Alex Thomas
12 March 2010
Professor William Stein
Math 414
Twelve Steps:
The Proofs for the Prime Number Theorem and Prime Number
Theorem in Arithmetic Progression by Newman(via Zagier) and
Soprounov Respectively
In this project I will explain the proof of the Prime Number Theorem(PNT) by D.J.
Newman, as described by D. Zagier, and the proof by Ivan Soprounov of prove the Prime
Number Theorem in Arithmetic Progression(PNTAP). Once I have explained the
asymptotic estimates of the prime counting function, and the prime counting function for
primes congruent to some number modulo another, I will test the latter estimate by means
of a program I have written in Sage.
We will begin with Newman’s proof for the PNT
Newman’s proof of the PNT begins with establishing Euler’s product formula. First,
let us show the Riemann zeta function. This function was first examined by Euler in a proof
of Euclid’s theorem, as well as by Dirichlet, but Riemann was the first to study in-depth as a
function of a complex variable1, hence its name.
Definition 1: the Riemann zeta function
∞
1
𝜁(𝑠) ≝ ∑ 𝑠 , 𝑠 ∈ ℂ
𝑛
𝑛=1
Let us now examine the expanded series.
Eq. 1
1
1
1
1
1
𝜁(𝑠) = ∑ 𝑠 = 1 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + ⋯
𝑛
2
3
4
5
∞
𝑛=1
Now we multiply both sides by the second term.
Eq. 2
∞
1
1
1
1
1
1
1
1
𝜁(𝑠) = 𝑠 ∑ 𝑠 = 𝑠 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + ⋯
𝑠
2
2
𝑛
2
4
6
8
10
𝑛=1
Next we subtract Eq. 2 from Eq, 1.
1
LeVeque, 154, 155
Eq. 3
1
1
1
1
1
1
𝜁(𝑠) − 𝑠 𝜁(𝑠) = (1 − 𝑠 ) 𝜁(𝑠) = 1 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + ⋯
2
2
3
5
7
9
Then let us multiply both sides by the second term again.
Eq. 4
1
1
1
1
1
1
1
(1 − 𝑠 ) 𝜁(𝑠) = 𝑠 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + ⋯
𝑠
3
2
3
9
15
21
27
As before, let us subtract Eq. 4 from Eq. 3.
Eq. 5
1
1
1
1
1
1
1
(1 − 𝑠 ) (1 − 𝑠 ) 𝜁(𝑠) = 1 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + ⋯
3
2
5
7
11
13
17
We do these same steps to obtain another factor.
Eq. 6
1
1
1
1
1
1
1
1
(1 − 𝑠 ) (1 − 𝑠 ) (1 − 𝑠 ) 𝜁(𝑠) = 1 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + 𝑠 + ⋯
5
3
2
7
11
13
17
19
If we continue in this way the right-hand side approaches 1, and this gives us
Eq. 7
𝜁(𝑠) ∏(1 − 𝑝−𝑠 ) = 1
𝑝
One can notice the similarity of this proof to the Sieve of Eratosthenes.
Now solving for 𝜁(𝑠), we get the Euler product formula as desired.
Result 1: Euler’s product formula
−1
𝜁(𝑠) = (∏(1 − 𝑝−𝑠 ))
𝑝
= ∏(1 − 𝑝−𝑠 )−1
𝑝
1
Next we will show that 𝜁(𝑠) − 𝑠−1 is analytic (Zagier uses the word ‘holomorphic’, which
means the same in this situation)2 on {𝑠 ∈ ℂ: ℜ(𝑠) > 0}, the right half of the complex plane.
We will transform the expression into something that will let us talk of analyticity. We can
1
treat 𝑠−1 as the result of an improper integral, and then finesse the summand in the
Riemann zeta function underneath the integral. The way this is done becomes obvious if
you transform the rightmost side into the middle one.
2
Zagier, 1
Eq. 8
∞
∞
𝑛+1
1
1
1
1
1
𝜁(𝑠) −
= ∑ 𝑠 − ∫ 𝑠 𝑑𝑥 = ∑ ∫
− 𝑠 𝑑𝑥
𝑠
𝑠−1
𝑛
𝑛
𝑥
1 𝑥
𝑛
∞
𝑛=1
𝑛=1
Now we examine the absolute value of the summed integral, whence we obtain
1
Result 2: 𝜁(𝑠) − 𝑠−1 is analytic on {𝑠 ∈ ℂ: ℜ(𝑠) > 0}
∞
∞ 𝑥
|𝑠|
1
1
1
𝑠
|∫ 𝑠 − 𝑠 𝑑𝑥| = |𝑠 ∫ ∫ 𝑠+1 𝑑𝑢 𝑑𝑥| ≤ max | 𝑠+1 | = ℜ(𝑠)+1
n≤u≤n+1 𝑢
𝑥
𝑛
1 𝑛
1
𝑛 𝑢
This shows that the summand is less than that of a convergent power series times a
constant, |𝑠|, so by the comparison test, it too must be convergent.
Now we will show ∑𝑝≤𝑥 log(𝑝) = 𝑂(𝑥), where 𝑂(𝑥) is big O notation, but first let us
officially define this series as a function.
Definition 2: 𝜃(𝑥)
𝜃(𝑥) ≝ ∑ log(𝑝)
𝑝≤𝑥
First, we will try and get a handle on how fast 𝜃(𝑥) changes with respect to 𝑥.
Eq. 9
𝑛
2𝑛
2
= (1 + 1)
2𝑛
2𝑛
2𝑛
) ≥ ( ) ≥ ∏ 𝑝 = 𝑒 𝜃(2𝑛)−𝜃(𝑛)
𝑖
𝑛
= ∑(
𝑖=0
𝑛<𝑝≤2𝑛
This has given us something which we can manipulate by taking the log of both sides.
Eq. 10
2𝑛 log(2) ≥ 𝜃(2𝑛) − 𝜃(𝑛)
This gives us something with 𝜃(𝑥) on the left side, and something 𝑂(𝑥) on the right.
