Previous set of notes: Notes 3. Next set of notes: 246C Notes 1.

One of the great classical triumphs of complex analysis was in providing the first complete proof (by Hadamard and de la Vallée Poussin in 1896) of arguably the most important theorem in analytic number theory, the prime number theorem:

Theorem 1 (Prime number theorem)Let denote the number of primes less than a given real number . Then (or in asymptotic notation, as ).

(Actually, it turns out to be slightly more natural to replace the approximation in the prime number theorem by the logarithmic integral , which turns out to be a more precise approximation, but we will not stress this point here.)

The complex-analytic proof of this theorem hinges on the study of a key meromorphic function related to the prime numbers, the Riemann zeta function . Initially, it is only defined on the half-plane :

Definition 2 (Riemann zeta function, preliminary definition)Let be such that . Then we define

Note that the series is locally uniformly convergent in the half-plane , so in particular is holomorphic on this region. In previous notes we have already evaluated some special values of this function:

However, it turns out that the zeroes (and pole) of this function are of far greater importance to analytic number theory, particularly with regards to the study of the prime numbers.The Riemann zeta function has several remarkable properties, some of which we summarise here:

Theorem 3 (Basic properties of the Riemann zeta function)

- (i) (Euler product formula) For any with , we have where the product is absolutely convergent (and locally uniform in ) and is over the prime numbers .
- (ii) (Trivial zero-free region) has no zeroes in the region .
- (iii) (Meromorphic continuation) has a unique meromorphic continuation to the complex plane (which by abuse of notation we also call ), with a simple pole at and no other poles. Furthermore, the Riemann xi function is an entire function of order (after removing all singularities). The function is an entire function of order one after removing the singularity at .
- (iv) (Functional equation) After applying the meromorphic continuation from (iii), we have for all (excluding poles). Equivalently, we have for all . (The equivalence between the (5) and (6) is a routine consequence of the Euler reflection formula and the Legendre duplication formula, see Exercises 26 and 31 of Notes 1.)

*Proof:* We just prove (i) and (ii) for now, leaving (iii) and (iv) for later sections.

The claim (i) is an encoding of the fundamental theorem of arithmetic, which asserts that every natural number is uniquely representable as a product over primes, where the are natural numbers, all but finitely many of which are zero. Writing this representation as , we see that

whenever , , and consists of all the natural numbers of the form for some . Sending and to infinity, we conclude from monotone convergence and the geometric series formula that whenever is real, and then from dominated convergence we see that the same formula holds for complex with as well. Local uniform convergence then follows from the product form of the Weierstrass -test (Exercise 19 of Notes 1).The claim (ii) is immediate from (i) since the Euler product is absolutely convergent and all terms are non-zero.

We remark that by sending to in Theorem 3(i) we conclude that

and from the divergence of the harmonic series we then conclude Euler’s theorem . This can be viewed as a weak version of the prime number theorem, and already illustrates the potential applicability of the Riemann zeta function to control the distribution of the prime numbers.The meromorphic continuation (iii) of the zeta function is initially surprising, but can be interpreted either as a manifestation of the extremely regular spacing of the natural numbers occurring in the sum (1), or as a consequence of various integral representations of (or slight modifications thereof). We will focus in this set of notes on a particular representation of as essentially the Mellin transform of the theta function that briefly appeared in previous notes, and the functional equation (iv) can then be viewed as a consequence of the modularity of that theta function. This in turn was established using the Poisson summation formula, so one can view the functional equation as ultimately being a manifestation of Poisson summation. (For a direct proof of the functional equation via Poisson summation, see these notes.)

Henceforth we work with the meromorphic continuation of . The functional equation (iv), when combined with special values of such as (2), gives some additional values of outside of its initial domain , most famously

If one*formally*compares this formula with (1), one arrives at the infamous identity although this identity has to be interpreted in a suitable non-classical sense in order for it to be rigorous (see this previous blog post for further discussion).

From Theorem 3 and the non-vanishing nature of , we see that has simple zeroes (known as *trivial zeroes*) at the negative even integers , and all other zeroes (the *non-trivial zeroes*) inside the *critical strip* . (The non-trivial zeroes are conjectured to all be simple, but this is hopelessly far from being proven at present.) As we shall see shortly, these latter zeroes turn out to be closely related to the distribution of the primes. The functional equation tells us that if is a non-trivial zero then so is ; also, we have the identity

*critical line*. We have the following infamous conjecture:

Conjecture 4 (Riemann hypothesis)All the non-trivial zeroes of lie on the critical line .

This conjecture would have many implications in analytic number theory, particularly with regard to the distribution of the primes. Of course, it is far from proven at present, but the partial results we have towards this conjecture are still sufficient to establish results such as the prime number theorem.

