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Given: $\phi(s) = \sum_p \frac{\log p}{p^s}$

(summation is only over primes)

Why is: $\lim_{e \rightarrow 0} e \phi(1+e)$ = 1 ?


Section IV, Page 706

Some silly analysis by me:

So here's my dumb approach, which fails:

If we want $\lim_{e \rightarrow 0} e f(e)$ =1, a nice way would be $f(e) = 1/e$.

A way towards that would be:

$\int_1^\infty e^{-st} dt = 1/s$.

This is where I get stuck, since I want to do:

$\sum_p \frac{\log p}{p^s} \leq \int_{x=1}^\infty \frac{\log x}{x^s} dx$ ... but not wure where to go from here.

EDIT: Resolved.

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up vote 4 down vote accepted

I think you're approaching the question in the wrong way. The whole point is that you can show that for $\Re(s) > 1$, $$\Phi(s) = \sum_{p}{\frac{\log p}{p^s}} = \frac{1}{s - 1} + E(s),$$ where the function $E(s)$ is meromorphic on the open half-plane $\Re(s) > 1/2$ with poles possibly at the zeroes of $\zeta(s)$; in fact, Zagier shows that $$E(s) = - \frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s - 1} - \sum_{p}{\frac{\log p}{p^s (p^s - 1)}}.$$

Basically, this is saying that $\Phi(s)$ is meromorphic on an open neighbourhood of $\Re(s) \geq 1$ with a simple pole at $s = 1$, and the expansion above shows that the residue of $\Phi(s)$ at $s = 1$ is equal to $1$. This is precisely the same as saying that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1.$$ Indeed, we have that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = \lim_{\varepsilon \to 0} \frac{\varepsilon}{1 + \varepsilon - 1} + \lim_{\varepsilon \to 0} \varepsilon E(1 + \varepsilon),$$ and the first limit tends to $1$ (obviously) while the second limit tends to zero (as $E(1 + \varepsilon)$ tends to something finite).

If you don't understand this method at all (i.e. all about meromorphic extensions of functions, poles, and residues), then this is probably due to a lack of background in complex analysis. Seeing as this proof of the prime number theorem is all about complex analysis, I'd recommend reading up on all these basics beforehand.

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Where is it shown that $E(s)$ is holomorphic on $R(z) > 1/2$ ? To me, $E(s)$ is meromorphic on $R(z) > 1/2$, with possible poles at $s=1$ and the zeros of $\zeta(s)$. In (II), it is shown that $\zeta(s) - \frac{1}{z-1}$ extends holomorphicly to $R(s) > 0$, but I don't see how that helps $E(s)$ here. – LowerBounds Apr 17 '11 at 2:38
Whoops, sorry. In fact what I'd written first was equivalent to the Riemann Hypothesis. It should all be fixed now. The point is that the singularities of $E(s)$ occur whenever $\zeta'(s)/\zeta(s)$ has a singularity other than at $s = 1$, because this is cancelled out by the $1/(s - 1)$ term. – Peter Humphries Apr 17 '11 at 4:22

(IV) on page 706 shows that $g(s)=\Phi (s)-\frac{1}{s-1}$ is holomorphic for $Re(s)\geq 1$. Then $$\lim_{\epsilon \searrow 0} \epsilon \Phi (1+\epsilon) = \lim_{\epsilon \searrow 0} \epsilon g(1+\epsilon)+1=1$$

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