Eq. 11
𝑥
𝜃(𝑥) − 𝜃 ( ) ≤ 𝐶𝑥, 𝐶 > log 2
2
Now we will sum these differences for over increasing powers of 2.
Eq. 12
∞
∞
𝑥
𝑥
𝑥
∑ (𝜃 ( 𝑖 ) − 𝜃 ( 𝑖+1 )) ≤ 𝐶 ∑ 𝑖
2
2
2
𝑖=0
𝑖=0
Applying term by term cancellation on the left side, and the normal properties for a
geometric series on the right, we obtain
Eq. 13
∞
𝑥
𝜃(𝑥) ≤ 𝐶 ∑ 𝑖 = 2𝐶𝑥
2
𝑖=0
From this we see that
Result 3
𝜃(𝑥) = 𝑂(𝑥)
Now we will examine Φ(𝑠).
Definition 3: Φ(𝑠)
log(𝑝)
Φ(𝑠) ≝ ∑
𝑝𝑠
𝑝
We want to show that the Riemann zeta function has no zeroes for{𝑠 ∈ ℂ: ℜ(𝑠) ≥ 1}, also
1
we will show that Φ(𝑠) − 𝑠−1 is analytic there as well.
To examine 𝜁(𝑠) for zeroes, we will look at the logarithmic derivative.
Eq. 14
𝑑
𝑑
log(𝜁(𝑠)) = − log (∏(1 − 𝑝−𝑠 ))
𝑑𝑠
𝑑𝑠
𝑝
The rules of logarithms lead us to use the product form of 𝜁(𝑠); these rules will allow us
transform the logarithm of a product into a sum of logarithms.
Eq. 15
−
ζ′ (s)
𝑑
𝑑
= − log (∏(1 − 𝑝−𝑠 )−1 ) = − ∑ log((1 − 𝑝−𝑠 )−1 ) =
ζ(s)
𝑑𝑠
𝑑𝑠
𝑝
𝑝
−𝑠
𝑝 log(p)
log(𝑝)
(1 − 𝑝−𝑠 )2
−∑
=∑ 𝑠
−𝑠
−1
(1 − 𝑝 )
𝑝 −1
𝑝
𝑝
This leaves us with a suggestive form (the negative was appended to the logarithmic
derivative in order to get us here), which we can manipulate to achieve our intended result.
Eq. 16
log(𝑝)
log(𝑝) log(𝑝) log(𝑝)
log(𝑝)
∑ 𝑠
=∑ 𝑠
−
+
=
Φ(𝑠)
+
∑
𝑝 −1
𝑝 −1
𝑝𝑠
𝑝𝑠
𝑝 𝑠 (𝑝 𝑠 − 1)
𝑝
𝑝
𝑝
Now, if we establish analyticity for the rightmost part, we will have completed both of our
log(𝑝)
tasks. So now let us show that the series ∑𝑝 𝑝𝑠 (𝑝𝑠 −1) is convergent by the comparison test.
Eq. 17
|
log(𝑝)
log(𝑝)
| = ℜ(𝑠) 𝑠
− 1)
|(𝑝 − 1)|
𝑝
𝑝 𝑠 (𝑝 𝑠
Now we must get a handle on |(𝑝 𝑠 − 1)|, so let us separate off some of the series.
Eq. 18
𝑛
log(𝑝)
log(𝑝)
log(𝑝)
∑ 𝑠 𝑠
=∑ 𝑠 𝑠
+∑ 𝑠 𝑠
𝑝 (𝑝 − 1)
𝑝 (𝑝 − 1)
𝑝 (𝑝 − 1)
𝑝
𝑝=2
𝑝>𝑛
The term on the right of the plus sign is finite, so let us work on the one on the left, but we
will consider the value of 𝑝 first.
Eq. 19
𝑝ℜ(𝑠)
𝑝 > 𝑛 ⇒ 𝑝ℜ(𝑠) > 𝑛ℜ(𝑠) , ℜ(𝑠) > 1, ℜ(𝑠) > 0
𝑛
This will eventually give us a finite number with which to work.3
Eq. 20
ℜ(𝑠)
𝑝
1
𝑝ℜ(𝑠) − ℜ(𝑠) = 𝑝ℜ(𝑠) (1 − ℜ(𝑠) ) < 𝑝ℜ(𝑠) − 1
𝑛
𝑛
1
1
= 𝐾 > ℜ(𝑠)
1
𝑝
−1
𝑝ℜ(𝑠) (1 − ℜ(𝑠) )
𝑛
This gives us something to replace the troublesome |(𝑝 𝑠 − 1)|.
Eq. 21
log(𝑝)
log(𝑝)
<
ℜ(𝑠)
𝑠
|(𝑝 − 1)| 𝐾𝑝2ℜ(𝑠)
𝑝
1
Which is convergent for2ℜ(𝑠) − 1 > 0, or, ℜ(𝑠) > 2, as desired.
So this, along with Result 2, establishes that the rightmost side is at least meromorphic,
that is to say, analytic with a specific set of poles. We will examine Φ(𝑠) on the boundary of
{𝑠 ∈ ℂ: ℜ(𝑠) ≥ 1} for poles. If there are none, we will have succeeded. Let us examine five
points, and we will assume they are indeed poles of Φ(𝑠).
Eq. 22
𝑠 = 1,1 ± 𝑖𝛼, 1 ± 2𝑖𝛼
Let the complex ones’ poles be of order 𝜇, 𝜈 respectively (so 𝜇, 𝜈 are non-negative integers).
We have that order of the zeroes of a function, are the residues of the logarithmic
derivative, e.g.
3
Thanks to Prof. John Sylvester for this method of showing convergence for
1
ℜ(𝑠) > 2
Eq. 23
𝑓(𝑧) = 𝑧 , log(𝑓(𝑧)) = log(𝑧 𝑛 ) = 𝑛 log(𝑧)
𝑑
𝑛
𝑛 log(𝑧) =
𝑑𝑧
𝑧
𝑛
And since the residue is equal to the coefficient of the first negative-power term of the
Laurent series expansion of 𝑓(𝑧), we have 𝑛, as the residue at 𝑧 = 0, which happens to be
the order of the zero of 𝑓(𝑧) at 𝑧 = 0. Because of this, the residue of Φ(𝑠) at the points
above should equal the negative of the order of the zeroes of 𝜁(𝑠). Since Φ(𝑠) is not under
any power, so its pole at 𝑠 = 1 is a simple one (has an order of 1).
