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

Snapshot 3: the graphs of

and the formula using 100 pairs of zeta zeros

Here is the step function that is graphed when the left-hand button is clicked:

.

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)

.

,

,

,

,

where

is Euler's constant, and

is the first Stieltjes constant (

StieltjesGamma[1] in

*Mathematica*).

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 the only terms

through

.

Now consider the step function that is graphed when the right-hand button is clicked:

.

Here is the formula used to graph this function:

(2)

,

,

,

,

where

is the second Stieltjes constant (

StieltjesGamma[2] in

*Mathematica*), and

and

are the same as in equation (1). The second sum in equation (2) is too small to visibly affect the graphs, so it is not computed.

If you plot a graph using no zeta zeros, then the graph is computed with only the terms

through

.

**Where Do Equations (1) and (2) Come From?**To prove equation (1), we start with the following identity, which holds for

(see [3], equation 5.39):

(3)

.

Perron's formula (see reference [4]) takes an identity like equation (3) 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 (3), 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 3 at

due to the

. The expression

is the residue at

. The expression involving

,

, and

is very complicated because the pole is of third order.

*Mathematica* can compute these residues. For example, this calculation

Residue[Zeta[s]^3/ Zeta[2 s] x^s/s, {s, 1}] gives the complicated expression

, where

,

, and

have the values given above for equation (1).

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. (The

complex zero of

is

, so

is the

complex zero of

.)

In the same way, the second sum comes from the real zeros (i.e., at

) of

.

Equation (2) is proved the same way. We start with this identity, which holds for

(see [1], equation D-58, and [2], Theorem 305):

.

As with equation (1), we apply Perron's formula, this time using the integrand

.

[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] P. J. McCarthy,

*Introduction to Arithmetical Functions*, New York: Springer-Verlag, 1986, p. 237.

[4] H. L. Montgomery and R. C. Vaughan,

*Multiplicative Number Theory: I. Classical Theory*, Cambridge: Cambridge University Press, 2007, p. 397.