Abstract

For a general polynomial Euler product F(s) we define the associated Euler totient function φ(n, F) and study its asymptotic properties. We prove that for F(s) belonging to certain subclass of the Selberg class of L-functions, the error term in the asymptotic formula for the sum of φ(n, F) over positive integers n ≤ x behaves typically as a linear function of x. We show also that for the Riemann zeta function the square mean value of the error term in question is minimal among all polynomial Euler products from the Selberg class, and that this property uniquely characterizes ζ(s).

Highlights

  • Introduction and statement of resultsBy a polynomial Euler product we mean a function F(s) of a complex variable s = σ + it which for σ > 1 is defined by an absolutely convergent product of the formCommunicated by J

  • Corollary 1.5 For every polynomial Euler product F we have β(F) ≥ β(ζ ) = 1/(6π 2), where ζ is the Riemann zeta function. This leads to the following converse theorem characterizing ζ (s) as the only polynomial Euler product from the Selberg class with the minimal square mean of |E(x, F)|

  • Riemann zeta function is the only L-function F from the Selberg class with “easy” Dirichlet coefficients meaning that aF (n) = φ(log n) for certain entire function φ(z) of order 1 and a finite type i.e., satisfying φ(z) eα|z| for certain positive α and every complex z [8]

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Summary

Let us put

Corollary 1.5 For every polynomial Euler product F we have β(F) ≥ β(ζ ) = 1/(6π 2), where ζ is the Riemann zeta function. This leads to the following converse theorem characterizing ζ (s) as the only polynomial Euler product from the Selberg class with the minimal square mean of |E(x, F)|. 3. Riemann zeta function is the only L-function F from the Selberg class with “easy” Dirichlet coefficients meaning that aF (n) = φ(log n) for certain entire function φ(z) of order 1 and a finite type i.e., satisfying φ(z) eα|z| for certain positive α and every complex z [8]. Lemma 2.1 We have φ(n, F) n(log log n)d

Proof We have d
Suppose that y
We have
Integrating by parts we get x
Observe that for a polynomial Euler product F we have

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