# Proof of an identity of the logarithm of the zeta function

$$\log \zeta (s) = \sum_{n=2}^\infty \frac{\Lambda_1(n)}{n^s} \quad ( \sigma > 1) ,$$ where $\Lambda_1(n) = \frac{\Lambda(n)}{\log n }$, and $\Lambda (n) = \log p$ if $n$ is $p$ or a power of $p$. (p4, Theory of the Riemann Zeta Function, Titchmarsh)

The proof is given as follows:

\begin{align*} \frac{\zeta'(s)}{\zeta(s)} &= - \sum _p \frac{\log p}{p^s} \Big( 1 - \frac{1}{p^s} \Big)^{-1} \\ &= - \sum_p \log p \sum_{m=1}^\infty \frac{1}{p^{ms}} \\ &= - \sum_{n=2}^{\infty} \frac{\Lambda (n)}{n^s} \end{align*} Integrating both sides we obtain, $$\log \zeta(s) = \sum _{n=2}^{\infty} \frac{\Lambda_1(n)}{n^s}$$

What I don't understand is: There should be a constant factor under integration. How did the author get rid of it?

If we write$$\log\zeta(s)+K=\sum_{n=2}^\infty\frac{\Lambda_1(n)}{n^s},$$for some constant $K$, then$$\lim_{s\to+\infty}\log\zeta(s)+K=K\text{ and }\lim_{s\to+\infty}\sum_{n=2}^\infty\frac{\Lambda_1(n)}{n^s}=0.$$Therefore, $K=0$.