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I'm following this paper:

Define $\Phi(s) = \displaystyle\sum_p \frac{\log p}{p^s}$.

By taking a logarithm and differentiating the Euler product formula $\zeta(s) = \displaystyle\prod_p \frac{1}{1-p^{-s}}$, we derive

$$-\frac{\zeta'(s)}{\zeta(s)} = \Phi(s) + \displaystyle\sum_p \frac{\log p}{p^s(p^s-1)}.$$

Assume that $s_0 = 1 + ia$ is a zero of $\zeta$ of order $\mu$. Then we can write $\zeta(s) = (s-s_0)^{\mu} g(s)$ for some function $g(s)$ that is analytic and nonzero at $s_0$.

So $\zeta'(s) = \mu(s-s_0)^{\mu -1} g(s) + g'(s) (s-s_0)^{\mu}$ by the product rule.

With a little algebra you can derive

$$-\frac{\zeta'(s)}{\zeta(s)} = -\frac{\mu}{s-s_0} - \frac{g'(s)}{g(s)}.$$

This yields the formula

$$\Phi(s) = -\frac{\mu}{s-s_0} - \frac{g'(s)}{g(s)} - \displaystyle\sum_p \frac{\log p}{p^s(p^s-1)}.$$

Define $F(s) = -\frac{g'(s)}{g(s)} - \displaystyle\sum_p \frac{\log(p)}{p^s(p^s-1)}$. Now we can show that

$$\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon + ia) = \displaystyle\lim_{\epsilon \searrow 0} \epsilon F(1+\epsilon+ia) - \mu = - \mu.$$

through direct computation.

Now, they make a claim in the paper that "because $s=1$ is a simple pole of $\zeta$ of residue $1$", we have $\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon) = 1$. This is what I don't understand. Doing a similar process to what I did above gives me

$$\displaystyle\lim_{\epsilon \searrow 0} \epsilon \Phi(1+\epsilon) = \displaystyle\lim_{\epsilon \searrow 0} -\frac{\epsilon \mu}{\epsilon - ia} - \epsilon \frac{g'(1+\epsilon)}{g(1+\epsilon)} - \epsilon \displaystyle\sum_p \frac{\log p}{p^{1+\epsilon}(p^{1+\epsilon}-1)}$$

And I don't see any way I can get a $1$ as the result, since all terms should converge to zero. The second and third terms will go to zero because they did before and the first term will go to zero by simple limit laws.

I conclude I have made a mistake someplace or I am missing something obvious.

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up vote 2 down vote accepted

From your equation $$ -\frac{\zeta^\prime(s)}{\zeta(s)}=\Phi(s)+\sum_p\frac{\log p}{p^s(p^s-1)} $$ and the fact that $-\zeta^\prime/\zeta(s)$ has a simple pole of reside $1$, we deduce that the same holds for $\Phi(s)$, since the sum on $p$ converges at $s=1$. (Eg., by comparison to $\sum_n \log(n)/n^2$.) So $$ \Phi(s)=\frac{1}{s-1}+O(1), $$ substituting $s=1+\epsilon$ and multiplying by $\epsilon$ gives $$ \epsilon\,\Phi(1+\epsilon)=\epsilon\,\frac{1}{\epsilon}+O(\epsilon), $$ and the desired limit follows.

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Note your $g$ above is not a globally defined function; it depends on the zero $s_0$. And $s=1$ is not a zero of $\zeta(s)$. – stopple Dec 6 '11 at 17:45
I agree that $-\frac{\zeta'}{\zeta}$ has a simple pole of residue $1$ and I agree that the sum converges at $s=1$. So I also agree that $\Phi$ also must have a pole at $s=1$ as it's equal to the quotient of zeta functions. What I don't see is how these facts justify concluding $\Phi(s) = \frac{1}{s-1} + O(1)$: I get $\Phi(s) = \frac{1}{s-1} + ($stuff leftover after removing the term with the pole from $-\frac{\zeta'(s)}{\zeta(s)} +$ value of sum at $s)$, and I don't see why the second two terms yields $O(1)$. – tomcuchta Dec 6 '11 at 18:36
I think I got it. Since $-\frac{\zeta'(s)}{\zeta(s)}$ has a simple pole of residue $1$ at $s=1$, we can write $-\frac{\zeta'(s)}{\zeta(s)} = F(s) + \frac{1}{s-1}$ where $F(1)$ is a finite value. So you end up getting $$\epsilon \Phi(1+\epsilon) = \epsilon F(1+\epsilon) + 1 + \epsilon \displaystyle\sum_p \frac{\log p}{p^{1+\epsilon}(p^{1+\epsilon}-1)}$$ Which approaches $1$ as $\epsilon \rightarrow 0$ since $F(1+\epsilon) \rightarrow F(1)$ and the sum approaches some finite value. – tomcuchta Dec 6 '11 at 19:38
And to answer my question from two posts ago, you wrote $O(1)$: the statement $f=O(g)$ means $|f| \leq Cg$ for some constant $C$. Here you put $O(1)$ because $F(1) + \displaystyle\sum_p \frac{\log p}{p(p-1)}$ is a finite value, say $x_0$. So you would get $|F(1) + \displaystyle\sum_p \frac{\log p}{p(p-1)}| \leq x_0 \cdot 1$, which by definition means $F(1) + \displaystyle\sum_p \frac{\log p}{p(p-1)} = O(1)$. – tomcuchta Dec 6 '11 at 19:45
By $+O(1)$ I just mean all the terms in the Laurent series expansion of the form $(s-1)^n$, $n\ge 0$. This defines an analytic function in some neighborhood of $s=1$, hence bounded in that neighborhood. – stopple Dec 6 '11 at 19:52

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