# Logarithmic derivative of Riemann Zeta function

Given the logarithmic derivative of the zeta function $\dfrac{\zeta^\prime (s)}{\zeta(s)}$ how does it behave near $s=1$?

I mean if for $s=1$ the Laurent series for the logarithmic derivative becomes

$$\frac{\zeta^\prime (s)}{\zeta(s)}= A+ (s-1)^{-1}$$

where $A$ is a real number constant.

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The $A$ you have there is the Euler-Mascheroni constant, $\gamma$. Actually, your series goes like $\frac{\zeta^\prime (s)}{\zeta(s)}=-\frac1{s-1}+\gamma-(\gamma^2+2\gamma_1)(s-1)+\cdots$ where $\gamma_1$ is the first Stieltjes constant. – J. M. Oct 9 '11 at 12:52

Let's generalize and look at all of the constants in the Laurent expansion. Consider $$-\frac{\zeta^\prime}{\zeta}(s+1)=\frac{1}{s}+\sum_{j=0}^\infty \frac{(-1)^j}{j!}\eta_j s^j.$$ (I put in a negative sign so it corresponds to $\sum\limits_{n=1}^\infty \Lambda(n)n^{-s}$.) I claim that $\eta_0 =-\gamma_0$, Euler's Constant, and that in general $\eta_j$ will be a sum of products of the Stieltjes constants.

Constants arising from a limit: It is interesting to note that in a way analogous to the fact that the Stieltjes constants are given by the limit $$\gamma_n=\lim_{n\to\infty}{\left(\left(\sum_{k = 1}^m \frac{(\log k)^n}{k}\right) - \frac{(\log m)^{n+1}}{n+1}\right)},$$ we have that $$\eta_n=\lim_{n\to\infty}{\left(\left(\sum_{k = 1}^m \frac{(\log k)^{n}\Lambda(k)}{k}\right) - \frac{(\log m)^{n+1}}{n+1}\right)}$$ where $\Lambda(n)$ is the von Mangoldt Lambda Function. This can be proven by using exponential generating series, and then the Dirichlet series for $-\frac{\zeta^\prime}{\zeta}(s).$

Computing the Constants: We can compute $\eta_j$ in terms of the Stieltjes constants by using the relation $$\log\left(1+\sum_{i=0}^\infty \frac{(-1)^j}{j!}\gamma_j s^{j+1}\right)= \log (s\zeta(s+1))=-\sum_{j=0}^\infty \frac{(-1)^j}{j!}\eta_j s^j$$

and then expanding by using the identity $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$

From this we get the power series $$\gamma_0 s-\left(2\gamma_1+\gamma_0^2\right)\frac{s^2}{2}+\left(3\gamma_2+6\gamma_1\gamma_0+2\gamma_0^3\right)\frac{s^3}{6}+\cdots$$ so that $$\eta_0 =-\gamma, \ \ \ \eta_1 =-(2\gamma_1 +\gamma_0^2), \ \ \ \eta_2 =-(3\gamma_2 +6\gamma_1\gamma_0 +2\gamma_0^3) \qquad \text{et cetera}.$$

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THANK YOU very much to the people who answered :) – Jose Garcia Oct 9 '11 at 13:15

As an addendum to Eric's answer, Coffey gives a recursion relation for the coefficients $\dfrac{\eta_j}{j!}$ appearing in the expansion of the $\zeta$ function's logarithmic derivative. Letting $\gamma_k$ be the Stieltjes constants, and with the initial condition $\eta_0=-\gamma_0=-\gamma$, we have the recursion relation

$$\frac{\eta_j}{j!}=(-1)^{j+1}\left(\frac{j+1}{j!}\gamma_j+\sum_{k=0}^{j-1} \frac{(-1)^{k-1}}{(j-k-1)!}\frac{\eta_k}{k!}\gamma_{j-k-1}\right)$$

A similar formula also appears in Choudhury's paper.

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