A Direct Proof of the
Prime Number Theorem
Stephen Lucas
Department of Mathematics and Statistics
James Madison University, Harrisonburg VA
The primes . . . those exasperating, unruly integers that refuse to be divided evenly by any integers
except themselves and one.
Martin Gardner
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Outline
Why me?
What is the Prime Number Theorem.
History of Prime Number Theorem.
Transforms.
A direct proof.
Riemann’s and the “exact” forms.
Number of distinct primes.
Conclusion.
Thanks to Ken Lever (Cardiff), Richard Martin (London)
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Research Projects
Numerical Analysis
Accurately finding multiple roots.
Euler-Maclaurin-like summation for Simpson’s rule.
Placing a circle pack.
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Research Projects
Numerical Analysis
Accurately finding multiple roots.
Euler-Maclaurin-like summation for Simpson’s rule.
Placing a circle pack.
Applied Math
Deformation of half-spaces.
Micro-pore diffusion in a finite volume.
Froth Flotation.
Deformation of spectacle lenses.
Direct PNT – p. 3/28
Research Projects
Numerical Analysis
Accurately finding multiple roots.
Euler-Maclaurin-like summation for Simpson’s rule.
Placing a circle pack.
Applied Math
Deformation of half-spaces.
Micro-pore diffusion in a finite volume.
Froth Flotation.
Deformation of spectacle lenses.
Pure Math
Fractals in regular graphs.
Integral approximations to π.
The Hamiltonian cycle problem.
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Research Projects
Numerical Analysis
Accurately finding multiple roots.
Euler-Maclaurin-like summation for Simpson’s rule.
Placing a circle pack.
Applied Math
Deformation of half-spaces.
Micro-pore diffusion in a finite volume.
Froth Flotation.
Deformation of spectacle lenses.
Pure Math
Fractals in regular graphs.
Integral approximations to π.
The Hamiltonian cycle problem.
New
Improving kissing sphere bounds.
Counting cycles in graphs.
Integer-valued logistic equation.
Fitting data to predator-prey.
Representing reals using bounded
continued fractions.
Analysis of “Dreidel”.
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The Prime Number Theorem
A prime number is any integer ≥ 2 with no divisors except itself and one.
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The Prime Number Theorem
A prime number is any integer ≥ 2 with no divisors except itself and one.
Let π(x) be the number of primes less than or equal to x. (π(x) → ∞ as
x → ∞)
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The Prime Number Theorem
A prime number is any integer ≥ 2 with no divisors except itself and one.
Let π(x) be the number of primes less than or equal to x. (π(x) → ∞ as
x → ∞)
The prime number theorem states that
x
π(x) ∼
ln x
or
π(x)
lim
= 1.
x→∞ x/ ln x
Direct PNT – p. 4/28
The Prime Number Theorem
A prime number is any integer ≥ 2 with no divisors except itself and one.
Let π(x) be the number of primes less than or equal to x. (π(x) → ∞ as
x → ∞)
The prime number theorem states that
x
π(x) ∼
ln x
or
π(x)
lim
= 1.
x→∞ x/ ln x
Another form states that
Z x
dt
π(x) ∼ −
0 ln t
(= li(x))
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Plot of π(x), x/ ln x, li(x)
