# On the sign of the real part of the Riemann zeta function

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On the sign of the real part of theRiemann zeta function

Richard P. BrentANU

Joint work with

Juan Arias de Reyna (University of Seville)Jan van de Lune (Hallum, The Netherlands)

19 March 2012

In memory of Alf van der Poorten

Arias de Reyna, Brent and van de Lune Sign of

Abstract

The real part of the Riemann zeta function (s) is usuallypositive in the half-plane to the right of the critical line

Alf van der Poorten

Like almost all number theorists, Alf van der Poorten wasinterested in the Riemann zeta function. For example, Alf gavea very clear exposition of Aprys proof of the irrationality of(3), in his 1979 paper A proof that Euler missed . . . Aprysproof of the irrationality of (3).Thus, although my only joint work with Alf was on continuedfractions of algebraic numbers, I decided to talk today on atopic related to the Riemann zeta function.

Arias de Reyna, Brent and van de Lune Sign of

Summary

I Notation and definitions.I Motivation

Some notation and definitions

P is the set of primes and p P is a prime.

We always have s = + it C.log (s) denotes the main branch defined in the usual way onthe open set G equal to the complex plane C with cuts alongthe half-lines (+ i, + i] for each zero or pole + i of(s) with 1/2. Thus log (s) is real and positive in (1,+).For s G, we can define arg (s) by

log (s) = log |(s)|+ i arg (s).

|B| denotes the Lebesgue measure of a set B.Usually > 1/2 is fixed, t R is regarded as the independentvariable, and b = p.

Arias de Reyna, Brent and van de Lune Sign of

Motivation

Several authors (Gram, Titchmarsh, Edwards, . . .) haveobserved that

The Euler product

Recall the Euler product formula

(s) =

p

(1 ps

)1,

valid for > 1.Taking logarithms, we get

log (s) =

p

log(1 ps

)=

p

ps + O(1).

There is a sense in which log (s) can be approximated by ashort Dirichlet series

px p

s in the region 1/2 1, butwe wont discuss this today.

Arias de Reyna, Brent and van de Lune Sign of

RemarksLet 0 = 1.1923473371 be the real root in (1,+) of

p

arcsin(p) =

2.

It was shown by Jan van de Lune (1983) that( 0) 0 .

Also, for any (1, 0), there exist arbitrarily large t such that

Figure from Van de Lune (1983)

From the Figure, we see that the prime p contributes at mostarcsin(p) to arg ( + it). If ( + it) < 0 then we must have|arg ( + it)| > /2.

Arias de Reyna, Brent and van de Lune Sign of

Some numerical results

If we compute (s) for randomly chosen s with =

- The seventh interval where
Asymptotic densities

Fix > 1/2, and define

d+() := limT+

12T|{t [T ,+T ] : 0}| ,

d() := limT+

12T|{t [T ,+T ] :

Approximation of d() via arg (s)

It is easier to work with

d() = limT+

12T|{t [T ,+T ] : |arg ( + it)| > /2}| .

Observe that

A mean-value result

Using Chebyshevs inequality and a mean-value result3 for|

Intuition

Informally, the idea is that, for a prime p and large t ,

pit = exp(it log(p))

behaves like a random variable distributed uniformlyon the unit circle.Moreover, for different primes, the random variables areindependent (because the log p are independent over Q).A theorem of Bohr and Jessen (to be stated soon) justifies thisintuition.

Arias de Reyna, Brent and van de Lune Sign of

The probability space

Let T = {z C : |z| = 1} denote the unit circle with the usualprobability measure (that is d2 if we identify T with theinterval [0,2) via z = exp(i)).Define := TP with the product measure P = P .Each point of is a sequence = (xp)pP , with each xp T.This formalises the idea that the xp may be considered as arandom variables, uid4 on the unit circle.

4Uniformly and independently distributed.Arias de Reyna, Brent and van de Lune Sign of

The measure P of Bohr and Jessen

Before stating the theorem of Bohr and Jessen we need todefine a measure P.Fix > 1/2. The sum of random variables

S = pP

log(1 pxp) :=pP

k=1

1k

pkxkp

converges almost everywhere, so S is a well-defined randomvariable.The measure P of Bohr and Jessen is defined to be thedistribution of S, i.e. for each Borel set B C we have

P(B) := P{ = (xp) : S() B}.

Arias de Reyna, Brent and van de Lune Sign of

A theorem of Bohr and Jessen

In modern language, Bohr and Jessen (1930/31) showed that,for each rectangle B with sides parallel to the axes,

P(B) = limT

12T|{t [T ,+T ], log ( + it) B}|

(and the limit exists).It is easy to deduce that the same result holds forJordan-measurable sets B C.Bohr and Jessen also showed that P is absolutely continuouswith respect to the Lebesgue measure on C.