Return now to the original region where . To take more advantage of the Euler product formula (3), we take complex logarithms to conclude that

for suitable branches of the complex logarithm, and then on taking derivatives (using for instance the generalised Cauchy integral formula and Fubini’s theorem to justify the interchange of summation and derivative) we see that From the geometric series formula we have and so (by another application of Fubini’s theorem) we have the identity for , where the von Mangoldt function is defined to equal whenever is a power of a prime for some , and otherwise. The contribution of the higher prime powers is negligible in practice, and as a first approximation one can think of the von Mangoldt function as the indicator function of the primes, weighted by the logarithm function.The series and that show up in the above formulae are examples of Dirichlet series, which are a convenient device to transform various sequences of arithmetic interest into holomorphic or meromorphic functions. Here are some more examples:

Exercise 5 (Standard Dirichlet series)Let be a complex number with .

- (i) Show that .
- (ii) Show that , where is the divisor function of (the number of divisors of ).
- (iii) Show that , where is the Möbius function, defined to equal when is the product of distinct primes for some , and otherwise.
- (iv) Show that , where is the Liouville function, defined to equal when is the product of (not necessarily distinct) primes for some .
- (v) Show that , where is the holomorphic branch of the logarithm that is real for , and with the convention that vanishes for .
- (vi) Use the fundamental theorem of arithmetic to show that the von Mangoldt function is the unique function such that for every positive integer . Use this and (i) to provide an alternate proof of the identity (8). Thus we see that (8) is really just another encoding of the fundamental theorem of arithmetic.

Given the appearance of the von Mangoldt function , it is natural to reformulate the prime number theorem in terms of this function:

Theorem 6 (Prime number theorem, von Mangoldt form)One has (or in asymptotic notation, as ).

Let us see how Theorem 6 implies Theorem 1. Firstly, for any , we can write

The sum is non-zero for only values of , and is of size , thus Since , we conclude from Theorem 6 that as . Next, observe from the fundamental theorem of calculus that Multiplying by and summing over all primes , we conclude that From Theorem 6 we certainly have , thus By splitting the integral into the ranges and we see that the right-hand side is , and Theorem 1 follows.

Exercise 7Show that Theorem 1 conversely implies Theorem 6.

The alternate form (8) of the Euler product identity connects the primes (represented here via proxy by the von Mangoldt function) with the logarithmic derivative of the zeta function, and can be used as a starting point for describing further relationships between and the primes. Most famously, we shall see later in these notes that it leads to the remarkably precise Riemann-von Mangoldt explicit formula:

Theorem 8 (Riemann-von Mangoldt explicit formula)For any non-integer , we have where ranges over the non-trivial zeroes of with imaginary part in . Furthermore, the convergence of the limit is locally uniform in .

Actually, it turns out that this formula is in some sense *too* precise; in applications it is often more convenient to work with smoothed variants of this formula in which the sum on the left-hand side is smoothed out, but the contribution of zeroes with large imaginary part is damped; see Exercise 22. Nevertheless, this formula clearly illustrates how the non-trivial zeroes of the zeta function influence the primes. Indeed, if one formally differentiates the above formula in , one is led to the (quite nonrigorous) approximation

Comparing Theorem 8 with Theorem 6, it is natural to suspect that the key step in the proof of the latter is to establish the following slight but important extension of Theorem 3(ii), which can be viewed as a very small step towards the Riemann hypothesis:

Theorem 9 (Slight enlargement of zero-free region)There are no zeroes of on the line .

It is not quite immediate to see how Theorem 6 follows from Theorem 8 and Theorem 9, but we will demonstrate it below the fold.

Although Theorem 9 only seems like a slight improvement of Theorem 3(ii), proving it is surprisingly non-trivial. The basic idea is the following: if there was a zero at , then there would also be a different zero at (note cannot vanish due to the pole at ), and then the approximation (9) becomes

But the expression can be negative for large regions of the variable , whereas is always non-negative. This conflict eventually leads to a contradiction, but it is not immediately obvious how to make this argument rigorous. We will present here the classical approach to doing so using a trigonometric identity of Mertens.In fact, Theorem 9 is basically equivalent to the prime number theorem:

Exercise 10For the purposes of this exercise, assume Theorem 6, but do not assume Theorem 9. For any non-zero real , show that as , where denotes a quantity that goes to zero as after being multiplied by . Use this to derive Theorem 9.

This equivalence can help explain why the prime number theorem is remarkably non-trivial to prove, and why the Riemann zeta function has to be either explicitly or implicitly involved in the proof.

This post is only intended as the briefest of introduction to complex-analytic methods in analytic number theory; also, we have not chosen the shortest route to the prime number theorem, electing instead to travel in directions that particularly showcase the complex-analytic results introduced in this course. For some further discussion see this previous set of lecture notes, particularly Notes 2 and Supplement 3 (with much of the material in this post drawn from the latter).

