0
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Apr 17, 2011 at 2:38
  • $\begingroup$ 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. $\endgroup$ Apr 17, 2011 at 4:22
1
$\begingroup$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.