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I came across two remarks that I would appreciate help in making the connections.

I) In Riemann's Explicit Formula: for $x > 1$

$\Pi = Li(x) - \sum_{\rho:\zeta(\rho)=0}Li (x^{\rho})- \log(2) +$ term relating to trivial zeros of $\zeta$

the $Li (x)$ term comes from the pole of $\zeta(s)$ at $s = 1$.

I know there is a pole at $s = 1$ (a.k.a. the harmonic series), but how does one show that the $Li(x)$ comes from this?

II) I understand the proof showing the non-trivial zeros $\rho$ of $\zeta(s)$ satisfy

$0 < Re(\rho)< 1$.

How is this equivalent to the Prime Number Theorem?

(The essence of the questions come from remarks in Stopple's "Primer of Analytic Number Theory.")

I have a feeling that these questions are asking a lot, so I appreciate any help.

$EDIT$: The link provided in the comment below by Raymond Manzoni provides the answer to part I.

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For the first part I hope that my answer here will help. –  Raymond Manzoni Jul 4 '13 at 23:10
    
@RaymondManzoni Thanks very much. +1 at least for such an endeavor. I wrote some hasty comments to you which I deleted because your link deserves the utmost thought for me before I take your time with any further questions. But I am looking forward to really studying it. Regards, –  Andrew Jul 4 '13 at 23:38
    
I am glad it interested you @Andrew. Of course I'll try to answer any question. Regards, –  Raymond Manzoni Jul 5 '13 at 8:41
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To prepare for Edwards' book you'll have... to get better at complex analysis of course ! :-) I can't really propose english introduction books (I'm French) but other recommendations may be helpful. An excellent and old book is Whittaker & Watson's 'A course of modern analysis' with many powerful things about complex analysis and a chapter about the gamma function before the zeta function itself (and other useful special functions like theta or li). –  Raymond Manzoni Jul 6 '13 at 0:13
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Concerning the question II I'll have to look at the details of all this in ch 2 of Hardy's book about Ramanujan and/or Landau's thick book 'Primzahlen' (Hardy was convinced that zeta was needed to get the PNT using the equivalence PNT $\leftrightarrow$ no zero of real part $1$) but not now (it's late and from what I saw the answer won't be quick...). Good evening, –  Raymond Manzoni Jul 6 '13 at 0:19

2 Answers 2

up vote 3 down vote accepted
+50

Concerning question I the explicit formulas link (I updated the $(2)-(4)$ part for clarification) should help to find the origin of the $\operatorname{li}(x^{\rho})$ for the Riemann explicit formula starting from the Euler product formula ($\zeta$ as function of the primes) up to the explicit formula for $\pi(x)$ (the primes as function of the $\zeta$ zeros).

A detailed exposition of von Mangoldt's proof for this formula is in pages $62$ to $65$ of Edwards' hard to replace book 'Riemann's Zeta Function'.


(Notation: in ANT it is usual to note '$\sigma$' the real part of the complex number $\,s:=\sigma+it\,$)

Concerning question II I'll reproduce verbatim Ingham's demonstration ( page $37$) that
$\qquad\qquad$PNT $\implies$ {no zeros on the line $\sigma=1$}

