The Von Mangoldt function is defined as follows: $$\Lambda (n)={\begin{cases}\log p&{\text{if }}n=p^{k}{\text{ for some prime }}p{\text{ and integer }}k\geq 1,\\0&{\text{otherwise.}}\end{cases}}$$

On the Wikipedia page over the Von Mangoldt function the identity below is listed: $$\ln\zeta(s)=\sum _{{n=2}}^{\infty}{\frac{\Lambda (n)}{\ln(n)}}{\frac{1}{n^{s}}}\qquad {\text{Re}}(s)>1$$ Question

How do you prove this? I'm not familiar with the Von Mangoldt function so pardon me if the identity has a completely trivial derivation, but I couldn't find it anywhere so I had to ask here.


1 Answer 1


If $n=p^k$ with $p$ prime and $k\geqslant 1$ then $\frac{\Lambda(n)}{\log(n)}=\frac{\log(p)}{\log(p^k)}=\frac{1}{k}$, thus $$ \sum_{n=1}^{+\infty}\frac{\Lambda(n)}{\log(n)n^s}=\sum_{p^k}\frac{1}{kp^{ks}} $$ Moreover, the Euler's product formula for $\zeta$ states that $$ \zeta(s)=\prod_{p\text{ prime}}\left(1-\frac{1}{p^s}\right)^{-1} $$ Thus $$ \log\zeta(s)=-\sum_{p\text{ prime}}\ln\left(1-\frac{1}{p^s}\right)=\sum_{p\text{ prime}}\sum_{k=1}^{+\infty}\frac{1}{kp^{ks}}=\sum_{p^k}\frac{1}{kp^{ks}}=\sum_{n=1}^{+\infty}\frac{\Lambda(n)}{\log(n)n^s} $$

  • $\begingroup$ Thanks for the answer. Just a quick question, how does $$-\sum_p\ln\left(1-\frac{1}{p^s}\right) = \sum_p\sum_{k=1}^{\infty}\frac{1}{kp^{ks}}$$ Are you using the taylor series for $\ln(x+1)$ or? $\endgroup$ Apr 28, 2020 at 14:33
  • $\begingroup$ Yes, I'm using $$\ln(1-x)=-\sum_{k=1}^{+\infty}\frac{x^k}{k}$$ $\endgroup$
    – Tuvasbien
    Apr 28, 2020 at 14:34
  • $\begingroup$ Thank you very much! $\endgroup$ Apr 28, 2020 at 14:35

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