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I am looking for the references on Taylor series expansion of Riemann xi function at $\frac{1}{2}$.

$$ \xi (s)=\sum_0^{\infty}a_{2n}(s-\frac{1}{2})^{2n}$$ where $$a_{2n}=4\int_1^{\infty}\frac{d[x^{3/2}\psi'(x)]}{dx}\frac{(\frac{1}{2}ln(x))^{2n}}{(2n)!}x^{-1/4}dx$$ and $$\psi(x)=\sum_{m=1}^{\infty}e^{-m^2\pi x}=\frac{1}{2}[\theta_3(0,e^{-\pi x})-1]$$

Specifically I would like to know how fast $a_{2n}$ goes to zero.

Has anyone proved that $$a_0>a_2>a_4>...>a_{2n}>...>a_{\infty}=0$$

Thanks a lot!

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I have a short proof that a_2n greater than a_2n+2 for all n. Are you interested? – user85015 Jan 7 at 18:56
Of course he is^^ – Matthias Ludewig Jan 7 at 22:44
knowing that $\xi(s)$ is an entire function, there is no doubt that $a_{2n}$ goes to $0$ quite fast. for the sign, it should be more complicated, at least than $\frac{\xi(s)}{s(s-1)} = \int_0^\infty x^{s-1}\Phi(x) dx$ which is much simpler, also even around $\Re(s) = 1/2$, but has a radius of convergence of $1/2$ at $1/2$. – user1952009 Jan 8 at 7:50
I have a pdf file of the proof that a_2n > a_2n+1. I can send you a copy of this proof if you are still interested. Send request to Unfortunately, MathOverflow does not input pdf files. I do not have time to covert the file to MathJax format. – Richard Katnik Mar 10 at 19:51

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