Some gems from analytic number theory

I am currently attending the Counting Arithmetic Objects summer school at Centre de Recherches Mathematiques in Montreal. During the second day of lectures I heard a remarkable talk from Professor Andrew Granville. He spoke with outstanding clarity on the basic aim of analytic number theory, and I wish to remark on some of the epiphanies I had during the talk.

Without a doubt, one of the most important identities in analytic number theory is the following:

\int_0^1 e^{2\pi i n t}dt = \begin{cases} 1, & \text{ if } n = 0 \\    0, & \text{ if } n \ne 0 \end{cases}

The key insight is that the above identity serves as an “indicator function” for whether a quantity is zero or not. Indeed, this basic observation is behind behind the solutions to Waring’s problem and the proof of Vinogradov’s theorem on the sum of three primes. This “smooth” indicator function allows the use of methods in analysis to deal with arithmetic problems.

A related identity is Perron’s formula. Here again we start with an “indicator” function, defined for c > 0

\frac{1}{2\pi i} \int_ {c-i\infty}^{c+i\infty} \frac{y^s}{s} ds = \begin{cases} 1, & \text{ if } y > 1 \\ 0, & \text{ if } y < 1 \end{cases}

This allows to handle sums of the shape

\displaystyle \sum_{n \geq 1} a_n.

To see this, we start with the observation that for each n in the range of summation we have (provided x is not an integer) that x/n > 1. Hence we have

\displaystyle \sum_{n \leq x} a_n = \sum_{n \leq x} a_n \frac{1}{2 \pi i } \int_{c - i\infty}^{c + i\infty} \left(\frac{x}{n} \right)^s \frac{1}{s} ds.

The outer sum is finite, but we would not change the sum at all if we included all positive integers n > x because of the indicator function nature. For well behaved a_n‘s and taking c sufficiently large to ensure absolute convergence, we have that

\displaystyle \sum_{n \geq 1} a_n \frac{1}{2 \pi i} \int_{c - i\infty}^{c + i\infty} \frac{1}{s} \left(\frac{x}{n}\right)^s ds = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \left(\sum_{n\geq1} \frac{a_n}{n^s}\right) \frac{x^s}{s} ds.

We note that g(s)= \displaystyle \sum_{n \geq 1} \frac{a_n}{n^s} is the Dirichlet series associated to the a_n‘s. Summarizing, we obtain the equation

\displaystyle \sum_{n \leq x} a_n = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} g(s) \frac{x^s}{s}ds.

Now we can apply this to the familiar Von Mangoldt function \Lambda(n) defined by

\displaystyle \Lambda(n) = \begin{cases} \log p, & \text{ if } n = p^m \\ 0, & \text{ otherwise.} \end{cases}

where as usual p denotes a prime. Now using Perron’s formula, we have

\displaystyle \sum_{n \leq x} \Lambda(n) = \sum_{p^m \leq x} \log p = \frac{1}{2 \pi i} \int_{c - i\infty}^{c + i\infty} - \frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s}ds

Where \zeta(s) denotes the Riemann zeta function. This gives rise, after evaluating the integral on the right using residue theory and noting that the poles of \zeta'(s)/\zeta(s) are 1 and the zeroes of \zeta, we have

\displaystyle \sum_{p^m \leq x} \log p = x - \sum_{\rho: \zeta(\rho) =0} \frac{x^\rho}{\rho}+\frac{\zeta'(0)}{\zeta(0)}.

This is likely one of the most striking equations in mathematics as the left hand side is a discrete sum while the right hand side involved values of a meromorphic function. Somehow, the zeroes of the zeta function “knows” where the primes are.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s