Let's start with the equation $(2.1)$ from the explicit formulas link : $$\tag{1}f(s):=-\frac{\zeta'(s)}{\zeta(s)}=s\int_1^{\infty}\frac{\psi(x)}{x^{s+1}}dx\quad (s>1)$$ "we have, for $\,\sigma>1$, $$\tag{2}\phi(s):=\int_1^{\infty}\frac{\psi(x)-x}{x^{s+1}}dx=-\frac{\zeta'(s)}{s\;\zeta(s)}-\frac 1{s-1}$$ say; $\phi(s)$ is regular in $\sigma>0$ except (possibly) for simple poles at zeros of $\zeta(s)$. Now suppose the PNT true, i.e. $\psi(x)=x+o(x)$. Then, given $\epsilon>0$, we have $\,|\psi(x)-x|<\epsilon x\;$ for $\;x>x_0=x_0(\epsilon)\;(>1)$. Hence, for $\sigma>1$, $$|\phi(s)|<\int_1^{x_0}\frac{|\psi(x)-x|}{x^2}dx+\int_{x_0}^\infty\frac{\epsilon}{x^\sigma}dx<K+\frac{\epsilon}{\sigma-1},$$ where $K=K(x_0)=K(\epsilon)$. Thus $$|(\sigma-1)\phi(\sigma+ti)|<K(\sigma-1)+\epsilon<2\epsilon$$ for $\,1<\sigma<\sigma_0=\sigma_0(\epsilon,K)=\sigma_0(\epsilon)$. Hence, for any fixed $t$, $$\tag{3}(\sigma-1)\,\phi(\sigma+ti)\to 0$$ as $\sigma\to 1+0$. This shows that the point $1+ti\,$ cannot be a zero of $\zeta(s)$, for in that case $(\sigma-1)\phi(\sigma+ti)$ would tend to a limit different from $0$, namely the residue of $\phi(s)$ at the simple pole $1+ti$."


The converse implication : $\qquad${no zeros on the line $\sigma=1$} $\implies$ PNT

is not so direct since the usual proofs require {no zeros on the line $\sigma=1$} but also a theorem on the order of magnitude of $\frac{\zeta'(s)}{\zeta(s)}$ to imply the PNT (c.f. the discussion page $37-39$ of Ingham).

This subsidiary theorem is something like Hardy & Littlewood's $f(\sigma+it)=O(|t|^\alpha)$ with $\alpha<1$, for $\sigma\ge 1$ and large $|t|$.
In fact $\zeta(s)=O(\ln|t|)$ and $\zeta'(s)=O\bigl(\ln^2|t|\bigr)$) were obtained as well as $\frac{\zeta'(s)}{\zeta(s)}=O\bigl(\ln^9|t|\bigr)$ for $\rho>1-A\,\ln^{-9}|t|$ allowing to find a large 'zero-free region' along $\sigma=1$. I think that this additional requirement came from the infinite bounds of : $$\tag{4}\psi^*(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}f(s)\frac{x^s}s\,ds$$ This initial point of view (de la Vallée-Poussin and Hadamard's) is well exposed in Titchmarsh reference book 'The theory of the Riemann Zeta-function' (around page $50$). $$-$$ An important progress was made when Wiener, combining his work about Fourier transforms and Lambert series with Ikehara's Theorem, obtained the Tauberian theorems he exposed in two books : 1932 : 'Tauberian theorems' (a $100$ pages paper accessible after free registration at JStor) and 1933 : 'The Fourier integral and certain of its applications' (ch.$19$ 'Ikehara's Theorem').

The 'Wiener–Ikehara theorem' theorem asserts (Chandrasekharan) :

If $A(t)$ is a non-negative, non-decreasing function of $t$, defined for $t\ge 0$ and if the integral $$\int_0^\infty A(t)\,e^{-ts}\,dt$$ is convergent for $\sigma>1$ to the function $f(s)$ analytic for $\sigma\ge 1$ except for a simple pole at $s=1$ with residue $1\,$ then : $$\lim_{t\to\infty} \ e^{-t}\,A(t)=1$$ (for a proof and many more informations concerning the PNT see Montgomery and Vaughan's book on 'Multiplicative NT' page $259$ or Chandrasekharan's 'Introduction to ANT' p.$124$ or Wiener's work)

After setting $x:=e^t$ in equation $(1)$ we get (for $\sigma>1$) : $$-\frac{\zeta'(s)}{s\;\zeta(s)}=\int_0^{\infty}\psi(e^t)\,e^{-ts}\,dt$$

