Using Zeta Zeros to Tally Sigma Times Tau

In number theory, the number of divisors of an integer is usually denoted by . ( is the lowercase Greek letter tau.) For example, 4 has three divisors (namely, 1, 2, and 4), so . The sum of the divisors of is denoted by . ( is the lowercase Greek letter sigma.) So, .
Suppose . The sum of for is an irregular step function that jumps up at every integer . For example, for , this sum is = . For , this sum is = .
This Demonstration shows how we can approximate this step function with a sum that involves zeros of the Riemann zeta () function.


  • [Snapshot]
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Snapshot 1: the graphs of the step function and the formula using no zeta zeros
Snapshot 2: the graphs of and the formula using 100 pairs of zeta zeros
After you use the slider to choose (the number of pairs of zeta zeros to use), this Demonstration uses the following formula to calculate :
(1) .
In this formula,
where is Euler's constant.
In equation (1), is the complex zero of the Riemann zeta function with positive real part. The first three complex zeros of the zeta function are approximately , , and . These zeros occur in conjugate pairs, so if is a zero, then so is .
If you use the slider to choose, say, (one pair of zeta zeros), then the first sum in equation (1) adds the two terms that correspond to the first pair of conjugate zeros, and . These terms are conjugates of each other. When these terms are added, their imaginary parts cancel while their real parts add. So, the applied to the first sum is merely an efficient way to combine the two terms for each pair of zeta zeros.
Notice that the second sum has the same form as the first, except that the second sum extends over the real zeros of the zeta function, namely, . However, the second sum is too small to visibly affect the graphs, so this sum is not computed here.
If you plot a graph using no zeta zeros, then the graph is computed with only the terms through .
Where Does Equation (1) Come From?
To prove equation (1), we start with the following identity, which holds for (see [1], equation D-58, and [2], Theorem 305):
(2) .
Perron's formula (see reference [3]) takes an identity like equation (2) and gives a formula for the sum of the numerators as a function of , in this case, .
When we apply Perron's formula to equation (2), we get equation (1). To apply Perron's formula, we integrate this integrand
around a contour in the complex plane. Each part of equation (1) is the residue at one of the poles of this integrand. The residue at the pole at is . At , has a pole (of order 1), so the integrand has a pole of order 2 at due to the . The residue at is . Similarly, the integrand also has a pole of order 2 at . The expression is the residue at .
Mathematica can compute these residues. For example, this calculation Residue[Zeta[s]^2 Zeta[s-1]^2 / Zeta[2 s - 1] x^s/s, {s, 2}] gives , where and have the values given above.
Finally, the integrand has a pole at each complex zero of . The first sum in equation (1) is just the sum of the residues at these complex zeros of zeta. Each complex zero gives rise to one term in the sum. ( is the complex zero of , so is the complex zero of ).
In the same way, the second sum arises from the real zeros () of . ( is the real zero of , so is the real zero of ).
The Demonstration "Using Zeta Zeros to Tally Squarefree Divisors" gives additional details on Perron's formula.
[1] H. W. Gould and T. Shonhiwa, "A Catalog of Interesting Dirichlet Series."
[2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 4th ed., Oxford: Oxford University Press, 1965, p. 256.
[3] H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory: I. Classical Theory, Cambridge: Cambridge University Press, 2007, p. 397.
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