Eq. 24
res(Φ(𝑠), 𝑠 = 1) = 𝜖Φ(1 + 𝜖) = 1
res(Φ(𝑠), 𝑠 = 1 ± 𝑖𝛼) = 𝜖Φ(1 + 𝜖 ± 𝑖𝛼) = −𝜇
res(Φ(𝑠), 𝑠 = 1 ± 2𝑖𝛼) = 𝜖Φ(1 + 𝜖 ± 2𝑖𝛼) = −𝜈
𝛼 ∈ ℝ, 𝛼 ≠ 0
Now we will attempt to create a contradiction.
Eq. 25
2
4
∑ (
) Φ(1 + 𝑟𝑖𝛼) = Φ(1 − 2𝑖𝛼) + 4Φ(1 − 𝑖𝛼) + 6Φ(1) + 4Φ(1 + 𝑖𝛼) + Φ(1 + 2𝑖𝛼)
2+𝑟
𝑟=−2
Let us see if we can transform this unwieldy right-hand side into something more useful .
Eq. 26
log(𝑝)
log(𝑝)
log(𝑝)
log(𝑝)
log(𝑝)
= ∑ 1+𝜖−2𝑖𝛼 + 4 ∑ 1+𝜖−𝑖𝛼 + 6 ∑ 1+𝜖 + 4 ∑ 1+𝜖+𝑖𝛼 + ∑ 1+𝜖+2𝑖𝛼 =
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
log(𝑝)
log(𝑝)
log(𝑝)
log(𝑝)
log(𝑝)
∑ 1+𝜖−2𝑖𝛼 + 4 1+𝜖−𝑖𝛼 + 6 1+𝜖 + 4 1+𝜖+𝑖𝛼 + 1+𝜖+2𝑖𝛼 =
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
1
1
1
1
1
∑ log(𝑝) ( 1+𝜖−2𝑖𝛼 + 4 1+𝜖−𝑖𝛼 + 6 1+𝜖 + 4 1+𝜖+𝑖𝛼 + 1+𝜖+2𝑖𝛼 )
𝑝
𝑝
𝑝
𝑝
𝑝
𝑝
This looks like the expansion of a linear expression raised to the fourth power
Eq. 27
log(𝑝)
1
1
1
1
∑ 1+𝜖 ( −2𝑖𝛼 + 4 −𝑖𝛼 + 6 + 4 𝑖𝛼 + 2𝑖𝛼 ) =
𝑝
𝑝
𝑝
𝑝
𝑝
𝑖𝛼 4
log(𝑝)
∑ 1+𝜖 ((𝑝 2 )
𝑝
𝑝
𝑝
𝑖𝛼 3
𝑖𝛼 2
𝑖𝛼
𝑖𝛼 2
𝑖𝛼 3
𝑖𝛼
𝑖𝛼 4
+ 4 (𝑝 2 ) (𝑝− 2 ) + 6 (𝑝 2 ) (𝑝− 2 ) + 4 (𝑝 2 ) (𝑝− 2 ) + (𝑝− 2 ) ) =
4
4
𝑖𝛼
log(𝑝) 𝑖𝛼
log(𝑝)
𝛼
∑ 1+𝜖 (𝑝 2 + 𝑝− 2 ) = ∑ 1+𝜖 (2 cos ) ≥ 0
𝑝
𝑝
2
𝑝
𝑝
Now, multiply the right-hand side of Eq. 25 by 𝜖, let 𝜖 go to zero, and we get
Eq. 28
−𝜈 − 4𝜇 + 6 − 4𝜇 + −𝜈 = 6 − 8𝜇 − 2𝜈 ≥ 0
But here we must remember that 𝜇 is a non-negative integer, so in order for this statement
to be true, it must be zero. This means there is no pole there, so we have a contradiction.
This, with the Result 2, gives us the desired result.
Result 4:
𝜁(𝑠) ≠ 0, ∀𝑠 ∈ ℂ s. t. ℜ(𝑠) ≥ 1
∞ 𝜃(𝑥)−𝑥
Now we will examine ∫1 𝑥 2 𝑑𝑥 for convergence. First we will try to make the integral
that we are concerned with out of something with established properties.
Eq. 29
𝑛
∑q≤xi log(q) − ∑q≤xi−1 log(q)
log(𝑝)
log(𝑝)
Φ(𝑠) = ∑
=
lim
∑
=
lim
∑
, 𝑝 ∈]𝑥𝑖−1 , 𝑥𝑖 ]
𝑛→∞
𝑛→∞
𝑝𝑠
𝑝𝑠
𝑝𝑠
𝑛
𝑝
𝑝
𝑝
This gives us something with 𝜃(𝑥), which is promising.
Eq. 30
𝑛
∞
∑q≤xi log(q) − ∑q≤xi−1 log(q)
𝑑𝜃(𝑥)
lim ∑
=
∫
𝑠
𝑛→∞
𝑝
𝑥𝑠
1
𝑝
We are nearly where we want, let us now change this Stieltjes integral into the more
familiar Riemann integral with integration by parts for Stieltjes integrals.
Eq. 31
∞
∞
𝑑𝜃(𝑥)
𝜃(𝑟)
1
∫
=
lim
−
𝜃(1)
−
∫
𝜃(𝑥)𝑑
(
)
𝑟→∞ 𝑟 𝑠
𝑥𝑠
𝑥𝑠
1
1
𝜃(1) is obviously zero, so let us examine the limit.
Eq. 32
∑
𝜃(𝑟)
O(r)
p≤r log(p)
lim 𝑠 = lim
= lim 𝑠 = 0
𝑠
𝑟→∞ 𝑟
𝑟→∞
𝑟→∞ 𝑟
𝑟
This is obtained with Result 3. So now we are left with a friendlier Stieltjes integral.
