# The derivatives of Riemann xi function

What are the first few values of derivatives of Riemann xi function at zero? Is there any general formula for calculating the nth derivative of the riemann zeta function at zero? What happens to the value of the nth derivative of xi function as "n" tends to infinity? what is the Taylor series expansion of xi function around zero?

## migrated from mathoverflow.netFeb 7 '16 at 12:06

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## 1 Answer

Assuming you mean the version defined by Landau, namely

$$\xi(s) = \frac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta\left(s\right)$$

then any computer algebra system will be able to do this for you. For example, in Maple,

xi := s -> s/2*(s-1)*Pi^(-s/2)*GAMMA(s/2)*Zeta(s):
nth := proc(n) limit(diff(xi(s),s\$n),s=0) end


will give you what you want. For example nth(2) is

$$-\gamma \left( 1 \right) +\gamma/2+1/8\, \left( \ln \left( \pi \right) \right) ^{2}-1/2\,\gamma\,\ln \left( 2 \right) -1/4\,\ln \left( \pi \right) \gamma+1/2\,\ln \left( 2 \right) \ln \left( \pi \right) +1/2\, \left( \ln \left( 2 \right) \right) ^{2}-1/2\, \ln \left( \pi \right) -\ln \left( 2 \right) +1/16\,{\pi }^{2}-3/8 \,{\gamma}^{2}$$

Warning: these grow in size very fast! [Also, you'll need

$$\gamma(n) := \lim_{m\rightarrow\infty} \left(\sum_{k=1}^{m}\frac{\ln(k)^n}{k} - \frac{\ln(m)^{n+1}}{n+1}\right)$$ ]

I don't believe that many of your questions have actually 'known' answers.