Arias de Reyna, Brent and van de Lune Sign of

The measure on RWe can specialise5 to sets B of the form R B. For Jordansubsets B R, define

(B) := P(R B).

Thend() = (R\[/2, /2]) ,

d+() =

(kZ

(2k /2, 2k + /2)

).

The measure is the distribution function of the randomvariable =S:

(B) = P(R B) = P{ : S() R B}= P{ : =S() B}.

5Since R is not bounded, we need a limiting argument to justify this.Arias de Reyna, Brent and van de Lune Sign of

The characteristic function

Recall that the characteristic function (y) of a random variableX is defined by

(y) := E[exp(iXy)] .

This is just a Fourier transform; we omit a factor 2 in theexponent to agree with the statistical literature.The characteristic function associated with is thecharacteristic function of the random variable =S:

(x) :=

eix=S() d .

Arias de Reyna, Brent and van de Lune Sign of

as an infinite product over the primes

We have(x) =

p

I(p, x) ,

where (as usual) the product is over all primes p, and

I(b, x) :=1

2

20

eix arg(1b1eit ) dt .

Arias de Reyna, Brent and van de Lune Sign of

Sketch of the proof

By definition

(x) =

eix=S() d =

p

eix arg(1pxp) d.

By independence the integral of the product is the product ofthe integrals:

(x) =

p

eix arg(1pxp) d.

Each random variable xp is distributed as ei on the unitcircle.

Arias de Reyna, Brent and van de Lune Sign of

I(b, x) as an integral

Assume b > 1. Recall the definition

I(b, x) :=1

2

20

eix arg(1b1eit ) dt .

Then we have two equivalent expressions for I(b, x) as adefinite integral:

I(b, x) =1

0

cos(

x arctan(

sin b cos

))d

=2

10

cos(

x arcsinub

) du1 u2

.

Arias de Reyna, Brent and van de Lune Sign of

Connection with Bessel functions

Suppose b is large in the first representation

I(b, x) =1

0

cos(

x arctan(

sin b cos

))d .

Approximating arctan(sin /(b cos )) by sin()/b, we get

I(b, x) 1

0

cos(x

bsin

)d = J0(x/b)

from an integral representation of the Bessel function J0.A more detailed asymptotic analysis shows that I(b, x) hasinfinitely many real zeros near the points

{(3/4 + k)/arcsin(1/b) : k Z0}.

Arias de Reyna, Brent and van de Lune Sign of

What does look like?

(x) is a product:

(x) =

p

I(p, x) .

Each factor in the product has infinitely many real zeros, andthe same is true for (x).We have the bound

|(x)| C exp

(c x

1/

log x

)

for some positive constants C = C(), c = c() and x 2.The best possible constants are not known, but empirically wecan take C = c = 1 for all x 7 and > 1/2.

Arias de Reyna, Brent and van de Lune Sign of

Pictures of 1.02(x)Due to the exponential decay of |(x)| it is difficult to give asingle plot that shows its behaviour. Following are plots for = 1.02 and x in the intervals [0,5], [5,10], . . ., [25,30].

Arias de Reyna, Brent and van de Lune Sign of

x [5,10]

Arias de Reyna, Brent and van de Lune Sign of

x [10,15]

Arias de Reyna, Brent and van de Lune Sign of

x [15,20]

Arias de Reyna, Brent and van de Lune Sign of

x [20,25]

Arias de Reyna, Brent and van de Lune Sign of

x [25,30]

Arias de Reyna, Brent and van de Lune Sign of

Digression the hypergeometric differential equation

The hypergeometric differential equation is the second-orderlinear equation

z(1 z)w + (c (a + b + 1)z)w abw = 0,

where primes denote differentiation with respect to z. Herea,b, c are constants. In a neighbourhood of the origin, thehypergeometric function

2F1(a,b; c; z) =

n=0

(a)n(b)n(c)n

zn

n!

is a solution; a second solution is

z1c2F1(1 + a c,1 + b c; 2 c; z).

Arias de Reyna, Brent and van de Lune Sign of

cos(2x arcsin u) as a hypergeometric function

For |u| < 1 and all x C,

cos(2x arcsin u) = 1 +

n=1

(2u)2n

(2n)!

n1j=0

(j2 x2).

To prove this, let f (u) := LHS, g(u) := RHS above. Then

(1 u2)f (u) uf (u) + 4x2f (u) = 0.

Also, g(u) satisfies the same differential equation. Sinceg(0) = f (0) = 1 and g(0) = f (0) = 0, the two solutionscoincide.Notes. When x Z the series reduces to a polynomial.The differential equation can

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