🏛 A Project Cauchy Op-ed
Named after French mathematician Augustin-Louis Cauchy, Project Cauchy column is where I invite some of the HFI Programming club members to provide neat proofs or explanations about some number theory puzzles. I highly suggest that you read these articles with a pencil and paper so you can sketch things out and scribble solutions to exercises as you come across them. This time, Travor presented me one of the neatest proofs to the prime number theorem.
As the title suggests, today we are going to do some integral calculus. First, let's recall the definition of Riemann integral:
Riemann Integral and Its Formal Definition
Traditionally, an integral of some function over some interval is defined as the signed area of over the curve:
and Riemann integral is one way to define it rigorously. Particularly, we
define it as the sum of the areas of tiny rectangles:
where are sampled in and the increasing sequence is called a partition of the interval :
To formalize Riemann integral, let's define the as the length of the maximum of interval in a partition :
Then we say that some function is Riemann integrable if for all , there exists such that when we always have
Alternatively, if some function is Riemann-integrable on , then the limit
exists and converges to .
From Riemann Integral to Riemann-Stieltjes Integral
Although Riemann integral appears to be sufficient to integrate functions, it is not friendly to integrate piecewise continuous functions. Let's first look at its definition:
In order for this it to exist, we need to set up constraints on and :
Theorem 1: The Riemann-Stieltjes integral exists if is continuous on and is of bounded variation on
When we say is of bounded variation, we mean that the following quantity exists:
Proof. Let's define as another partition of such that is its subsequence and be the sampled abscissa in , so if we designate to be the set of such that 's are contained in interval :
then we have
Because is uniformly continuous within , we know that for every there exists such that when satisfy then . Accordingly, if we were to take the absolute values of (1) and (2), then
Now, let be another partition of , be its correponding samples abscissa and be the union of both partitions, then
which indicates the validness of this theorem.
Properties of Riemann-Stieltjes Integral
In addition to the constraint for the Riemann-Stieltjes integral to exist, we can also transform it into a Riemann integral at specific occasions:
Theorem 2: If exists and is continuous on then
Proof. Since is differentiable, we can use mean value theorem to guarantee the existence of such that , so
As a result, is of bounded variation, implying the existence of the left hand side of (3).
By the uniform continuity of , we know that for all there exists such that whenever , indicating and as . Accordingly, we arrive at the conclusion that
thus completing the proof.
In addition, we can also apply integration by parts on Riemann-Stieltjes integrals. Particularly, if we assume has a continuous derivative and is of bounded variation on , then
Riemann-Stieltjes Integral and Partial Summation
Let be some arithmetic function and be its summatory function
Let have continuous derivative on and then we have
where we require that . Recall (4), we observe that is a step function that only jumps at integer values, so we only need to sum over 's such that . Hence, this integral becomes a summation that sums over integers values in :
Since is uniformly continuous on , we have that when ever , thus the second summation is of , leaving us
Employing (5) in different situations can give us plentiful brilliant results. Let's have a look at some of them:
It was well-known that the harmonic series diverges, and we can provide a formal proof using Riemann-Stieltjes integral:
Since as , we conclude that the harmonic series diverges.
Evaluation of an Interesting Series
Now, let's first consider the finite case:
In fact, using Riemann-Stieltjes integral, we can show
Now, employing this obtained identity and the asymptotic formula for harmonic series yields:
Now, take the limit on both side gives
The Prime Number Theorem
If we were to define the prime indicator function
Then the prime counting function can be defined as
Now, let's also define Chebyshev's function :
Hence, we have
It is known that , so plugging it into the above equation gives
which is the prime number theorem.
To sum up, in this blog, we first define and explore the Riemann-Stieltjes integral, then use this integration technique to solve problems via asymptotic expansion. Lastly, we provide a proof for the prime number theorem.