1014
10
12
10
10
100
-1
10
x/ln x
-3
10
10
8
-4
10
10
6
10-5
Relative Error
Number of Primes
-2
10
li(x)
104
-6
10
10
2
-7
10
0
10 1
10
-8
10
3
10
5
7
10
x
10
9
10
11
13
10
10
15
10
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History of the PNT
Legendre (1798) conjectured π(x) =
x
.
A ln x + B
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History of the PNT
Legendre (1798) conjectured π(x) =
x
.
A ln x + B
Legendre (1808) used numerical evidence to claim
x
π(x) =
, where lim A(x) = 1.08366 . . ..
x→∞
ln x + A(x)
Direct PNT – p. 6/28
History of the PNT
Legendre (1798) conjectured π(x) =
x
.
A ln x + B
Legendre (1808) used numerical evidence to claim
x
π(x) =
, where lim A(x) = 1.08366 . . ..
x→∞
ln x + A(x)
Gauss (1849, as early as 1792) used numerical evidence to
conjecture that
Z x
dt
.
π(x) ∼ −
0 ln t
Direct PNT – p. 6/28
History of the PNT
Legendre (1798) conjectured π(x) =
x
.
A ln x + B
Legendre (1808) used numerical evidence to claim
x
π(x) =
, where lim A(x) = 1.08366 . . ..
x→∞
ln x + A(x)
Gauss (1849, as early as 1792) used numerical evidence to
conjecture that
Z x
dt
.
π(x) ∼ −
0 ln t
Dirichlet (1837) introduced Dirichlet series:
∞
X
f (n)
b
.
f (s) =
s
n
n=1
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History of the PNT (cont’d)
Chebyshev (1851) introduced
X
ln p,
θ(x) =
ψ(x) =
X
ln p,
pm ≤x
p≤x
and showed that the PNT is equivalent to
θ(x)
= 1,
lim
x→∞ x
ψ(x)
lim
= 1.
x→∞ x
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History of the PNT (cont’d)
Chebyshev (1851) introduced
X
ln p,
θ(x) =
ψ(x) =
X
ln p,
pm ≤x
p≤x
and showed that the PNT is equivalent to
θ(x)
= 1,
lim
x→∞ x
ψ(x)
lim
= 1.
x→∞ x
Riemann (1860) introduced the Riemann zeta function:
−1
∞
Y
X
1
1
=
,
1− s
ζ(s) =
s
n
p
p
n=1
<(s) > 1.
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Riemann Zeta Function
Integrals extend to whole plane by analytic continuation.
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Riemann Zeta Function
Integrals extend to whole plane by analytic continuation.
1
+ γ + γ1 (s − 1) + · · ·.
Only singularity at s = 1, ζ(s) =
s−1
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Riemann Zeta Function
Integrals extend to whole plane by analytic continuation.
1
+ γ + γ1 (s − 1) + · · ·.
Only singularity at s = 1, ζ(s) =
s−1
Showed using residue calculus that
ψ(x) = x −
X xρ
ρ
ζ 0 (0) 1
−2
−
− ln 1 − x
,
ρ
ζ(0)
2
where ρ are the non-trivial zeros of ζ(s), so the PNT is equivalent to
1 X xρ
= 0.
lim
x→∞ x
ρ
ρ
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Riemann Zeta Function
Integrals extend to whole plane by analytic continuation.
1
+ γ + γ1 (s − 1) + · · ·.
Only singularity at s = 1, ζ(s) =
s−1
Showed using residue calculus that
ψ(x) = x −
X xρ
ρ
ζ 0 (0) 1
−2
−
− ln 1 − x
,
ρ
ζ(0)
2
where ρ are the non-trivial zeros of ζ(s), so the PNT is equivalent to
1 X xρ
= 0.
lim
x→∞ x
ρ
ρ
Riemann hypothesis: <(ρ) =
1
.
2
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First Proof of the PNT
Hadamard & de la Valèe Poussin (1896, independently) showed ∃a, t0
1
, |t| ≥ t0 , so
such that ζ(σ + it) 6= 0 if σ ≥ 1 −
a log |t|
−c(log x)1/14
.
ψ(x) = x + O xe