Eq. 33
∞
∞
∞
𝑑𝜃(𝑥)
1
𝜃(𝑥)
∫
=
−
∫
𝜃(𝑥)𝑑
(
)
=
𝑠
∫
𝑑𝑥
𝑠+1
𝑥𝑠
𝑥𝑠
1
1
1 𝑥
Now we will apply a change of variable to this in order to get the equation that will let us
obtain the desired result.
Eq. 34
∞
∞
𝜃(𝑥)
𝜃(𝑒 𝑡 )
Φ(𝑠) = 𝑠 ∫ 𝑠+1 𝑑𝑥 = 𝑠 ∫
𝑑𝑡
𝑒 𝑠𝑡
1 𝑥
0
Next we will make some definitions in order to use the Analytic Theorem.
Definition 3
𝜃(𝑒 𝑡 ) − 𝑒 𝑡
𝑓(𝑡) ≝
𝑒𝑡
∞
𝑔(𝑧) ≝ ∫ 𝑓(𝑡)𝑒 −𝑧𝑡 𝑑𝑡 , ℜ(𝑧) > 0
0
Now we will transform 𝑔(𝑧) into something from which we can determine whether it is
analytic on ℜ(𝑧) > 0.
Eq. 35
∞
∞
∞
𝜃(𝑒 𝑡 ) − 𝑒 𝑡 −𝑧𝑡
𝜃(𝑒 𝑡 ) − 𝑒 𝑡
𝑔(𝑧) = ∫ 𝑓(𝑡)𝑒 −𝑧𝑡 𝑑𝑡 = ∫
𝑒
𝑑𝑡
=
∫
𝑑𝑡 =
𝑒𝑡
𝑒 𝑡(𝑧+1)
0
0
0
∞
∞
𝜃(𝑒 𝑡 )
𝑒𝑡
Φ(𝑧 + 1) 1
∫ 𝑡(𝑧+1) 𝑑𝑡 − ∫ 𝑡(𝑧+1) 𝑑𝑡 =
−
𝑧+1
𝑧
0 𝑒
0 𝑒
This shows that 𝑔(𝑧) is analytic by Result 3, and by Result 2 we have that 𝑓(𝑡) is bounded,
and by definition it resembles a step function and so is continuous almost everywhere. So
∞
the Analytic Theorem provides us that ∫0 𝑓(𝑡)𝑑𝑡 converges.
Analytic Theorem:
Let 𝑓(𝑡), (𝑡 ≥ 0) be a bounded and locally integrable function and
∞
suppose that the function (𝑧) = ∫0 𝑓(𝑡)𝑒 −𝑧𝑡 𝑑𝑡 , ℜ(𝑧) > 0 extends
∞
holomorphically to ℜ(𝑧) ≥ 0. Then ∫0 𝑓(𝑡)𝑑𝑡 exists (and equals
𝑔(0)).4
Zagier goes onto show the final step here, and then proves the Analytic Theorem; we will
instead not prove the Analytic Theorem as it strays too far afield into complex analysis for
the purposes of this project.
So now we will apply a change of variable to 𝑓(𝑡) to get the integral in which we were first
interested.
Result 5:
𝑑𝑥
𝑥 = 𝑒𝑡,
= 𝑑𝑡
𝑥
∞
∞
∞
𝜃(𝑒 𝑡 ) − 𝑒 𝑡
𝜃(𝑥) − 𝑥
∫ 𝑓(𝑡)𝑑𝑡 = ∫
𝑑𝑡
=
∫
𝑑𝑥 convergent
𝑒𝑡
𝑥2
0
0
1
Now that we have established these five results, we can go on to finish the PNT by showing
𝜃(𝑥)~𝑥.
First we will assume that there is some 𝜆 > 1 such that 𝜃(𝑥) ≥ 𝜆𝑥, for an arbitrarily large 𝑥,
from this we would have
Eq. 36
4
Zagier, 4
𝜆𝑥
∫
𝑥
𝑛−1
𝜆𝑥
𝜃(𝑥) − 𝑡
𝜆𝑥 − 𝑡
𝜆𝑥 − 𝑡𝑖∗
(𝑡𝑖+1 − 𝑡𝑖 )
𝑑𝑡 ≤ ∫
𝑑𝑡 = lim ∑
𝑛→∞
𝑡2
𝑡2
𝑡𝑖∗ 2
𝑥
𝑖=0
𝑡𝑖∗ = 𝜆∗𝑖 𝑥, 𝑡𝑖 = 𝜆𝑖 𝑥, 𝑡𝑖+1 = 𝜆𝑖+1 𝑥, 1 ≤ 𝜆∗𝑖 , 𝜆𝑖 , 𝜆𝑖+1 ≤ 𝜆
𝑛−1
lim ∑
𝑛→∞
𝑖=0
𝜆𝑥 − 𝑡𝑖∗
𝑡𝑖∗ 2
𝑛−1
(𝑡𝑖+1 − 𝑡𝑖 ) = lim ∑
𝑛→∞
𝑛−1
𝜆 − 𝜆∗𝑖
lim ∑ ∗ 2 (𝜆𝑖+1
𝑛→∞
𝜆𝑖
𝑖=0
𝜆𝑥 − 𝜆∗𝑖 𝑥
𝜆∗𝑖 2 𝑥 2
𝑖=0
𝜆
− 𝜆𝑖 ) = ∫
1
(𝜆𝑖+1 𝑥 − 𝜆𝑖 𝑥) =
𝜆−𝑡
𝑑𝑡 > 0
𝑡2
But this presents a problem, if any interval with bounds being some integer to 𝜆 times that
integer is equal to the integral from 1 to 𝜆, then we have
Eq. 37
𝑛
𝑛
∞
𝜆𝑖+1
𝜆
𝜆
𝜃(𝑥) − 𝑥
𝜆−𝑡
𝜆−𝑡
𝜆−𝑡
∫
𝑑𝑥
=
lim
∑
∫
𝑑𝑡
=
lim
∑
∫
𝑑𝑡
=
lim
𝑛
∫
𝑑𝑡 → ∞
𝑛→∞
𝑛→∞
𝑛→∞
𝑥2
𝑡2
𝑡2
𝑡2
1
𝜆𝑖
1
1
𝑖=0
𝑖=0
This contradicts Result 5. By an equivalent argument, we can show that if instead we have
𝜆 < 1 such that 𝜃(𝑥) ≤ 𝜆𝑥, we run into the same contradiction (except this time with −∞).
