On 4/14/12 6:34 PM, email@example.com wrote: > But you have a formula > > LerchPhi(z,1,a) - LerchPhi(1/z,1,-a) > = 1/a - (-1/z)^a*Pi/sin(Pi*(1+a)) > > that seems to work for all z except real z with z>1. Replacing z by 1/z > and a by -a you get a formula > > LerchPhi(z,1,a) - LerchPhi(1/z,1,-a) > = 1/a + (-z)^(-a)*Pi/sin(Pi*(1-a)) > > that should work for all z except real z with 0<z<1. Isn't such a pair > good enough? The situation for the relation between polylogarithm and > Hurwitz zeta function is very similar (see the Wikipedia polylogarithm > page). > > Martin. > > PS: And why does the Mathematica simplifier produce (I*Pi + 2*ArcSinh > - 2*ArcTanh[Sqrt]) / Sqrt instead of Sqrt*I*Pi if that's what > it is for its definitions of ArcSinh and ArcTanh?
Actually, MMA FullSimplify does simplify it completely:
that should be valid for all complex z and s. This is the equivalent formula for Lerch Phi I'm looking for. Why? Because my goal is not to compute Phi, instead I have to compute many times something like:
f(z) = Principal Value of Integral of [ Phi(exp(I*(z-u)),s,a) - Phi(1/exp(I*(z-u)),s,-a) ] g(u) du
integral over several paths (not known a priori but crossing z) and I want to avoid the computation of Phi (if possible, but that seems to be the case at least for s=1).