did
Posts:
73
Registered:
9/14/05
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Re: Inversion Lerch Phi
Posted:
Apr 14, 2012 1:32 PM
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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*(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[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(0,1-s,log(z)/I/2/Pi) -
Heaviside(-argument(z-1))*I*2*Pi*(log(z))^(s-1)/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*(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).
Did
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