Therefore, we get our final result .
Result 6:
𝜃(𝑥)~𝑥
Now all that is left is to relate our findings to the prime counting function 𝜋(𝑥).
Eq. 38
𝜃(𝑥) = ∑ log(𝑝) ≤ ∑ log(𝑥) = 𝜋(𝑥) log(𝑥)
𝑝≤𝑥
𝜃(𝑥) = ∑ log(𝑝) ≥
𝑝≤𝑥
𝑝≤𝑥
∑
𝑥 1−𝜖 ≤𝑝≤𝑥
log(𝑝) ≥
∑
(1 − 𝜖) log(𝑥) =
𝑥 1−𝜖 ≤𝑝≤𝑥
(1 − 𝜖) log(𝑥) (𝜋(𝑥) + 𝑂(𝑥1−𝜖 )) → 𝜋(𝑥) log(𝑥) , as ϵ → 0
So we have the long sought after result.
PNT
𝜋(𝑥) log(𝑥) ~𝑥
x
𝜋(𝑥)~
log(𝑥)
And so, we have established the PNT by Newman’s Proof.
Soprounov’s proof follows in almost the exact same way, except for a few differences.
Now we will explain Soprounov’s proof of the PNTAP. First though, we must define two
functions: the Dirichlet character, and the Dirichlet L-series.
Defintion 4: Dirichlet Characters5
𝜒𝑘 mod 𝑚 : ℤ → ℂ
𝜒𝑘 (𝑎) = 𝑧
𝜒𝑘 (𝑎)𝜒𝑘 (𝑏) = 𝜒𝑘 (𝑎𝑏)
𝑎 ≡ 𝑏 (𝑚) ⇒ 𝜒𝑘 (𝑎) = 𝜒𝑘 (𝑏)
0 if gcd(𝑛, 𝑚) > 1
𝜒𝑘 (𝑛) mod 𝑚 ≝ { 2𝜋𝑖𝑟
} , 𝑘 ∈ ℤ𝜙(𝑚) , 𝑟 ∈ ℤ
𝑒 𝜙(𝑚) if gcd(𝑛, 𝑚) = 1
𝜒𝑘 (1) = 1
𝜒0 (𝑛) = 𝜒0 = 1 principle character
An important thing to note from the definition is that |𝜒| = 1 or 0 regardless of the integers
used in its construction. Now that we have this, we will define the Dirichlet L-series.
Definition 5: Dirichlet L-series
∞
𝜒𝑘 (𝑛)
𝐿(𝑠, 𝜒𝑞 ) ≝ ∑
𝑛𝑠
𝑛=1
Dirichlet used these two functions along with the Riemann zeta function to show that the
number of primes in an arithmetic progression is infinite.6 We must now begin our work on
the PNTAP. First we will establish Euler’s product formula for the Dirichlet L-series.
As before, let us examine the expanded series
Eq. 39
∞
𝜒𝑘 (𝑛)
𝜒𝑘 (𝑛2 ) 𝜒𝑘 (𝑛3 ) 𝜒𝑘 (𝑛4 ) 𝜒𝑘 (𝑛5 )
𝐿(𝑠, 𝜒𝑘 ) = ∑
= 1+
+
+
+
+⋯
𝑠
𝑛
𝑛2 𝑠
𝑛3 𝑠
𝑛4 𝑠
𝑛5 𝑠
𝑛=1
𝑛𝑖 ∤ 𝑚, 𝜒𝑘 mod 𝑞
Now we will attempt the same step as before, we will find the first prime in the list and
obtain
Eq. 40
∞
𝜒𝑘 (𝑝)
𝜒𝑘 (𝑝)
𝜒𝑘 (𝑛)
)
𝐿(𝑠,
𝜒
=
∑
=
𝑘
𝑝𝑠
𝑝𝑠
𝑛𝑠
𝑛=1
𝜒𝑘 (𝑝) 𝜒𝑘 (𝑝)𝜒𝑘 (𝑛2 ) 𝜒𝑘 (𝑛3 )𝜒𝑘 (𝑝) 𝜒𝑘 (𝑛4 )𝜒𝑘 (𝑝) 𝜒𝑘 (𝑛5 )𝜒𝑘 (𝑝)
+
+
+
+
+⋯=
𝑝𝑠
𝑝 𝑠 𝑛2 𝑠
𝑝 𝑠 𝑛3 𝑠
𝑝 𝑠 𝑛4 𝑠
𝑝 𝑠 𝑛5 𝑠
𝜒𝑘 (𝑝) 𝜒𝑘 (𝑝𝑛2 ) 𝜒𝑘 (𝑝𝑛3 ) 𝜒𝑘 (𝑝𝑛4 ) 𝜒𝑘 (𝑝𝑛5 )
+ 𝑠 𝑠 + 𝑠 𝑠 + 𝑠 𝑠 + 𝑠 𝑠 +⋯
𝑝𝑠
𝑝 𝑛2
𝑝 𝑛3
𝑝 𝑛4
𝑝 𝑛5
Now subtracting Eq. 40 from Eq. 39
5
6
LeVeque, 159
LeVeque, 155
Eq. 41
𝐿(𝑠, 𝜒𝑘 ) −
𝜒𝑘 (𝑝)
𝜒𝑘 (𝑝)
𝐿(𝑠, 𝜒𝑘 ) = (1 −
) 𝐿(𝑠, 𝜒𝑘 )
𝑠
𝑝
𝑝𝑠
This removes all the multiples of 𝑝 from the right-hand side, we would then pick another
prime and repeat, and in this way we will sieve out all the primes and solve for 𝐿(𝑠, 𝜒𝑞 ),
and obtain
Result 7a:
𝐿(𝑠, 𝜒𝑘 ) = ∏(1 − 𝜒𝑘 (𝑝)𝑝−𝑠 )−1
𝑝
Furthermore, in the case of 𝜒𝑘 = 𝜒0
Result 7b:
𝐿(𝑠, 𝜒0 ) = ∏(1 − 𝑝−𝑠 )−1 = ∏(1 − 𝑝−𝑠 ) ∏(1 − 𝑝−𝑠 )−1 = ∏(1 − 𝑝−𝑠 ) 𝜁(𝑠)
𝑝∤𝑚
𝑝|𝑚
𝑝
Now we will show that 𝐿(𝑠, 𝜒𝑘 ), and 𝐿(𝑠, 𝜒0 ) −
𝜙(𝑞) 1
𝑞
𝑠−1
𝑝|𝑚
is analytic on {𝑠 ∈ ℂ: ℜ(𝑠) > 0}. Let
Soprounov takes a partial sum from the Dirichlet L-series and changes into the following. I
was unable to figure out how he produced this change, so I am forced to take on faith that it
is true.
