On 2015-04-12 13:57, Dan Asimov wrote:

What's now called Riemann's xi function is (up to a constant factor):

xi(s) := s(s-1) zeta(s) gamma(s/2) pi^(-s/2),

which is an entire function of s, and its zeroes (the values of s for which xi(s) = 0) are exactly the mysterious "nontrivial" zeroes of zeta(s).

Another function (that Riemann called "xi(t)") is now usually called Landau's capital-xi(t) function: Ξ(t).

But since Ξ may not appear correctly in your e-mail, I'll call it X(t):

X(t) := xi(1/2 + i t).

It's easy to check that X(t) is real on the real axis and satisfies

X(t) == X(-t),

which is equivalent to the "functional equation of zeta".

I would like to know the power series coefficients of X(t) about 0.

Mathematica's attempt to do this results in unbelievably complicated expressions, and I've had very little luck simplifying them.

Wikipedia gives an expression for this power series — but each coefficient is expressed as an infinite series in terms of the mysterious nontrivial zeroes of the zeta function.

QUESTION (mostly to you-know-who): Is there a more or less straightforward expression for the power series coefficients of X(t) ???

(Of course, since X(t) == X(-t), only the even powers of t have nonzero coefficients.)

--Dan

For the first two terms, I get Ξ[t]== -((Gamma[1/4]*Zeta[1/2]*(1 + (1/2)*t^2* (8 - (1/4)*PolyGamma[1, 1/4] + (Derivative[1][Zeta][1/2]^2 - Zeta[1/2]* Derivative[2][Zeta][1/2])/Zeta[1/2]^2)))/ (4*Pi^(1/4))) == (1/(128 \[Pi]^(1/4))) Gamma[1/4] (-(32 + t^2 (128 - 32 Catalan + 4 EulerGamma^2 - 3 \[Pi]^2 + 4 \[Pi] Log[8 \[Pi]] + 4 Log[8 \[Pi]]^2 + 4 EulerGamma (\[Pi] + Log[64 \[Pi]^2]))) Zeta[1/2] + 16 t^2 Zeta''[1/2]) with no help from http://isc.carma.newcastle.edu.au/standard and the promise of worse to come. Let's hope you-know-who has better luck. --rwg