A Small Taste from My New Book: Season 2 Episode 4

 

In this episode, we continue our exploration of arithmetic functions from an analytic perspective. 

Building on ideas introduced earlier in the season, we focus on techniques that reveal how analytic tools can encode deep number‑theoretic information. These methods will serve as a bridge between theory and computation, preparing us for a more algorithmic approach in the episodes to come.

 

In “The Riemann Hypothesis Revealed”, the reader encounters a number of exercises/tasks that can be worked out without difficulty. Beyond completeness, the results obtained below play a central role in the narrative that follows. In particular, they prepare the ground for the classical explicit formula:

toward which we have been gradually building since the previous episode. This “crude” form is the Explicit Formula for the Chebyshev function , which is the fundamental bridge between the prime numbers and the zeros of the Riemann zeta function. It relates the distribution of primes to a sum over complex zeros .

The main term  is the average growth of the prime-counting function. It comes from the prime counting theorem which states that . If primes were perfectly distributed without “noise” this would be the only term.

The Oscillatory Term:  is the most important part of the formula. It represents the fluctuations or "music" of the primes.

The Zeros  are the non-trivial zeros of the zeta function.

The Constant Term arises from the value of the logarithmic derivative of the zeta function at  (see The Riemann Hypothesis Revealed).

The Smallest Term  is often called “trivial” term and comes from the trivial zeros of the zeta function located at the negative even integers

The function  will be defined next. Pay attention!

The following tasks are taken from: “The Riemann Hypothesis Revealed”

Task 1: There is a well-known arithmetic function called the Mangoldt function, denoted by ,  which returns  when  (i.e., a power of a prime number), and zero otherwise.

For example, we can compute:

This motivates the identity

Provide detailed justification for each step. Of course,  means  divides .

Work: In general, for ,

Since the only nonzero terms in the sum  come from the divisors  of the form ,   and  , it follows

We can group the sum by each prime factor :

For each internal sum, since  for every :

Substituting this back into the total sum:

As established, this expression is exactly equal to . Thus:

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Task 2: Next, recall Euler’s identity: 

Taking logarithms of both sides, and using the identity

for , show that:

Assuming term-by-term differentiation is valid, show the next identity

This formula is central in analytic number theory and is frequently used in the study of the distribution of primes.

Work out the intermediate steps to fully establish the above result.

Work: To establish these identities, we will follow the analytic path from the Euler Product Formula to the Dirichlet series of the logarithmic derivative of the Riemann zeta function.

Starting with the identity

Taking the logarithm of both sides (valid for ):

Now use the provided Taylor series  with . Since  for , the expansion is valid:

To find  we take the derivative with respect to :

The double sum covers every integer  that is a prime power. Since for integers  that are not prime powers, , we can rewrite it as a single sum over all :

______

Next

Perron’s formula.

Let  be an arithmetic function and

be the corresponding Dirichlet series. Let  be absolutely convergent for , where  is known as the abscissa of convergence. We also consider , . Then we define

The prime on the summation indicates that the last term of the sum must be multiplied by  when  is an integer.

If  we have:

This is known as Perron’s formula. Its proof can be found in “The Riemann Hypothesis Revealed” at the end of The Mellin Transform section.

Task 3: Assuming that you’ve already completed Task 1 & Task 2, an easy task is to show that the following identity holds:

 

Work: Clearly, specialize to . For ,

so here , and . Take  in the general identity (or just rename  as ),

and chose .

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This identity (Task 3) is nothing more than Perron’s formula applied to the Dirichlet series for , and may be interpreted as follows:

where

Indeed

At non-integer   is constant in the neighborhood of , so

At integer   has a jump of size  at :

Then

which is exactly

This is the starting point of the explicit formula.

Why this midpoint value?

Because  is a step function with jumps at primes and prime powers. The integral on the right is continuous in .  So the only way to match a discontinuous function with a continuous integral is to take the average of the left and right limits.

This is exactly the same phenomenon as:

• Fourier series converging to the midpoint at jump discontinuities,
• inverse Laplace transforms of step functions,

So the primed sum is not an arbitrary trick — it is the correct inverse Mellin transform of .

Where does  get used next?

Right after establishing the identity

we shift the contour to the left. To justify this, we first truncate the integral by fixing  and consider

We then close the contour by adjoining a semicircle in the left half–plane with diameter joining the points  and .  Finally, letting  shifts the contour to the left while capturing the residues of the enclosed poles.

The integrand

has poles from:

Pole of  at . In The Riemann Hypothesis Revealed we’ve shown that

so (simply factor the denominator: )

Thus:

Nontrivial zeros

Summing up

Trivial zeros

From the local behavior of a function  near a zero

Thus

Applying this to

Therefore

and

Thus

Summing up we have

Since

Setting

Pole at

We use

The residue contributes

 Finally combining all contributions

In a nutshell, when we shift the contour to the left the poles of  contribute residues:

·      a pole at  gives the main term ,

·      poles at the nontrivial zeros  give oscillatory terms ,

·         poles at negative even integers give small correction terms.

This produces the classical explicit formula:

with the midpoint interpretation at integers.

So  is the function whose Mellin transform is , and therefore the function that naturally appears when you invert that transform.

Let’s write a C++ program that calculates the von Mangoldt function and its representation for the next episode.

Epilogue (Episode 4)

In this episode, we explored analytical tools that reveal the structure hidden within arithmetic functions. These ideas prepare the ground for a more computational perspective, where theory meets implementation.

If the mystery of the Riemann Hypothesis has ever intrigued you, The Riemann Hypothesis Revealed might be the most practical doorway you’ll find into this fascinating world. Rather than burying the reader under the formal weight typical of academic texts, the book offers a direct and intuitive kickstart into the ideas that make the problem so captivating. Supported by more than 150 pages of worked examples available on this blog, it invites you to explore the theory through concrete calculations and guided insight. If you’re curious about one of mathematics’ greatest unsolved problems and want a starting point that actually gets you moving, this book was written with you in mind.