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Improvements to the PNT Proof
Mertens (1898) gave a short proof that ζ(s) 6= 0, <(s) = 1.
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Improvements to the PNT Proof
Mertens (1898) gave a short proof that ζ(s) 6= 0, <(s) = 1.
Ikehara (1930) & Wiener (1932) used Tauberian theorems to prove
Ikehara’s theorem:
∞
X
an
b
Let f (s) =
, {ai } real and nonnegative.
s
n
n=1
If fb(s) converges for <(s) > 1, and ∃A > 0 s.t. for all t ∈ R,
P
A
+
ˆ
→ finite limit as s → 1 + it, then n≤x an ∼ Ax.
f (s) −
(s − 1)
This can be used to show lim ψ(x)/x = 1 directly.
x→∞
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Improvements (cont’d)
Erdös & Selberg (1949) produced an “elementary” proof – no complex
analysis.
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Improvements (cont’d)
Erdös & Selberg (1949) produced an “elementary” proof – no complex
analysis.
Newman (1980) showed lim ψ(x)/x = 1 using straightforward
x→∞
contour integration.
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Improvements (cont’d)
Erdös & Selberg (1949) produced an “elementary” proof – no complex
analysis.
Newman (1980) showed lim ψ(x)/x = 1 using straightforward
x→∞
contour integration.
Lucas, Martin & Lever – current work, looks at π(x) directly.
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Laplace Transforms
L{f (x)} = f¯(s) =
Z
∞
e−sx f (x) dx.
0
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Laplace Transforms
L{f (x)} = f¯(s) =
Z
∞
e−sx f (x) dx.
0
1
−1 ¯
L {f (s)} = H(x)f (x) =
2πi
H is the unit step function.
Z
+i∞
f¯(s)esx ds.
−i∞
to the right of any singularities in f¯(s).
assume f¯(s) → 0 as |s| → ∞.
Direct PNT – p. 12/28
Laplace Transforms
L{f (x)} = f¯(s) =
Z
∞
e−sx f (x) dx.
0
1
−1 ¯
L {f (s)} = H(x)f (x) =
2πi
H is the unit step function.
Z
+i∞
f¯(s)esx ds.
−i∞
to the right of any singularities in f¯(s).
assume f¯(s) → 0 as |s| → ∞.
Choose ḡ(s) such that f¯(s) − ḡ(s) is analytic at the rightmost
singularity in f¯(s).
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Laplace Transforms
L{f (x)} = f¯(s) =
Z
∞
e−sx f (x) dx.
0
1
−1 ¯
L {f (s)} = H(x)f (x) =
2πi
H is the unit step function.
Z
+i∞
f¯(s)esx ds.
−i∞
to the right of any singularities in f¯(s).
assume f¯(s) → 0 as |s| → ∞.
Choose ḡ(s) such that f¯(s) − ḡ(s) is analytic at the rightmost
singularity in f¯(s). If ḡ(s) → 0 as |s| → ∞, then shift the integration
contour to the left, apply the Cauchy integral theorem, and get that
f (x) = g(x) + O(xc ), where c is the new position of . (e.g. Smith,
1966)
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Arithmetic Functions and Dirichlet Series
An arithmetic function is a map f : N → C.
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Arithmetic Functions and Dirichlet Series
An arithmetic function is a map f : N → C.
∞
X
f (n)
b
,
Its Dirichlet series is f (s) =
s
n
n=1
convergent for <(s) > c if f (n) = O(nc−1 ) as n → ∞.
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Arithmetic Functions and Dirichlet Series
An arithmetic function is a map f : N → C.
∞
X
f (n)
b
,
Its Dirichlet series is f (s) =
s
n
n=1
convergent for <(s) > c if f (n) = O(nc−1 ) as n → ∞.
Given that
1
2πi
Z
+i∞
−i∞
then
1
2πi
Z
+i∞
−i∞