Let's apply the WIT to the Chebyshev function $\;A(x):=\psi\bigl(e^x\bigr)$ then :

  • $\psi$ is non-decreasing and $\psi\bigl(e^x\bigr)\ge 0$
  • $\zeta(s)$ and $\zeta'(s)$ are analytic for $\sigma>0$ except at $s=1$ where $\,\frac{\zeta'(s)}{s\;\zeta(s)}$ admits a simple pole
  • $\zeta(s)$ does not vanish in the half-plane $\sigma\ge 1$ (this is where the hypothesis $\zeta(s)\not = 0$ for $\sigma=1$ appears since the other cases are well known)

All this implies that $\,\psi\bigl(e^t\bigr)\sim e^t\;$ or $\;\psi(x)\sim x\;$ as $\;x\to \infty$ (i.e. the PNT). Of course all the 'machinery' is in the WIT here ! ( or the reverse ? ;-) )

and we got the wished direct implication : $\qquad${no zeros on the line $\sigma=1$} $\implies$ PNT
adding only continuity in the close half-plane $\sigma\ge1$ of $\;\displaystyle \zeta(s)-\frac 1{s-1}$ and $\;\displaystyle\zeta'(s)+\frac 1{(s-1)^2}$. $$-$$ In $1980$ Newman proposed another proof of the PNT using (in Korevaar's words) a 'poor man' version of Wiener–Ikehara's. His proof required only the analyticity and non-vanishing of $(s-1)\zeta(s)$ on the closed half-plane $\{s:\Re(s)\ge1\}$) (i.e. well known facts except on the $\Re(s)=1$ line).

Newman's theorem may be rewritten with the Laplace integral replacing the Dirichlet series (Korevaar and Zagier) :
Let $A(t)\;$ be a bounded on $(0,\infty)$ and locally integrable function and suppose that the function $$g(s):=\int_0^\infty A(t)\,e^{-ts}\,dt,\quad \Re(s)>0$$ extends holomorphically to $\,\Re(s)\ge 0$ then the limit as $s\to 0$ exists and
$$\int_0^\infty A(t)\,dt=g(0)$$

Newman proposed different proofs of the PNT. The more direct is to use the formula from inversion of Dirichlet series for $\Re(s)>1$ : $\,\displaystyle\frac 1{\zeta(s)}=\sum \frac{\mu(n)}{n^s}$. Since further $(s-1)\zeta(s)$ is analytic and zero free over $\Re(s)\ge 1$ the theorem (in its Dirichlet form) applies and we get convergence to $\,\displaystyle\sum \frac{\mu(n)}{n}=0$ which, according to Landau is equivalent to (for Hardy 'as deep as') the prime number theorem.

For his proof Newman modified a theorem of Ingham (using Fourier analysis) and came back to contour integral with the idea of replacing $(4)$ by a finite $C_R$ contour integral. The Cauchy formula gave directly (concluding with the limit as $R\to\infty\;$ that $\;\lim_{T\to\infty} \{\text{left part}\}=0$) : $$g(0)-g_T(0)=\frac1{2\pi i}\int_{C_R}\bigl(g(s)-g_T(s)\bigr)\,x^s\left(\frac 1s+\frac s{R^2}\right)\,ds,\quad g_T(s):=\int_0^T A(t)\,e^{-ts}\,dt$$ (the complete and very short proof of Zagier should be examined !)

This work is described in a few nice papers :

You may enjoy these last references as well as the history of all this exposed by Bateman and Diamond in 'A hundred years of Prime Numbers'.