** — 1. Meromorphic continuation and functional equation — **

We now focus on understanding the meromorphic continuation of , as well as the functional equation that that continuation satisfies. The arguments here date back to Riemann’s original paper on the zeta function. The general strategy is to relate the zeta function for to some sort of integral involving the parameter , which is manipulated in such a way that the integral makes sense for values of outside of the halfplane , and can thus be used to define the zeta function meromorphically in such a region. Often the Gamma function is involved in the relationship between the zeta function and integral. There are many such ways to connect to an integral; we present some of the more classical ones here.

One way to motivate the meromorphic continuation is to look at the continuous analogue

of (1). This clearly extends meromorphically to the whole complex plane. So one now just has to understand the analytic continuation properties of the residual For instance, using the Riemann sum type quadrature one can write this residual as since , it is a routine application of the Fubini and Morera theorems to establish analytic continuation of the residual to the half-plane , thus giving a meromorphic extension of to the region . Among other things, this shows that (the meromorphic continuation of) has a simple pole at with residue .

Exercise 11Using the trapezoid rule, show that for any in the region with , there exists a unique complex number for which one has the asymptotic for any natural number , where . Use this to extend the Riemann zeta function meromorphically to the region . Conclude in particular that and .

Exercise 12Obtain the refinement to the trapezoid rule when are integers and is continuously three times differentiable. Then show that for any in the region with , there exists a unique complex number for which one has the asymptotic for any natural number , where . Use this to extend the Riemann zeta function meromorphically to the region . Conclude in particular that .

One can keep going in this fashion using the Euler-Maclaurin formula (see this previous blog post) to extend the range of meromorphic continuation to the rest of the complex plane. However, we will now proceed in a different fashion, using the theta function

that made an appearance in previous notes, and try to transform this function into the zeta function. We will only need this function for imaginary values of the argument in the upper half-plane (so ); from Exercise 7 of Notes 2 we have the modularity relation In particular, since decays exponentially to as , blows up like as .We will attempt to apply the Mellin transform (Exercise 11 from Notes 2) to this function; formally, we have

There is however a problem: as goes to infinity, converges to one, and the integral here is unlikely to be convergent. So we will compute the Mellin transform of : The function decays exponentially as , and blows up like as , so this integral will be absolutely integrable when . Since we can write By the Fubini–Tonelli theorem, the integrand here is absolutely integrable, and hence From the Bernoulli definition of the Gamma function (Exercise 29(ii) of Notes 1) and a change of variables we have and hence by (1) we obtain the identity whenever . Replacing by , we can rearrange this as a formula for the function (4), namely whenever .Now we exploit the modular identity (12) to improve the convergence of this formula. The convergence of is much better near than near , so we use (13) to split

and then transform the first integral using the change of variables to obtain Using (12) we can write this as Direct computation shows that and thus whenever . However, the integrand here is holomorphic in and exponentially decaying in , so from the Fubini and Morera theorems we easily see that the right-hand side is an entire function of ; also from inspection we see that it is symmetric with respect to the symmetry . Thus we can define as an entire function, and hence as a meromorphic function, and one verifies the functional equation (6).It remains to establish that is of order . From (11) we have so from the triangle inequality

From the Stirling approximation (Exercise 30(v) from Notes 1) we conclude that for (say), and hence is of order at most as required. (One can show that has order exactly one by inspecting what happens to as , using that in this regime.) This completes the proof of Theorem 3.

Exercise 13 (Alternate derivation of meromorphic continuation and functional equation)

- (i) Establish the identity whenever .
- (ii) Establish the identity whenever , is not an integer, , where is the branch of the logarithm with real part in , and is the contour consisting of the line segment , the semicircle , and the line segment .
- (iii) Use (ii) to meromorphically continue to the entire complex plane .
- (iv) By shifting the contour to the contour for a large natural number and applying the residue theorem, show that again using the branch of the logarithm to define .
- (v) Establish the functional equation (5).

Exercise 14Use the formula from Exercise 12, together with the functional equation, to give yet another proof of the identity .

Exercise 15 (Relation between zeta function and Bernoulli numbers)

- (i) For any complex number with , use the Poisson summation formula (Proposition 3(v) from Notes 2) to establish the identity
- (ii) For as above and sufficiently small, show that Conclude that for any natural number , where the Bernoulli numbers are defined through the Taylor expansion Thus for instance , , and so forth.
- (iii) Show that for any odd natural number . (This identity can also be deduced from the Euler-Maclaurin formula, which generalises the approach in Exercise 12; see this previous post.)
- (iv) Use (14) and the residue theorem (now working inside the contour , rather than outside) to give an alternate proof of (15).