Eq. 42
𝑥
𝑥
𝜒𝑘 (𝑛) 𝐴(𝑥)
𝐴(𝑡)
∑
=
+
𝑠
∫
𝑑𝑡
𝑠
𝑠
𝑠+1
𝑛
𝑥
1 𝑡
𝑛=1
Definition 6:
𝑥
𝐴(𝑥) ≝ ∑ 𝜒𝑘 (𝑛)
𝑛=1
If 𝜒𝑘 (𝑛) is not principal then 𝐴(𝑥) is bounded by a finite number of roots of unity, as the
sum over the mth roots of unity hits zero every 𝜙(𝑚) times. So we will let 𝑥 → ∞ to obtain
Eq. 43
∞
∞
𝐴(𝑡)
𝐴(𝑡)
𝐿(𝑠, 𝜒𝑘 ) = 𝑠 ∫ 𝑠+1 𝑑𝑡 = 𝑠 ∫ 𝑠+1 𝑑𝑡
1 𝑡
1 𝑡
This is convergent, and therefore analytic for ℜ(𝑠) > 0. For the case of 𝜒𝑘 (𝑛) being
principal we can look at Result 7b, or even the definition of 𝐿(𝑠, 𝜒𝑘 ), and see that it is
meromorphic, with only a simple pole at 𝑠 = 1 with residue being
Eq. 44
𝑝−1
𝜙(𝑝)
𝜙(𝑚)
∏(1 − 𝑝−1 ) = ∏ (
) = ∏(
)=
𝑝
𝑝
𝑚
𝑝|𝑚
𝑝|𝑚
𝑝|𝑚
This gives us all the information needed to say
Result 8:
𝜙(𝑞) 1
𝐿(𝑠, 𝜒𝑘 ), 𝐿(𝑠, 𝜒0 ) −
are analytic on {𝑠 ∈ ℂ: ℜ(𝑠) > 0}
𝑞 𝑠−1
Now we will show that 𝜃𝑞 (𝑥) = 𝑂(𝑥), but first we must define it
Definiton 7:
𝜃𝑞 (𝑥) ≝ 𝜙(𝑚) ∑ log(𝑝)
𝑝≤𝑥
𝑝≡𝑎 (𝑚)
This will be easy enough to show.
Result 9:
𝜃𝑞 (𝑥) ≤ 𝜙(𝑚)𝜃(𝑥) = 𝑂(𝑥) ⇒ 𝜃𝑞 (𝑥) = 𝑂(𝑥)
Now we will examine
Definition 8:
𝜒𝑘 (𝑝) log 𝑝
𝜙(𝑠, 𝜒𝑘 ) ≝ ∑
𝑝𝑠
𝑝
Φ𝑚 (𝑠) ≝ ∑ 𝜙(𝑠, 𝜒𝑘 )
𝜒𝑘
Φ𝑚,𝑎 (𝑠) ≝ ∑ ̅̅̅̅̅̅̅̅
𝜒𝑘 (𝑎)𝜙(𝑠, 𝜒𝑘 ) , ̅̅̅̅̅̅̅̅
𝜒𝑘 (𝑎) conjugate of 𝜒𝑘 (𝑎)
𝜒𝑘
We want to show that the Dirichlet L-series has no zeroes for {𝑠 ∈ ℂ: ℜ(𝑠) ≥ 1} for any 𝜒𝑘 ,
1
also we will show that Φ𝑚 (𝑠) − 𝑠−1 is analytic there as well.
In order to guarantee our desired result for any 𝜒, we will look at the a new function.
Definition 9:
ℒ(𝑠) ≝ ∏ 𝐿(𝑠, 𝜒𝑘 )
𝜒
Soprounov takes as given, as will we, that 𝐿(1, 𝜒𝑘 ) ≠ 0 provided that 𝜒𝑘 is not principal.
So let us now again to the logarithmic derivative of ℒ(𝑠).
Eq. 45
𝐿′(𝑠, 𝜒𝑘 ) ∏𝜒 𝐿(𝑠, 𝜒𝑘 )
𝑑
∑
(
)
𝜒
𝐿(𝑠, 𝜒𝑘 )
ℒ ′ (𝑠) 𝑑𝑠 ∏𝜒 𝐿(𝑠, 𝜒𝑘 )
𝐿′(𝑠, 𝜒𝑘 )
=
=
=
∏𝜒 𝐿(𝑠, 𝜒𝑘 )
∏𝜒 𝐿(𝑠, 𝜒𝑘 )
ℒ(𝑠)
𝐿(𝑠, 𝜒𝑘 )
Therefore, we need only examine the logarithmic derivative of the Dirichlet L-series.
Eq. 46
𝑑
𝑑
log(𝐿(𝑠, 𝜒𝑘 )) = − log (∏(1 − 𝜒𝑘 (𝑝)𝑝−𝑠 )−1 )
𝑑𝑠
𝑑𝑠
𝑝
We will use the rules of logarithms as before.
