A pseudo zeta function and the distribution of primes

Abstract

The Riemann zeta function is given by:

ζ(s) may be analytically continued to the entire s-plane, except for a simple pole at s = 0. Of great interest are the complex zeros of ζ(s). The Riemann hypothesis states that the complex zeros all have real part 1/2. According to the prime number theorem, p_n ≈ n logn, where p_n is the nth prime. Suppose that p_n were exactly nlogn. In other words, in the Euler product above, replace the nth prime by nlogn. In this way, we define a pseudo zeta function C(s) for Re s > 1. One can show that C(s) may be analytically continued at least into the half-plane Re s > 0 except for an isolated singularity (presumably a simple pole) at s = 0. It may be shown that the pseudo zeta function C(s) has no complex zeros whatsoever. This means that the complex zeros of the zeta function are associated with the irregularity of the distribution of the primes.

For Re s > 1, the Riemann zeta function is defined by the series ζ(s) = ∑_n=1^∞ 1/n^s. Euler (1,2) observed that

1

This is a simple consequence of unique factorization.

Riemann (1,2) showed that ζ(s) continues analytically to the entire s-plane, except for a simple pole at s = 1. The ζ function has real zeros at s = −2, −4, −6, … . Of great interest are the complex zeros of ζ(s). The celebrated Riemann hypothesis states that the complex zeros all have real part 1/2.

From Eq. 1 above,

so the Riemann hypothesis says that 1/ζ(s) has no poles with Re s ≠ 1/2 in the critical strip 0 ≤ Re s ≤ 1. Because of the symmetry implied by Riemann's functional equation (see An Unsolved Problem below), it is equivalent to show that 1/ζ(s) has no poles with Re s > 1/2.

Now, for Re s > 1, we have (2)

If we write

2

then we have

3

Note that the series ∑_k=2^∞ 1/kϕ(ks) is uniformly convergent, hence analytic, in any half-plane Re s ≥ (½) + δ, δ > 0, for then Re(ks) ≥ 1 + 2δ, and |ϕ(ks)| ≤ Σ_pp^−k(1/2+δ). This series is uniformly convergent, because

Accordingly, ∑_k=2^∞ 1/k|ϕ(ks)| is dominated by the constant series M Σ_k=2^∞ 1/k² < ∞.

So only the first term ϕ(s) in Eq. 2 can cause trouble; hence the Riemann hypothesis is equivalent to the claim that ϕ(s) can be analytically continued into the strip 1/2 < Re s < 1. (NB ϕ will have a logarithmic singularity at s = 1.)

Motivated by the prime number theorem, which states that the nth prime p_n ≈ n log n, we consider the “comparison series”

4

Can ψ be continued analytically into the critical strip? This question is at least vaguely related to the Riemann hypothesis, because presumably the properties of ψ are related to those of ϕ.

An Integral Formula

From Eq. 4, we have the Stieltjes integral formula

Write {x} = x − [x], the fractional part of x. Then the integral above is

5

Note that the second integral in Eq. 5 defines an analytic function for Re s > 0.

Denote the first integral in Eq. 5 by I(s). Then I(s) = J(s) + K(s), where

and

If we integrate K by parts, we get

whence

Therefore,

6

Next, we integrate J(s) by parts:

We obtain

In the latter integral, make the substitution x = e^t; the integral becomes

which equals

7

The second integral Y(s) in Eq. 7 is an analytic function if Re(s − 1) < 1, i.e., Re s < 2. As for the first integral W(s), it is convergent for 1 < Re s < 2, and we have the explicit formula

Now (s − 1)^{−(s − 1)} = exp[−(s − 1) log(s − 1)] is a single-valued analytic function on the (simply connected) slit plane ℂ\[1, ∞). Then for s ∉ [1, ∞), we have the relation

8

whence J(s) can be analytically continued into the entire s plane minus the point {1}.

The Main Theorem

From the results of the preceding section,

also has an analytic continuation into the critical strip. That is, in particular, it has no singularities there.

Hence

has an analytic continuation into the critical strip 0 < Re s ≤ 1 in which it has no singularities. [This is in contrast with ϕ(s), which must be singular at the complex zeros of J(s).]

Conclusion.

Theorem. Let C(s) = ∏_n=2^∞ (1 − (n log n)^−s)⁻¹. Then C(s) continues analytically into the critical strip and has no zeros there.

Significance of the theorem: If the primes were distributed more regularly (i.e., if p_n ≡ n log n), then the Riemann hypothesis would be trivially true. In reality, the zeros of J(s) are related to the irregularities in the distribution of the primes. [Of course, the latter fact was known to Riemann; see his “explicit formula” (1) for π(x), the number of primes less than x.]

Some Related Pseudo ζ Functions

It is, of course, impossible for the nth prime p_n to equal n log n, for the latter is not even an integer. But what if p_n is replaced by [n log n], the greatest integer in n log n? More generally, what if p_n is replaced by n log n + ɛ_n, where (ɛ_n)₂^∞ is a bounded sequence? It turns out that the corresponding pseudo zeta function also has no complex zeros.

To see this, consider the difference

This series converges uniformly for Re s > 0. Indeed, writing σ = Re s, we have

so that our series is dominated by

where ɛ = sup_n|ɛ_n|. Interchanging the order of summation and integration, we have

which converges uniformly for σ > 0.

Accordingly, the two series

have the same analytic behavior in the critical strip. Hence ∑_n=2^∞ 1/(n log n+ɛ_n)^s has no complex singularities in the critical strip. But this means that the associated pseudo zeta function

has no zeros in the critical strip.

An Unsolved Problem

The Riemann zeta function satisfies the functional equation (1, 2)

where f(s) = Γ(s/2)ζ(s)π^−s/2.

Question: Does the pseudo zeta function (say with p_n replaced by n log n) also satisfy some sort of functional equation? This would be quite interesting if true.