Exercise 16 (Convexity bounds)It is possible to improve the bounds (iii) in the region ; such improvements are known as

- (i) Establish the bounds for any and with .
- (ii) Establish the bounds for any and with . (Hint: use the functional equation.)
- (iii) Establish the bounds for any and with . (Hint: use the Phragmén-Lindelöf principle, Exercise 19 from Notes 2, after dealing somehow with the pole at .)
subconvexity estimates. For instance, it is currently known that for any and , a result of Bourgain; the Lindelöf hypothesis asserts that this bound in fact holds for all , although this remains unproven (it is however a consequence of the Riemann hypothesis).

Exercise 17 (Riemann-von Mangoldt formula)Show that for any , the number of zeroes of in the rectangle is equal to . (Hint:apply the argument principle to evaluated at a rectangle for some that is chosen so that the horizontal edges of the rectangle do not come too close to any of the zeroes (cf. the selection of the radii in the proof of the Hadamard factorisation theorem in Notes 1), and use the functional equation and Stirling’s formula to control the asymptotics for the horizontal edges.)We remark that the error term , due to von Mangoldt in 1905, has not been significantly improved despite over a century of effort. Even assuming the Riemann hypothesis, the error has only been reduced very slightly to (a result of Littlewood from 1924).

Remark 18Thanks to the functional equation and Rouche’s theorem, it is possible to numerically verify the Riemann hypothesis in any finite portion of the critical strip, so long as the zeroes in that strip are all simple. Indeed, if there was a zero off of the critical line , then an application of the argument principle (and Rouche’s theorem) in some small contour around but avoiding the critical line would be capable of numerically determining that there was a zero off of the line. Similarly, for each simple zero on the critical line, applying the argument principle for some small contour around that zero and symmetric around the critical line would numerically verify that there was exactly one zero within that contour, which by the functional equation would then have to lie exactly on that line. (In practice, more efficient methods are used to numerically verify the Riemann hypothesis over large finite portions of the strip, but we will not detail them here.)

** — 2. The explicit formula — **

We now prove Riemann-von Mangoldt explicit formula. Since is a non-trivial entire function of order , with zeroes at the non-trivial zeroes of (the trivial zeroes having been cancelled out by the Gamma function), we see from the Hadamard factorisation theorem (in the form of Exercise 35 from Notes 1) that

away from the zeroes of , where ranges over the non-trivial zeroes of (note from Exercise 11 that there is no zero at the origin), and is some constant. From (4) we can calculate while from Exercise 27 of Notes 1 we have and thus (after some rearranging) whereOne can compute the values of explicitly:

Exercise 19By inspecting both sides of (16) as , show that , and hence .

Jensen’s formula tells us that the number of non-trivial zeroes of in a disk is at most for any and . One can obtain a local version:

Exercise 20 (Local bound on zeroes)

- (i) Establish the upper bound whenever and with . (
Hint:use (10). More precise bounds are available with more effort, but will not be needed here.)- (ii) Establish the bounds uniformly in . (
Hint:use the Euler product.)- (iii) Show that for any , the number of non-trivial zeroes with imaginary part in is . (
Hint:use Jensen’s formula and the functional equation.)- (iv) For , , and , with not a zero of , show that (
Hint:use Exercise 9 of Notes 1.)

Meanwhile, from Perron’s formula (Exercise 12 of Notes 2) and (8) we see that for any non-integer , we have

We can compute individual terms here and then conclude the Riemann-von Mangoldt explicit formula:

Exercise 21 (Riemann-von Mangoldt explicit formula)Let and . Establish the following bounds:(

- (i) .
- (ii) .
- (iii) For any positive integer , we have
- (iv) For any non-trivial zero , we have
- (v) We have .
- (vi) We have .
Hint:for (i)-(iii), shift the contour to for an that gets sent to infinity, and using the residue theorem. The same argument works for (iv) except when is really close to , in which case a detour to the contour may be called for. For (vi), use Exercise 20 and partition the zeroes depending on what unit interval falls into.)

- (viii) Using the above estimates, conclude Theorem 8.

The explicit formula in Theorem 8 is completely exact, but turns out to be a little bit inconvenient for applications because it involves all the zeroes , and the series involving them converges very slowly (indeed the convergence is not even absolute). In practice it is preferable to work with a smoothed version of the formula. Here is one such smoothing:

Exercise 22 (Smoothed explicit formula)

- (i) Let be a smooth function compactly supported on . Show that is entire and obeys the bound (say) for some , all , and all .
- (ii) With as in (i), establish the identity with the summations being absolutely convergent by applying the Fourier inversion formula to , shifting the contour to frequencies for some , applying (8), and then shifting the contour again (using Exercise 20 and (i) to justify the contour shifting).
- (iii) Show that whenever is a smooth function, compactly supported in , with the summation being absolutely convergent.
- (iv) Explain why (iii) is
formallyconsistent with Theorem 8 when applied to the non-smooth function .

