Ok it seems that li(x)=Ei(log(x)) and this works for complex values. Furthermore Li(x)=li(x)-li(2).
It seems that Li(x) has been artificially curtailed to take only real arguments. Michel On Nov 2, 3:00 pm, Michel <[EMAIL PROTECTED]> wrote: > Hi, > > Is there a method in sage to compute Li(z) for z complex? I am trying > to illustrate > Riemann prime number formula but now it seems Li(z) accept only real > arguments. > > Michel --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to sage-support@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sage-support URLs: http://www.sagemath.org -~----------~----~----~----~------~----~------~--~---