Eq. 47
−
𝐿′(𝑠, 𝜒𝑘 )
𝑑
𝑑
= − log (∏(1 − 𝜒𝑘 (𝑝)𝑝−𝑠 )−1 ) = − ∑ log((1 − 𝜒𝑘 (𝑝)𝑝−𝑠 )−1 ) =
𝐿(𝑠, 𝜒𝑘 )
𝑑𝑠
𝑑𝑠
𝑝
𝑝
𝜒𝑘 (𝑝)𝑝−𝑠 log(p)
𝜒𝑘 (𝑝) log(𝑝)
(1 − 𝜒𝑘 (𝑝)𝑝−𝑠 )2
−∑
=∑ 𝑠
−𝑠
−1
(1 − 𝜒𝑘 (𝑝)𝑝 )
𝑝 − 𝜒𝑘 (𝑝)
𝑝
𝑝
We now use the exact same methods as in the proof for the PNT.
Eq. 48
𝜒𝑘 (𝑝) log(𝑝)
𝜒𝑘 (𝑝) log(𝑝) 𝜒𝑘 (𝑝) log(𝑝) 𝜒𝑘 (𝑝) log(𝑝)
∑ 𝑠
=∑ 𝑠
−
+
=
𝑝 − 𝜒𝑘 (𝑝)
𝑝 − 𝜒𝑘 (𝑝)
𝑝𝑠
𝑝𝑠
𝑝
𝑝
Φm (𝑠) + ∑
𝑝
𝜒𝑘2 (𝑝) log(𝑝)
𝑝 𝑠 (𝑝 𝑠 − 𝜒𝑘 (𝑝))
The analyticity for the term on the right is essentially the same as before, so like both
authors, we will take it as given.
Combined with Result 8, this establishes that the rightmost side is at least meromorphic, that is
to say, analytic with a specific set of poles. The rest of the proof of this claim follows
indentically to the proof for Result 4.
Result 10:
𝜁(𝑠) ≠ 0, ∀𝑠 ∈ ℂ s. t. ℜ(𝑠) ≥ 1
Result 8 and Result 10 give
1
Φ𝑚,𝑎 (𝑠) − 𝑠−1
∞ 𝜃𝑞 (𝑥)−𝑥
Now we will show that ∫1
𝑥2
Result 11:
as analytic on {𝑠 ∈ ℂ: ℜ(𝑠) ≥ 1}.
𝑑𝑥 is convergent.
Eq. 49
𝜒𝑘
̅̅̅̅̅̅̅̅
Φ𝑚,𝑎 (𝑠) = ∑ ̅̅̅̅̅̅̅̅
𝜒𝑘 (𝑎)𝜙(𝑠, 𝜒𝑘 ) = ∑ (𝜒
𝑘 (𝑎) ∑
𝜒𝑘
𝜒𝑘
̅̅̅̅̅̅̅̅
∑ (𝜒
𝑘 (𝑎) ∑
𝜒𝑘
𝑝
𝑝
(𝑝) log 𝑝
)=
𝑝𝑠
𝜒𝑘 (𝑝) log 𝑝
)
𝑝𝑠
The ̅̅̅̅̅̅̅̅
𝜒𝑘 (𝑎) can also be written as 𝜒𝑘 (𝑎−1 mod 𝑚), by the multiplicative property of 𝜒𝑘 , so it
works like a filter with 𝜒𝑘 (𝑝) leaving us with only the primes congruent to 𝑎 modulo m.
Eq. 50
∑ ̅̅̅̅̅̅̅̅
𝜒𝑘 (𝑎)𝜒𝑘 (𝑝) =
𝜒𝑘
∑
1 = 𝜙(𝑚)
𝑎−1 𝑝≡1(𝑚)
So now we are left with turning Φ𝑚,𝑎 (𝑠) into something familiar.
Eq. 51
log 𝑝
log 𝑝
∑ 𝜙(𝑚) 𝑠 = ∑ 𝜙(𝑚) 𝑠
𝑝
𝑝
−1
𝑎
𝑝≡𝑎(𝑚)
𝑝≡1(𝑚)
This is reminiscent of Φ(𝑠) from the proof of the PNT, and indeed it can be transformed
into an integral in the same manner. So now we have, with a familiar change of variable
Eq. 52
∞ 𝜃 (𝑥)
∞ 𝜃 (𝑒 𝑡 )
𝑞
𝑞
Φ𝑚,𝑎 (𝑠) = 𝑠 ∫
𝑑𝑥
=
𝑠
∫
𝑑𝑥
𝑠+1
𝑥
𝑒 𝑠𝑡
1
1
From here we proceed in the exact same way as in the proof for the PNT, using the Analytic
Theorem we get that indeed
Result 12:
∞ 𝜃 (𝑥) − 𝑥
𝑞
∫
𝑑𝑥 convergent
𝑥2
1
From this point on, the Soprounov proof says nothing new from the Newman proof, so we
will cut to the chase and get to the finale.
PNTAP
𝜋(𝑥, 𝑞) =
𝑥
𝜙(𝑞) log 𝑥
Now I explain the program by which I will test these results.
def PNTAP_test1(x,q,a):
assert gcd(q,a) == 1
finish_q = int(Mod(x,q))
tries = (x-finish_q)/q + 1
hits = 0
for k in range(tries) :
y = q*k+a
if y > x:
break
if is_prime(y):
hits += 1
estimate = N(x/(euler_phi(q)*log(x)))
abs_error = abs(hits-estimate)
rel_error = abs_error/hits
return [hits,estimate.n(digits=5), \
abs_error.n(digits=5), \
rel_error.n(digits=5)]
Step 1: Make sure that 𝑎 and 𝑞 are coprime
The range of numbers to be tested will be partitioned into testing grounds by the
arithemetic progression, so we will see how much is cropped off by the cap of the range.