** — 3. Extending the zero free region, and the prime number theorem — **

We now show how Theorem 9 implies Theorem 6. Let be parameters to be chosen later. We will apply Exercise 22 to a function which equals one on , is supported on , and obeys the derivative estimates

for all and , and for all and . Such a function can be constructed by gluing together various rescaled versions of (antiderivatives of) standard bump functions. For such a function, we have On the other hand, we have and and hence We split into the two cases and , where is a parameter to be chosen later. For , there are only zeros, and all of them have real part strictly less than by Theorem 9. Hence there exists such that for all such zeroes. For each such zero, we have from the triangle inequality and so the total contribution of these zeroes to (17) is . For each zero with , we integrate parts twice to get some decay in : and from the triangle inequality and the fact that we conclude Since is convergent (this follows from Exercise 20 we conclude (for large enough depending on ) that the total contribution here is . Thus, after choosing suitably, we obtain the bound and thus whenever is sufficiently large depending on (since depends only on , which depends only on ). A similar argument (replacing by in the construction of ) gives the matching lower bound whenever is sufficiently large depending on . Sending , we obtain Theorem 6.

Exercise 23Assuming the Riemann hypothesis, show that for any and , and that for any and . Conversely, show that either of these two estimates are equivalent to the Riemann hypothesis. (Hint:find a holomorphic continuation of to the region in a manner similar to how was first holomorphically continued to the region ).

It remains to prove Theorem 9. The claim is clear for thanks to the simple pole of at , so we may assume . Suppose for contradiction that there was a zero of at , thus

for sufficiently close to . Taking logarithms, we see in particular that Using Lemma 5(v), we conclude that Note that the summands here are oscillatory due to the cosine term. To manage the oscillation, we use the simple pole at that gives for sufficiently close to one, and on taking logarithms as before we get These two estimates come close to being contradictory, but not quite (because we could have close to for most numbers that are weighted by . To get the contradiction, we use the analytic continuation of to to conclude that and hence Now we take advantage of the*Mertens inequality*(which is a quantitative variant of the observation that if is close to then has to be close to ) as well as the non-negative nature of to conclude that and hence This leads to the desired contradiction by sending , and proves the prime number theorem.

Exercise 24Establish the inequality for any and .

Remark 25There are a number of ways to improve Theorem 9 that move a little closer in the direction of the Riemann hypothesis. Firstly, there are a number ofzero-free regionsfor the Riemann zeta function known that give lower bounds for (and in particular preclude the existence of zeros) a small amount inside the critical strip, and can be used to improve the error term in the prime number theorem; for instance, theclassical zero-free regionshows that there are no zeroes in the region for some sufficiently small absolute constant , and lets one improve the error term in Theorem 6 to (with a corresponding improvement in Theorem 1, provided that one replaces with the logarithmic integral ). A further improvement in the zero free region and in the prime number theorem error term was subsequently given by Vinogradov. We also mention a number of importantzero density estimateswhich provide non-trivial upper bounds for the number of zeroes in other, somewhat larger regions of the critical strip; the bounds are not strong enough to completely exclude zeroes as is the case with zero-free regions, but can at least limit the collective influence of such zeroes. For more discussion of these topics, see the various lecture notes to this previous course.

## 48 comments

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12 February, 2021 at 9:53 am

pcaceres00I would appreciate if someone could comment on the attached paper. Fascinating subject.

Many thanks in advance

Pedro Caceres

(763) 412-8915

12 February, 2021 at 9:56 am

pcaceres00It would help if I add the link!

https://www.researchgate.net/publication/340296254_An_Engineer's_Approach_to_the_Riemann_Hypothesis_and_why_it_is_true

12 February, 2021 at 10:16 am

Raphael“In this paper we will provide a proof of the RH.” Possibly one or more such papers are published every day. Here to see quickly if possibly could make sense to have a look on one of them https://www.scottaaronson.com/blog/?p=304 .

12 February, 2021 at 11:12 am

AnonymousIn the second line, “la” is missing from the name de la Vallee Poussin

12 February, 2021 at 11:16 am

AnonymousIn (2), 645 should be 945

12 February, 2021 at 12:21 pm

AnonymousIn theorem 3, it may be added that Euler product (3) is locally uniformly convergent (which may be used later to justify the term-wise logaritmic derivative to obtain (8) – without using Fubini theorem)

12 February, 2021 at 12:47 pm

Rex“…from the divergence of the harmonic series we then conclude Eulerās theorem . This can be viewed as a weak version of the prime number theorem,…”

I believe I heard somewhere the heuristic that a subset of the natural numbers has divergent reciprocal sum if and only if, roughly speaking, its density declines at rate or slower. Is this true?

