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The derivation of Von Mangoldt’s explicit formula $$ψ(x)=x -log⁡(2π) +∑_n\frac{x^{-2n}}{2n} -∑_ρ\frac{x^ρ}{ρ}$$ can be achieved by applying the residue theorem to the integral below (Havil's book, p 202) $$ψ(x)=\frac{1}{2πi}\int_{a-i∞}^{a+i∞}-\frac{ζ'(s)}{ζ(s)} \frac{x^s}{s} ds$$ $[-\frac{ζ'(s)}{ζ(s)} \frac{x^s}{s}]$ has four singularities: when s is equal to 0 and 1 and when ζ(s) is equal to zero (trivial and non-trivial) and the question is when s=1 how the residue is found to be equal to x?

Arwashan
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1 Answers1

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In the limit $s\to1$ we have

\begin{eqnarray} \frac{\zeta'(s)}{\zeta(s)} &=& \frac{\left(\frac1{s-1}\right)'+O(1)}{\frac1{s-1}+O(1)} \\ &=& -\frac{\left(\frac1{s-1}\right)^2+O(1)}{\frac1{s-1}+O(1)} \\ &=& -\frac{\frac1{s-1}+O(s-1)}{1+O(s-1)} \\ &=& -\frac1{s-1}+O(1)\;. \end{eqnarray}

Thus the residue of the entire product at $s=1$ is simply minus the value of the remaining factors (which are analytic at $s=1$) at $s=1$, and thus $x$.

joriki
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  • Thanks, two questions: – Arwashan Apr 20 '20 at 14:25
  • Thanks, two questions: how did you obtain the remaining of zeta is O(1)? and how did you obtain the derivative of O(1) is also O(1)? – Arwashan Apr 20 '20 at 14:31
  • @Arwashan: You can find the Laurent series of the zeta function in various places on the net, e.g. Wikipedia and this question on this site. To your second question: The derivative of a function that is $O(1)$ is not necessarily $O(1)$, e.g. consider $\sin\frac1x$. Rather, what I used was that $\zeta(s)-\frac1{s-1}$ is analytic at $s=1$, which implies that it and all its derivatives are $O(1)$ as $s\to1$. – joriki Apr 20 '20 at 14:48
  • Thanks for the reference. But sorry I don't see how zeta(s)-1/(s-1) is analytic at s=1 – Arwashan Apr 20 '20 at 15:10