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Thanks very much, Raymond. As you can probably guess regarding my comment about complex analysis, it will take a lot of work for me to make the most of these. But I am pleased that this will be motivating. –  Andrew Jul 6 '13 at 21:01
    
@Andrew: once you'll know basic complex analysis you'll have to learn some number theory too (Analytic Number Theory of course! :-)). My favorite ANT book is Apostol : very useful as a reference too to know what a multiplicative, Möbius, Chebyshev and so on function is. Another venerable book is the Hardy & Wright with the missing index. Both books ask very little prerequisites but are rather dense! Fine studies –  Raymond Manzoni Jul 7 '13 at 17:28
    
Thanks for your advice. I have gone through Stopple's amazon.com/Primer-Analytic-Number-Theory-Pythagoras/dp/… which I really liked. That's how I got interested in this. –  Andrew Jul 7 '13 at 18:01
    
We should have a special mode for answers like yours, where the upvote button is at the bottom of the post...+1 –  draks ... Nov 20 '13 at 22:27
    
Thanks @draks and sorry for the extra work ! :-) –  Raymond Manzoni Nov 20 '13 at 22:43

To see why the Prime Number theorem is implied, by definition,

$$\Pi(z)=\sum_{r=1}^\infty\frac{1}{r}\pi(z^{1/r}).$$

Instead of $\Pi(x)$, consider Von Mangoldt's formula, using Chebyshev's prime counting function:

$$\psi(z):=\sum_{p \ \mbox{prime}, \ p^k\leq z}\ln p=\sum_{n\leq z}\Lambda(n),$$

where the sum is over all prime powers not exceeding $z$.

Then the $\Pi$ formula above is equivalent to:

$$\psi(z)=z-\ln(2\pi)-\frac{1}{2}\ln(1-z^{-2})-\sum_{\rho}\frac{z^\rho}{\rho}$$

where $\rho$ are the nontrivial zeros of the zeta function. The prime number theorem is equivalent to showing that $\psi(z)\approx z$. Write $\rho=a+i\sigma$. So $|z^\rho|=|z|^a$. Now try to show that if $0<a<1$, then the sum is of lower order than $z$. Making this argument rigorous is nontrivial but I believe it is in spirit with de la Valee Poussin's original proof of the prime number theorem.

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Hi Alex. (A minor typo in the 1/2 ln term - x should be z.) Thanks for your reply. I understand what you are saying. I would like to ponder: I thought "equivalent" meant that with only the zeta function and the knowledge that its non-trivial zeros lie entirely within the critical strip one can derive the PNT. What I think you are suggesting is that the $\Sigma$ term is not an impediment. Also, as you anticipated, I am trying to show that the sum is of lower order, clearly each term is of lower order, but am stuck on the sum (maybe need to change the sum to an integral?) –  Andrew Jul 4 '13 at 16:45
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Subsequently, I found a great deal of helpful material in Edwards's "Riemann's Zeta Function" which discusses the difficulty of showing the sum is of lower order. And it also gives a nice intuitive discussion of the conception of Von Mangoldt's formula. –  Andrew Jul 4 '13 at 20:01
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Actually, although this is a good heuristic, it is not easily made into a proof. The problem is that a sum of functions with a certain asymptotic decay need not have the same decay. For example, $\sum x^n\,e^{-x}/n!=1$ despite the summands' being of exponential decay, and, then $\sum x^{2n}\,e^{-x}/n!=e^x$... That is, possibly contrary to general understanding, as far as I know Riemann's explicit formula (with non-vanishing on $1+i\mathbb R$) does not give any Prime Number Theorem. Nevertheless, again, this is a great heuristic. (Thus, I fear that @Andrew's misgivings are justified.) –  paul garrett Jul 4 '13 at 23:26
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Since asking this question, I happily have learned a bit more and can see the value of this answer. Dividing both side of your $\psi$ relation by $z$ you get $\psi/z$ goes to 1 if the remaining terms on the RHS go to zero as x goes to infinity. Thus the only element of concern is the last summation. If you can take the limit of each term separately then $x^{Re(\rho) - 1}$ will do the trick with $Re(\rho)<1$. This is from Edward's excellent "Riemann's Zeta Function." Thanks and regards, –  Andrew Aug 27 '13 at 1:11

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