[Yes; this follows fairly readily from the Cauchy condensation test. For instance, density decaying faster than will ensure convergence, whereas density that stays above for all sufficiently large will guarantee divergence. -T]12 February, 2021 at 12:59 pm

RexSo then the divergence of prime reciprocals already gives us yet another simple proof of “the prime number theorem up to a constant”?

[Not quite; the density could be wildly fluctuating to be significantly denser than at some scales and significantly sparser at other scales. -T]12 February, 2021 at 6:39 pm

Lior SilbermanNot quite: the Cauchy condensation test applies to monotone series, and here you don’t know that the density of primes is monotone. Maybe there are long stretches with no primes (so the density falls low), follows by long sequences with many primes (so the density becomes large).

12 February, 2021 at 1:03 pm

Rex“We remark that by sending to in Theorem 3(i) we conclude that

”

It seems here we are implicitly using some Tauberian theorem or other, to assert that the divergence of as implies the divergence of the partial sums of ? And similarly on the product side?

[Monotone convergence suffices here. -T]12 February, 2021 at 3:24 pm

TheoristI think, contrary to the prime number theorem, the equivalent relation that p_n is asymptotic to n ln(n), where p_n is the n-th prime itself, is much less known. By the way, this relation in turn is equivalent to the nice result that the p_n-th root of n^n tends to Euler’s number e if n tends to infinity.

12 February, 2021 at 6:34 pm

AnonymousI have been looking for the original proof of the prime number theorem. Thank you!

12 February, 2021 at 6:43 pm

Lior SilbermanThe original proof is due to Riemann (1859) (conditioned on the Riemann hypothesis). The essential ideas are there except for the proof that there are no zeroes on the line , and for the more delicate handling this required compared with the case where the zeroes are well off that line.

12 February, 2021 at 8:31 pm

Aditya Guha RoyLior I doubt that. I think the first proof was given by Hadamard and Poussin independently around the same time. Their proofs were heavily based on analysing a result due to Chebyshev on Chebyshev functions, and both used complex analytic methods quite heavily.

Once this proof was found, people wanted to somehow eliminate the complex analytic ingredients in quest to obtain an “elementary” proof of the result. There are also reports of several debates among mathematicians during that time on whether any such non complex analytic proof can be found or not. It is also believed that Hardy carried the impression that the prime number theorem draws so heavily on analysis of the zeta function, that eliminating the complex analytic ingredients from any proof of the result is almost impossible.

Much later the first and also the only such “elementary” proof was found independently by Paul Erdos and Atle Selberg, both using an identity due to Selberg.

Even later a very short proof was found by Newman. It is perhaps the shortest known proof of the prime number theorem and uses only very basic level complex analysis, such as those of analytic continuation and evaluation of contour integrals.

Towards the end of 2020 I read a nice article on the timeline of the prime number theorem, with some reports on debates among mathematicians on non-complex analytic proofs of the result, as well as the issue of rewarding the Fields Medal to Atle Selberg and not Erdos for the elementary proof, but I cannot find it unfortunately (:( ). If someone find it or if I find it, do comment below.

13 February, 2021 at 12:23 am

UlrichThis one:

13 February, 2021 at 6:02 am

AnonymousBoth Selberg and Erdos contributed to the proof (Selberg initially thought that his important estimate was not sufficient to prove the PNT, but Erdos showed him how to prove the PNT from Selberg’s estimate)

Unfortunately, Selberg chose to publish a separate proof (based on another derivation of the PNT from his basic estimate) – thereby forcing Erdos to publish separately his original derivation of the PNT from Selberg’s basic estimate.

13 February, 2021 at 9:47 am

Lior SilbermanAditya: as I explained above, Riemann indeed proved the Prime Number Theorem, conditionally on the Riemann Hypothesis. Conditional proofs have a long history in mathematics — most famously Serre and Ribet proving Fermat’s Last Theorem conditionally on the Tanyiama–Shimura Conjecture, which was later substantially proved by Wiles.

Riemann’s memoir contains most of the key ideas of the proof. De la ValĆ©e Poussin and Hadamard completed the work by giving the first

unconditionalproofs, but Riemann’s contribution is still the greater.Mathematics is a race where often winners are declared by coming

last: whoever makes the final contribution gets the glory. Sometimes it’s right (the hardest part is often left for last, after all), but sometimes it’s wrong: until the earlier work there was no room for the later. Before Riemann there was no avenue to prove the PNT at all. After Riemann there was a conditional proof and a clear direction.For the same reason I credit FLT to Ribet, and the Modularity Theorem to Wiles. Here Wiles’s contribution is greater (the proof of the FLT is mainly of historical importance, while modularity is an essential part of mathematics), but still the FLT was proved by Ribet, not Wiles.

13 February, 2021 at 11:51 pm

Aditya Guha RoyAh, I see your point of view, Lior. If you like to call / view it that way then it’s fair enough.

By “original” I meant the first complete proof which doesn’t assume anything which is not known till date.

