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How do I evaluate $\xi(0)$ for the Riemann xi function?

I know $\xi(0) = \xi(1)$ and

  • $\xi(0) = \tfrac{1}{2} \cdot 0 \cdot (-1) \cdot \Gamma(0) \cdot \zeta(0)$
  • $\xi(1) = \tfrac{1}{2} \cdot 1 \cdot 0 \cdot \Gamma(\tfrac{1}{2}) \cdot \zeta(1)$

and $\zeta(0) = -\frac{1}{2}$, $\Gamma(\tfrac{1}{2}) = \sqrt{2\pi}$

but $\Gamma(0) = \infty$ and $\zeta(1) = \infty$ so I don't know how to evaluate it.

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Hint: look here for additional Riemann Zeta function expansions. Take a look at the equation (20) there. –  m0nhawk Feb 11 '13 at 16:06

1 Answer 1

up vote 4 down vote accepted

By definition, $$\xi(s)=\frac{1}{2}s(s-1)\pi^{-\frac{s}{2}}\Gamma\left(\frac{1}{2}s\right)\zeta(s).$$ Around $s=1$, $\zeta(s)=\frac{1}{s-1}+O(1),$ so that $$\xi(s)=\frac{1}{2}s\pi^{-\frac{s}{2}}\Gamma\left(\frac{1}{2}s\right)+O\left(s-1\right).$$ Letting $s\rightarrow1$, we find that $\xi(1)=\frac{1}{2}.$

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Thanks a lot for helping me –  user58512 Feb 11 '13 at 16:24

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