↑ Analytic Number Theory

Dirichlet Series

Motivation

Given a positive integer $n$, what is the number of its divisors? This problem appears constantly in coding challenges. Write this number of divisors into a condense form

and let us also define this as a function in Julia with Primes.jl and plot the first values with CairoMakie.jl

Seems quite random, however in this article we will discuss the following striking asymptoticity obtained by Dirichlet in 1849

as plotted

As shown above, the asymptotic formula is already evident for $n \leq 100$.

The correct way to study $\sum_{n \leq x}d(n)$ is through its Dirichlet series $L(s, d) \coloneqq \sum_{n > 0} d(n)/n^s$. I will present two reasons to introduce Dirichlet series: they capture the algebraic structure behind $d(n)$, and their poles and zeros reflects the behavior of $\sum_{n \leq x} d(n)$ as $x \to +\infty$. This topic will motivate the study of analytic number theory.

Arithmetic Functions and Dirichlet Series

Let us extract the obvious information from the definition $d(n) = \sum_{d \mid n} 1$: it is a function defined on $\mathbb Z_+$, and the sum is over divisors. This abstracts into the following definition.

Definition. The commutative ring of arithmetic functions is the set of functions

equipped with pointwise addition and multiplication given by convolution

The function

serves as unity.

For every arithmetic function $a(n)$ there is an associated Dirichlet series $\sum_{n > 0}a(n)/n^s$. For now just ignore the convergence issue and consider this as a foraml series. The ring of formall Dirichlet series is nothing but $\mathbb C \llbracket \Z_+ \rrbracket$, the complete monoid algebra of the multiplicative monoid $(\Z_+, \times)$. Our association defines a ring isomorphism

this would be clear once we notice that for the function

there is

Remark. Mapping $1/p^s$ to $x_p$ for each prime $p$ gives an isomorphism

where the right-hand side is the formal power series ring of independent variables indexed by prime numbers.

Now consider the constant function $\mathbf{1}(n) = 1$, we immediately check that $d = \mathbf{1} * \mathbf{1}$, which means $L(s, d) = L(s, \mathbf{1})^2$. This $L(s, \mathbf{1})$ is better known as the Riemann $\zeta$ function, the central object of analytic number theory. We will study the very basics of $\zeta(s)$, then $\zeta(s)^2$ comes for free. But that will not be helpful unless we understand how $L(s, a)$ can be used to study $\sum_{n \leq x} a(n)$.

Perron's Formula

If the Dirichlet series $L(s, a) = \sum_{n > 0} a(n)n^{-s}$ converges absolutely at some $s \in \mathbb C$, then convergence is guaranteed at every point to the right of $s$. If we define

the Dirichlet series will converge as long as $\Re s > \sigma_a$.

Assume $\Re s > \sigma_a$ and apply Abel's summation formula, we have

where $A(x) = \sum_{n \leq x} a(x)$. Absolute convergence is necessary since we need to interchange summation and integral.

After a change of variable $x = \mathrm e^t$, we see that $L(s, a)/s$ is the Laplace transform of $A(\mathrm e^t)$:

Apply the inverse Laplace transform and we get

where $c$ is any real number larger than $\sigma_a$. The left-hand side is often written as $\sum'_{n \leq x} a(n)$, the only difference from the usual $\sum$ is that when $x$ is an integer, the last term $a(x)$ is replaced by $a(x)/2$. The above equality is called Perron's formula.

Contour Shifting

The Perron formula is exact, but it gives no information unless we can find the dominant terms. The idea is to shift the integral contour left to get a smaller integral, while the singularities crossed should contribute to the dominant terms.

Take $a = \mathbf{1}$ for example, in this case $A(x) = \lfloor x \rfloor \sim x$ is obvious, so we expect to obtain the same result from Dirichlet series $L(s, \mathbf{1}) = \zeta(s)$. The series

converges for $\Re s > 1$, and has a pole at $s = 1$. But what is to the left of this pole? The hope is to find a meromorphic function extending $\sum_{n > 0}n^{-s}$ so that we can define $\zeta(s)$ on the points to the left of $s = 1$. This is called a meromorphic continuation. It turns out that $\zeta$ has a meromorphic continuation to the entired complex plane But for our purpose, we just need to extend $\zeta$ a little bit left, and the treatment is much easier.

Consider the alternating sum

which converges for $\Re s > 0$ (uniformly in compact subsets). The trick is

and therefore

for $\Re s > 0$. The possible poles are from the zeros of $1 - 2^{1-s}$, i.e. $s = 1 + \frac{2 \pi \mathrm i}{\log 2}m, m \in \Z$. Only $s = 1$ is a pole and to show this we consider another function

which gives a different expression

Now the possible poles are of the form $s = 1 + \frac{2\pi \mathrm i}{\log 3}m, m \in \Z$. Notice that $\gcd(2, 3) = 1$, the quotient $\frac{\log 3}{\log 2} = \log_2 3$ cannot be a rational number, so $s = 1$ is the only pole. We may plot the $\zeta$ function as follows, the color reflects argument, and brightness reflects absolute value. As we can see, the brightest point is the pole.

Consider the following contour, which we will shift.

References

• T. M. Apostol, Introduction to Analytic Number Theory, Undergraduate Texts in Mathematics, Springer, New York, 1976. DOI: 10.1007/978-1-4757-5579-4.

• C. Sorensen, From Classical L-Functions to Modern Reciprocity Laws, Graduate Texts in Mathematics 307, Springer, Cham, 2025. DOI: 10.1007/978-3-032-03035-1.