pi^{5} \zeta(3) + \frac{315}{8} \, \pi^{3} \zeta(5) + \frac{2835}{4} \, \pi \zeta(7) }}} {{{id=46| for n in [2..7]: view(ls(n,0,pi/3) == lstoli(ls(n,0,pi/3))) /// \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{2}(\frac{1}{3} \, \pi) = \textrm{Cl}_{2}(\frac{1}{3} \, \pi) \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{3}(\frac{1}{3} \, \pi) = -\frac{7}{108} \, \pi^{3} \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{4}(\frac{1}{3} \, \pi) = \frac{1}{2} \, \pi \zeta(3) + \frac{9}{2} \, \textrm{Cl}_{4}(\frac{1}{3} \, \pi) \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{5}(\frac{1}{3} \, \pi) = -\frac{1543}{19440} \, \pi^{5} + 6 \, \textrm{Gl}_{4,1}(\frac{1}{3} \, \pi) \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{6}(\frac{1}{3} \, \pi) = \frac{35}{36} \, \pi^{3} \zeta(3) + \frac{15}{2} \, \pi \zeta(5) + \frac{135}{2} \, \textrm{Cl}_{6}(\frac{1}{3} \, \pi) \newcommand{\Bold}[1]{\mathbf{#1}}\textrm{Ls}_{7}(\frac{1}{3} \, \pi) = -\frac{74369}{326592} \, \pi^{7} - \frac{15}{2} \, \pi \zeta(3)^{2} + 135 \, \textrm{Gl}_{6,1}(\frac{1}{3} \, \pi) }}} {{{id=42| lstoli(ls(6,3,pi/3)-2*ls(6,1,pi/3)) /// \newcommand{\Bold}[1]{\mathbf{#1}}\frac{313}{204120} \, \pi^{6} }}}The argument of the log-sine integral may be symbolic:
{{{id=13| assume(0