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According to Wikipedia, there is a global convergent series for Riemann Zeta function:

https://en.wikipedia.org/wiki/Riemann_zeta_function#Globally_convergent_series

Is there a similar global convergent series for Riemann Xi function ?

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Riemann Xi function can be defined via Riemann zeta function as:

$$\Xi(z)=\xi(1/2+iz)$$ $$\xi(s)=(1/2)s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$$

Since $\Xi(z)$ is an entire function of $z$, it can be Taylor-expanded at any point $z_0(\not=\infty)$ in the complex plane.

You can find out more in the following references:

(1)Riemann's Zeta FunctionJun 13, 2001 by Harold M. Edwards

(2)The Theory of the Riemann Zeta-Function (Oxford Science Publications)Feb 5, 1987 by E. C. Titchmarsh and D. R. Heath-Brown

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Since the Riemann Xi function is an entire function it can be expressed as a Taylor series that converges for all finite complex numbers. Go to mathworld.wolfram.com, search "Xi-function", and read the corresponding article.

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I believe the following two equivalent formulas for $\xi(s)$ are globally convergent. Note both formulas below are unchanged by the substitution $s=1-s$.

(1) $\quad\xi(s)=\frac{1}{2}-\frac{s\,(1-s)}{2}\sum\limits_{n=1}^\infty\left(\left(\sqrt{\pi}\,n\right)^{-s}\,\Gamma\left(\frac{s}{2},\pi\,n^2\right)+\left(\sqrt{\pi}\,n\right)^{-(1-s)}\,\Gamma\left(\frac{1-s}{2},\pi\,n^2\right)\right)$

(2) $\quad\xi(s)=\frac{1}{2}-\frac{s\,(1-s)}{2}\sum\limits_{n=1}^\infty\left(E_{\frac{1+s}{2}}\left(\pi\,n^2\right)+E_{\frac{1+(1-s)}{2}}\left(\pi\,n^2\right)\right)$

The two formulas above were derived from the Jacobi theta functional equation which was used by Riemann to prove the functional equation $\xi(s)=\xi(1-s)$.

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