By the way just some nitpicking : given that the truth of Riemann hypothesis is not known till date, would you still say that Riemann “proved” the prime number theorem?

(Of course now that we know the PNT is true so ex falso quodlibet plays the game.)

13 February, 2021 at 9:58 am

RexI’ve wondered whether Riemann was the very first person to introduce complex methods into number theory. Do you know if Dirichlet’s proof of arithmetic progressions used it in any essential way?

13 February, 2021 at 7:18 pm

Lior SilbermanIndeed Riemann completely revolutionized the field, which is why I credit him with a proof of the PNT. He both had a rigorous conditional proof (subject to RH) and charted the path for the first unconditional proofs.

Dirichletās proof of his theorem is based on Eulerās proof of the infinitude of primes (discussed above after Theorem 3), and lives entirely on the real axis. He originally only used the pole to show that there were infinitely many primes in each admissible arithmetic progression, and later observed that the proof is actually quantitative: generalizing Mertensās Theorem

Dirichlet shows (purely by real-variable methods, even if some later proofs use complex methods)

13 February, 2021 at 11:55 pm

Aditya Guha RoyRex, if you go through Dirichlet’s proof of Dirichlet’s theorem, you will find that his proof did involve complex analytic ingredients in it in analysing the L functions which were crucially used in proving the result.

14 February, 2021 at 11:35 am

Lior SilbermanAditya: Dirichlet’s proof (I link to an English translation) did not use complex analysis at all.

Today(with the benefit of hindsight) we can see that his arguments prove that his L-series have analytical continuation to the half-plane . But if you read the paper carefully you’ll see that the parameter only takesrealvalues. The values are complex (because the Dirichlet characters take complex values), but that does not make theproofinvolve complex analysis. In fact Dirichlet is careful to argue about real and complex parts separately.The paper is interesting also for Dirichlet’s calling attention to the distinction between absolutely convergent and conditionally convergent series, and showing he was aware that conditionally convergent series could be rearranged to sum to different values (what we call today “Riemann’s Rearrangement Theorem”).

14 February, 2021 at 11:36 am

Lior SilbermanHere’s the link to the proof since it didn’t work in the comment above (a double quotes symbol is off in the HTML)

https://arxiv.org/abs/0808.1408

14 February, 2021 at 8:30 pm

Aditya Guha RoyAh, thanks Lior. I didn’t know of this. I read the proof from Davenport’s text on MNT, and I developed the impression that the proof discussed in the text is what Dirichlet actually did himself. Thank you for the link.

13 February, 2021 at 3:33 am

Aditya Guha RoyReblogged this on Aditya Guha Roy's weblog.

13 February, 2021 at 6:25 am

Rex“By splitting the integral into the ranges and we see that the right-hand side is , and Theorem 1 follows."

I've seen this strategy of decomposing an integral or sum into ranges above and below in several places, to bound the pieces separately. But I've never really understood the basic principle behind why it works. Why is a good cutoff to use?

14 February, 2021 at 10:24 am

Terence TaoThe most difficult term to understand in the integral is the term, which varies in . But within the range , is constant (up to bounded multiplicative error), so this term can be simplified. In general, when trying to estimate an integral, it can often be beneficial to break up into pieces where one or more of the terms in the integral is approximately constant, as one can then often replace that term by the constant and get an easier expression to estimate which is still equivalent to the original expression up to constants. One could have replaced here by other small powers of , but this is as good a choice as any here as the contribution of the very small ends up being negligible and can be controlled satisfactorily by virtually any estimate.

13 February, 2021 at 8:38 am

Rex“From (6) we certainly have , thus

”

It seems the reference (6) wants to refer to the (unnumbered) formula five lines up, but it’s not actually directing to this.

[Corrected, thanks – T.]14 February, 2021 at 1:27 pm

Rex“or (writing )

”

I think here you want the bottom exponent to be ?

[Corrected, thanks – T.]14 February, 2021 at 2:44 pm

Avi LevyJust as the “ as Mellin transform of ” point of view explains how the Poisson summation formula is the source of the critical line reflection symmetry, it brings up a (vague and ill-posed) question about the nature of the square root fluctuations in the explicit formula.

On the one hand, the exponent is a fixed point of this symmetry. On the other hand, there are “randomness of the primes” heuristics where the comes from Brownian fluctuations, which traces its source to linearity of variance for sums of independent random variables.

Are these two explanations of the square root fluctuations somehow connected? Or is it akin to phenomena in statistical physics models where the self-dual point happens to coincide with the critical point, as in https://arxiv.org/abs/1006.5073 for example?

