What is the Method for Finding the Derivative of a Zeta Function?

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SUMMARY

The method for finding the derivative of the zeta function, \(\zeta(s)\), involves differentiating term by term when the real part of \(s\) is greater than 1. The derivative is expressed as \(\frac{\partial}{\partial s} \zeta(s) = -\sum_{k=1}^{\infty} k^{-s} \log k\). For cases where Re[s] is less than or equal to 1, analytic continuation or integral definitions can be employed. The Dirichlet series is absolutely convergent in the right half-plane Re(s) ≥ 1 + ε, allowing for valid term-by-term differentiation.

PREREQUISITES
  • Understanding of the Riemann zeta function and its properties
  • Familiarity with series convergence and differentiation techniques
  • Knowledge of analytic continuation in complex analysis
  • Basic understanding of the Euler-Maclaurin summation formula
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  • Study the properties of the Riemann zeta function in detail
  • Learn about analytic continuation methods for complex functions
  • Explore the Euler-Maclaurin summation formula and its applications
  • Investigate numerical methods for approximating zeta function derivatives
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Mathematicians, physicists, and students studying complex analysis or number theory, particularly those interested in the properties and applications of the Riemann zeta function.

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Hello,
I was trying recently to find a derivative of a zeta function but finally I failed. Can anyone show me a way to find it, I'm more interested in the way of finding it rather than clear aprox. solution. Thanks,
 
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If Re > 1 then,

[tex]\zeta(s) = \sum_{k=1}^{\infty} k^{-s}[/tex]

Can't we just differentiate term by term?

[tex]\frac{\partial}{\partial s} k^{-s} = -k^{-s} \log k[/tex]

Then we have:

[tex]\frac{\partial}{\partial s} \zeta(s) = -\sum_{k=1}^{\infty} k^{-s} \log k[/tex]

If Re <= 1 then we could use analytic continuation or we could use an integral definition for the zeta function.
 
Differentiating term by term is valid, the Dirichlet series is absolutely convergent in any right half plane Re(s)>=1+e, where e>0. You can differentiate the usual functional equation to get involving the derivative, though it has more terms than the usual one.

The same method of analytic continuation for zeta via Euler-Maclaurin summation will work here as well for example, if you're after an expression you'd like to numerically work with.
 

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