 0, x < 1,
xs
ds =
 1, x > 1,
s
ds
b
x f (s)
s
s
> 0,

∞
 0, x < n,
X
=
f (n) ×
 1, x > n,
n=1
=
X
f (n).
1≤n≤x
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The Transform
We can recognize this as the inversion of a Laplace-like transform
Z ∞
(x = et ):
f (x)x−s−1 dx,
(Mf )(t) =
1
H(x − 1)f (x)
=
1
2πi
Z
+i∞
(Mf )(t)xt dt.
−i∞
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The Transform
We can recognize this as the inversion of a Laplace-like transform
Z ∞
(x = et ):
f (x)x−s−1 dx,
(Mf )(t) =
1
H(x − 1)f (x)
=
1
2πi
Z
+i∞
(Mf )(t)xt dt.
−i∞
Some useful transform pairs are:
M
←→
M(f )
1
1
log
γ + ln(ln x)
s
s
c
1
a
c
log
, <(s) > c > 0
x li(x )
s−a
s−a−c
1
, <(s) > c; k > 0
Γ(k)−1 xc (ln x)k−1
k
(s − c)
f
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Average Order Asymptotics
1 X
e
The average order of an arithmetic function is f (x) =
f (n).
x
1≤n≤x
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Average Order Asymptotics
1 X
e
The average order of an arithmetic function is f (x) =
f (n).
x
1≤n≤x
1
e
Then xf (x) =
2πi
Z
+i∞
−i∞
fb(s)xs−1 ds,
Direct PNT – p. 15/28
Average Order Asymptotics
1 X
e
The average order of an arithmetic function is f (x) =
f (n).
x
1≤n≤x
1
e
Then xf (x) =
2πi
Z
+i∞
−i∞
fb(s)xs−1 ds,
b
b
M f (s)
M f (s + 1)
e
e
and f (x) ←→
.
and xf (x) ←→
s
s+1
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Average Order Asymptotics
1 X
e
The average order of an arithmetic function is f (x) =
f (n).
x
1≤n≤x
1
e
Then xf (x) =
2πi
Z
+i∞
−i∞
fb(s)xs−1 ds,
b
b
M f (s)
M f (s + 1)
e
e
and f (x) ←→
.
and xf (x) ←→
s
s+1
So, to find the asymptotic form of an average order:
find its Dirichlet series fb(s) in closed form,
fb(s)
,
identify the position and form of the singularities in
s
sum the inverse transforms of the singular parts.
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Riemann and Prime Zeta-Functions
The Riemann zeta-function is the Dirichlet series of f (n) = 1:
∞
Y
X
1
1
=
.
ζ(s) =
s
−s
n
1
−
p
p
n=1
Direct PNT – p. 16/28
Riemann and Prime Zeta-Functions
The Riemann zeta-function is the Dirichlet series of f (n) = 1:
∞
Y
X
1
1
=
.
ζ(s) =
s
−s
n
1
−
p
p
n=1
ζ(s) only singularity ∼
1
at s = 1, and ζ(s) 6= 0 for <(s) ≥ 1.
s−1
Direct PNT – p. 16/28
Riemann and Prime Zeta-Functions
The Riemann zeta-function is the Dirichlet series of f (n) = 1:
∞
Y
X
1
1
=
.
ζ(s) =
s
−s
n
1
−
p
p
n=1
ζ(s) only singularity ∼
1
at s = 1, and ζ(s) 6= 0 for <(s) ≥ 1.
s−1
The prime zeta-function is the Dirichlet series of

 1, n prime,
X 1
.
f (n) =
: P (s) =
s
 0, otherwise,
p
p
Direct PNT – p. 16/28
The Prime Zeta-Function
Now
log ζ(s) =
X
p
log
1
1 − p−s
=
∞
XX
p
1
ms
mp
m=1
∞
X
1
P (ms).
=
m
m=1
Direct PNT – p. 17/28
The Prime Zeta-Function
Now
log ζ(s) =
X
p
log
1
1 − p−s
=
∞
XX
p
1
ms
mp
m=1
∞
X
1
P (ms).
=
m
m=1
Using the Möbius inversion formula,
∞
X
µ(n)
log ζ(ns),
P (s) =
n
n=1

 0,
n contains a square factor,
where µ(n) =
 (−1)m , n a product of m distinct primes.
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The Prime Number Theorem
Let

 1, n prime,
f (n) =
 0, otherwise,
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The Prime Number Theorem
Let
Then

 1, n prime,
f (n) =
 0, otherwise,
fb(s) = P (s)
and
xfe(x) = π(x).
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The Prime Number Theorem
Let
Then
So

 1, n prime,
f (n) =
 0, otherwise,
fb(s) = P (s)
and
xfe(x) = π(x).
P (s)
.
π(x) ←→
s
M
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The Prime Number Theorem
Let
Then
So

 1, n prime,
f (n) =
 0, otherwise,
fb(s) = P (s)
and
xfe(x) = π(x).
P (s)
.
π(x) ←→
s
M
1
1
P (s)
is log
at s − 1, whose inverse
The rightmost singularity of
s
s
s−1
transform is li(x).
Direct PNT – p. 18/28
The Prime Number Theorem
Let
Then
So