14 February, 2021 at 3:03 pm

AnonymousIt is related to the idea of “square root cancellation”

18 February, 2021 at 3:20 pm

Terence TaoHmm, interesting question. Certainly probabilistic methods can be used without any appeal to the functional equation to give some statistical control on say the Riemann zeta function in the region , for instance one can show without much difficulty (basically the mean value theorem) that almost all of the zeroes avoid the region for any fixed . Here the “1/2” is the same 1/2 that comes from Brownian motion, or more generally in applications of the second moment method. This is only consistent with a functional equation if the critical line of that functional equation is at or to the left of . On the other hand I don’t see a probabilistic argument that explains why the critical line has to be exactly at ; this would involve some way to understand zeta to the left of the critical line that does not involve the functional equation. Still, this sort of coincidence between the abscissa coming from second moment considerations and the abscissa coming from the functional equation does seem to be pretty universal amongst all the other L-functions in number theory so presumably there is a more robust explanation of this phenomenon.

14 February, 2021 at 2:55 pm

AnonymousIn the Wikipedia article on the zeta function it is remarked that the functional equation (5) was discovered and conjectured by Euler (for integer values of ) in 1749.

15 February, 2021 at 12:45 am

justderivingWhat would happen to the field of math if the Riemann hypothesis were proven false or true?

15 February, 2021 at 12:59 pm

Rex“from Exercise 11 that there is no zero at the origin), and is some constant. From (4) we can calculate

”

This formula is missing on the right.

[Corrected, thanks – T.]18 February, 2021 at 4:13 pm

RexIs it just me, or is it still missing?

[Corrected, thanks – T.]23 February, 2021 at 5:34 am

RexAlso, there is now an extraneous in the formula immediately above this one. Probably from the first attempt to fix it.

[Corrected, thanks – T.]15 February, 2021 at 9:15 pm

Joseph SugarRiemann states in his paper that he is interested in the prime numbers’ density distribution, not in finding more precise prime number theorems. His formula for x = 2000.000 is exact, so finding better and better approximations was probably already not very exciting.

15 February, 2021 at 11:41 pm

anthonyquasIn the proof of Theorem 3(i), you have used monotone convergence, but that’s only good for $s$ real and positive, right?

16 February, 2021 at 3:03 pm

Lior SilbermanYes. The proof above uses monotone convergence for real and positive (actually for ), and then dominated convergence to go from there to the halfplane .

17 February, 2021 at 10:50 am

AnonymousI believe in Exercise 12, it should be rather than ? (The current version seems to give ).

[Corrected, thanks – T.]18 February, 2021 at 8:57 am

Aditya Guha RoyThought this would be relevant: https://mathoverflow.net/questions/58004/how-does-one-motivate-the-analytic-continuation-of-the-riemann-zeta-function/380502#380502 here prof. Tao discusses a nice way to motivate (/ see naturally) the meromorphic continuation of the zeta function.

18 February, 2021 at 2:21 pm

AnonymousI believe the exponent in equation (12) should be +1/2, not -1/2. Thank you for posting these great notes!

[Corrected, thanks – T.]19 February, 2021 at 6:21 am

AnonymousDear respected Pro.Tao,

Is there any relationship between you and my night dream?

In 2019, I dreamed about numbers but not clear. A month later, you claimed that you proved Collatz conjecture. In 2020, I also dreamed about many numbers but not clear. Strangely, a month later, you announced that you proved Sendov conjecture. But this time, with the third night dream one month ago in 2021, I had seen many numbers, many equations x, y,z , many maths signals very long and densly . I saw them very clearly like at noon, different from two previous dreams. I am happy myself and I wonder whether Pro.Tao is going to claim proving one clay millenium problem successfully. I hope my third dream is right certainly. My guess that they are Riemann hypothesis, Navier Stokes existence and smoothness, Birch Swinerton Dyer conjecture.

Bye Pro Tao.( I always wait good news from you)

27 February, 2021 at 9:14 am

Course announcement: 246B, complex analysis | What's new[…] (if time permits) the Riemann zeta function and the prime number theorem. […]

27 February, 2021 at 9:38 am

246B, Notes 3: Elliptic functions and modular forms | What's new[…] Previous set of notes: Notes 2. Next set of notes: Notes 4. […]

27 February, 2021 at 9:47 am

246C notes 1: Meromorphic functions on Riemann surfaces, and the Riemann-Roch theorem | What's new[…] set of notes: 246B Notes 4. Next set of notes: Notes 2. The fundamental object of study in real differential geometry are the […]

27 February, 2021 at 9:12 pm

Frank VegaGood notions (Thanks…)!!! It’s great to obtain for free just already summarized many things here (into a single page instead of trying to find them from many reference papers). I didn’t see the Lagarias and Robin inequalities here. Maybe, they are in the previous posts related to this one. Proving the RH by proving a simple inequality looks very attractive. However, behind these inequalities, we could find a deep estimates on the distribution of primes. I have moved forward in this tempting direction:

https://doi.org/10.5281/zenodo.3648511

What do you think?