Calculation of the residue of $[-\frac{ζ'(s)}{ζ(s)} \frac{x^s}{s}]$ when s=1

Question

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?

Answer 1

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 $x$.