Step 2: Get the distance between the beginning of the final interval and the cap
Step 3: Calculate the number of prime tests that will be conducted
Step 4: Get the value, y of the arithmetic progression for the current test
Step 5: If we have gone too far skip to Step 8
Step 6: If y is prime add one to the hit count
Step 7: Repeat Steps 4-6 until all the tests are done
Step 8: Calculate the estimate according to the PNTAP
Step 9: Calculate the absolute error
Step 10: Calculate the relative error
Return: Number of hits, estimate, absolute error, relative error
def PNTAP_test2(x,q):
phi_q = euler_phi(q)
cong_classes = q.coprime_integers(q)
hit_data = []
estimate = N(x/(phi_q*log(x)))
abs_error_data = []
rel_error_data = []
for k in range(phi_q):
a = cong_classes[k]
data = PNTAP_test1(x,q,a)
hit_data += [data[0]]
abs_error_data += [data[2]]
rel_error_data += [data[3]]
hit_mean = N(sum(hit_data)/phi_q)
abs_error_mean = N(sum(abs_error_data)/phi_q)
rel_error_mean = N(sum(rel_error_data)/phi_q)
hit_stddev = 0
abs_error_stddev = 0
rel_error_stddev = 0
for k in range(phi_q):
hit_stddev += N(sqrt((phi_q-1)^-1*(hit_data[k]-hit_mean)^2))
abs_error_stddev += N(sqrt((phi_q-1)^-1*\
(abs_error_data[k]-abs_error_mean)^2))
rel_error_stddev += N(sqrt((phi_q-1)^-1*\
(rel_error_data[k]-rel_error_mean)^2))
return [[hit_mean.n(digits=5),hit_stddev.n(digits=5)], \
[estimate.n(digits=5)], \
[abs_error_mean.n(digits=5), \
abs_error_stddev.n(digits=5)], \
[rel_error_mean.n(digits=5), \
rel_error_stddev.n(digits=5)]]
Step 1: Calculate the necessary values, 𝜙(𝑞), the coprime integers, and the estimate
Step 2: Pick a congruency class
Step 3: Obtain data on the given arithmetic progression for all the congruency classes
Step 4: Distribute the data among the hit data, absolute errors, and relative errors
Step 5: Calculate the mean for the hit data, absolute errors, and relative errors
Step 6: Calculate the standard deviation for the hit data, absolute errors, and relative errors
Return: hit mean, hit standard deviation, absolute error mean, absolute error standard
deviation, relative error mean, and relative error standard deviation
First, we will test the estimate with a particular congruency class (PNTAP_test1).
I will use the following arithmetic progressions to explore how far out we need to go to get
close results.
TEST 1
4x+1, x=[103, 105, 107]
[hits, estimate, mean, absolute error, relative error]
[80, 72.382, 7.6176, 0.095220] x= 103
[4783, 4342.9, 440.06, 0.092004] x= 105
[332180, 310210., 21970., 0.066138] x= 107
TEST 2
27x+8, cap=[103, 105, 107, 109]
[hits, estimate, mean, absolute error, relative error]
[8, 8.0425, 0.042490, 0.0053113] x= 103
[543, 482.55, 60.451, 0.11133] x= 105
[36912, 34468., 2444.2, 0.066216] x= 107
[2824735, 2.6808e6, 143900., 0.050945] x= 109
TEST 3
90x+37, x=[103, 105, 107, 109]
[hits, estimate, mean, absolute error, relative error]
[8, 6.0319, 1.9681, 0.24602] x= 103
[405, 361.91, 43.088, 0.10639] x= 105
[27734,25851, 1883.1, 0.067900] x= 107
[2119211, 2.0106e6, 108590., 0.051240] x= 109
TEST 4
105x+16, x=[103, 105, 107, 109]
[hits, estimate, mean, absolute error, relative error]
[3, 3.0159, 0.015934, 0.0053113] x= 103
[405, 361.91, 43.088, 0.10639] x= 105
[13835,12925., 909.57, 0.065744] x= 107
[1059463, 1.0053e6, 54152., 0.051112] x= 109
We see that, as expected, the relative error decreases as 𝑥 grows. 27x+8 is odd in that for x
= 103 it is unusually accurate, and unlike 105x+16, it can’t be explained as it being a very
small number.
Now we will try the second fact obtained from the PNTAP (PNTAP_test2) with two
arithmetic progressions.
TEST 5
m = 27, x=[103, 105, 107]
[[hit mean,hit stddev],[estimate],
[abs error mean,abs error stddev],
[rel error mean,rel error stddev]]
x=103 [[9.2778, 5.0932], [8.0425],
[1.4858, 3.9994],
[0.14958, 0.34342]]
x=105 [[532.83, 22.798], [482.55],
[50.284, 22.798],
[0.094220, 0.039111]]
x=107 [[36921., 170.26], [34468.],
[2453.2, 170.26],
[0.066443, 0.0043041]]
TEST 6
m = 42, x=[103, 105, 107]
[[hit mean,hit stddev],[estimate],
[abs error mean,abs error stddev],
[rel error mean,rel error stddev]]
x=103 [[13.750, 4.0704], [12.064],
[1.7181, 3.9359],
[0.11735, 0.25734]]
x=105 [[799.08, 24.473], [723.82],
[75.259, 24.473],
[0.094098, 0.027691]]
x=107 [[55381., 93.870], [51702.],
[3679.6, 93.874],
[0.066441, 0.0015834]]
The results for this test are less satisfactory, but still we see the means behave like we saw
the specific congruency classes from tests 1 through 4.
This result is subtly amazing. According to the PNTAP, the number of primes
congruent to a number modulo another depends only on the modulus, which means that
the primes are evenly distributed among the congruency classes of the multiplicative group
of the modulus. This hints at a structure to what otherwise appears to be a randomly
distributed set of numbers, the primes.
Bibliography
LeVeque, William J. Fundamentals of Number Theory. New York: Dover, 1996.
Soprounov, Ivan. A Short Proof of the Prime Number Theorem for Arithmetic Progressions.
[http://www.math.umass.edu/~isoprou/pdf/primes.pdf]. 3 March 2010
Zagier, D. Newman’s Short Proof of the Prime Number Theorem.
[http://mathdl.maa.org/images/upload_library/22/Chauvenet/Zagier.pdf]. 4 March
2010