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I'm following this paper: http://mathdl.maa.org/images/upload_library/22/Chauvenet/Zagier.pdf

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.

1 Answers 1

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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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    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.2011-12-06