did
Posts:
80
Registered:
9/14/05


Re: Inversion Lerch Phi
Posted:
Apr 14, 2012 1:32 PM


On 4/14/12 6:34 PM, clicliclic@freenet.de 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*(1a)) > > 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[1] >  2*ArcTanh[Sqrt[2]]) / Sqrt[2] instead of Sqrt[2]*I*Pi if that's what > it is for its definitions of ArcSinh and ArcTanh?
Actually, MMA FullSimplify does simplify it completely:
FullSimplify[ HurwitzLerchPhi[1/2, 1, 1/2]  HurwitzLerchPhi[2, 1, 1/2] + 2  (1/z)^(1/2)*Pi/Sin[Pi*3/2]]
\[Pi] (I/Sqrt[2] + Sqrt[(1/z)])
I'm not a MMA guru (nor Maple) so I can't say why Simplify does not do it completely.
For the polylogarithm, the inversion formulas in the Wikipedia page are not valid for z in ]0;1] or for z in [1;inf[.
By trials and errors and numerical tests (so not via a rigorous mathematical proof), I derived the single formula (in Maple notations):
polylog(s,z) + (1)^s*polylog(s,1/z) = (2*I*Pi)^nu/GAMMA(s)*Zeta(0,1s,log(z)/I/2/Pi)  Heaviside(argument(z1))*I*2*Pi*(log(z))^(s1)/GAMMA(s)
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*(zu)),s,a)  Phi(1/exp(I*(zu)),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).
Did

