Problem:


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 ?


Context: <http://mathdl.maa.org/images/upload_library/22/Chauvenet/Zagier.pdf>


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.



Given: $\phi(s) = \sum_p \frac{\log p}{p^s}$ Why is: $\lim_{e \rightarrow 0} e \phi(1+e)$ = 1 ?
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.


