Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have two relations:


2)$\psi(x)=\sum_{n\leq x}\Lambda(n)$.

From these two how does it follow that $-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$, for $s>1$. Can anyone explain how does it follow?

share|cite|improve this question
1… – Erick Wong May 19 '13 at 18:31

Let's start with the definition of the von Mangoldt function ($p$ will always mean a prime) : $$\ \Lambda(n):=\begin{cases} \log\, p & \text{if}\ n=p^k\ \text{and}\ k>0\\ 0 & \text{else} \end{cases}$$

We may use the Euler product in $\ \displaystyle\log \zeta(s)=-\sum_{p\ \text{prime}}\log(1-p^{-s})=\sum_p\sum_{k=1}^\infty \frac{p^{-ks}}k\ $ to obtain your first relation (after derivation) :

$$\tag{1}-\frac{\zeta'(s)}{\zeta(s)}=\sum_p\sum_{k=1}^\infty \frac{\log\,p}{p^{ks}}=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}\quad\text{for}\ \ \Re(s)>1$$

After that we want to use the definition of the second Chebyshev function : $$\tag{2} \psi(x)=\sum_{n\leq x}\Lambda(n)$$

But, using Abel's sum formula with $a(n):=\Lambda(n)$ and $\phi(n):=n^{-s}$, we get : $$\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}=s\int_1^\infty \frac {\sum_{n\leq x}\Lambda(n)}{x^{s+1}}\;dx$$ that is, using $(1)$ and $(2)$, the wished formula : $$-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx$$

Inverting this Mellin transform to express $\psi(x)$ would produce Perron's formula as shown in this derivation of the 'explicit formula' with better handling of the discontinuities.

share|cite|improve this answer
+1 very nice, Raymond... – draks ... May 19 '13 at 22:41
Thanks @draks even if you should know much about this now ! :-) This is only a small (edited) part of this expanded answer after all... – Raymond Manzoni May 19 '13 at 22:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.