 1, n prime,
f (n) =
 0, otherwise,
fb(s) = P (s)
and
xfe(x) = π(x).
P (s)
.
π(x) ←→
s
M
1
1
P (s)
is log
at s − 1, whose inverse
The rightmost singularity of
s
s
s−1
transform is li(x).
So π(x) ∼ li(x).
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Riemann’s Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
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Riemann’s Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
1
P (s)
at s = , µ(n) 6= 0.
Singularity in ζ(s) at s = 1 means singularities in
s
n
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Riemann’s Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
1
P (s)
at s = , µ(n) 6= 0.
Singularity in ζ(s) at s = 1 means singularities in
s
n
1
µ(n)
, whose inverse
At these points, singularities take the form s log
n
ns − 1
li(x1/n )
.
transforms are µ(n)
n
Direct PNT – p. 19/28
Riemann’s Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
1
P (s)
at s = , µ(n) 6= 0.
Singularity in ζ(s) at s = 1 means singularities in
s
n
1
µ(n)
, whose inverse
At these points, singularities take the form s log
n
ns − 1
li(x1/n )
.
transforms are µ(n)
n
So
∞
X
µ(n) 1/n li x
= R(x).
π(x) ∼ li(x) +
n
n=2
Direct PNT – p. 19/28
“Exact” Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
Direct PNT – p. 20/28
“Exact” Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
Zero of
ζ(s) ats = ρm means a singularity in P (s)/s at s = ρm of the form
s
1
log
− 1 , whose inverse is −li(xρm ).
s
ρm
Direct PNT – p. 20/28
“Exact” Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
Zero of
ζ(s) ats = ρm means a singularity in P (s)/s at s = ρm of the form
s
1
log
− 1 , whose inverse is −li(xρm ).
s
ρm
As for Riemann’s form, singularity in ζ(s) at s = ρm also means
singularities inP (s) at s= ρm /n, µ(n) 6= 0. These singularities are of the
ns
µ(n)
ρm /n
log
− 1 , whose inverses are −µ(n)li x
n.
form
ns
ρm
Direct PNT – p. 20/28
“Exact” Form of π(x)
∞
X
µ(n)
log ζ(ns).
Recall P (s) =
n
n=1
Zero of
ζ(s) ats = ρm means a singularity in P (s)/s at s = ρm of the form
s
1
log
− 1 , whose inverse is −li(xρm ).
s
ρm
As for Riemann’s form, singularity in ζ(s) at s = ρm also means
singularities inP (s) at s= ρm /n, µ(n) 6= 0. These singularities are of the
ns
µ(n)
ρm /n
log
− 1 , whose inverses are −µ(n)li x
n.
form
ns
ρm
The contributions of all singularities related to s = ρm contribute −R(xρm )
to π(x), and so
k
X
R(xρm ).
π(x) = lim Rk (x) where Rk (x) = R(x) −
k→∞
m=−k
Direct PNT – p. 20/28
Number of Distinct Primes
Define ω(n) to be the number of distinct primes in the prime
decomposition of n, and Ω(n) to be the total number of primes.
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Number of Distinct Primes
Define ω(n) to be the number of distinct primes in the prime
decomposition of n, and Ω(n) to be the total number of primes.
αk
1 α2
If n = pα
1 p2 . . . pk (pi ’s some primes), then ω(n) = k and
Ω(n) = α1 + α2 + · · · + αk .
Direct PNT – p. 21/28
Number of Distinct Primes
Define ω(n) to be the number of distinct primes in the prime
decomposition of n, and Ω(n) to be the total number of primes.
αk
1 α2
If n = pα
1 p2 . . . pk (pi ’s some primes), then ω(n) = k and
Ω(n) = α1 + α2 + · · · + αk .
It is well known (e.g.Hardy
& Wright)
that ω
e (n) = ln(ln n) + B1 + o(1),
X
1
1
+
= 0.26149 72128 47642 . . .,
where B1 = γ +
ln 1 −
p
p
p
X
1
e
=ω
e (n) + B2 − B1 ,
and Ω(n) = ω
e (n) +
p(p
−
1)
p
where B2 = 1.03465 38818 97438 . . ..
Direct PNT – p. 21/28
Asymptotics for ω
We can show that
ω
b (s) = ζ(s)P (s),
M
and ω
e (x) ←→
ζ(s + 1)
P (s + 1).
(s + 1)
Direct PNT – p. 22/28
Asymptotics for ω
We can show that
ω
b (s) = ζ(s)P (s),
M
and ω
e (x) ←→
ζ(s + 1)
P (s + 1).
(s + 1)
Expanding around the rightmost singularity at the origin,
∞
1 1 X µ(m)
γ−1
1
1
log +
ln ζ(m) +
log + O(1).
s
s s m=2 m
s+1
s
Direct PNT – p. 22/28
Asymptotics for ω
We can show that
ω
b (s) = ζ(s)P (s),
M
and ω
e (x) ←→
ζ(s + 1)
P (s + 1).
(s + 1)
Expanding around the rightmost singularity at the origin,
∞
1 1 X µ(m)
γ−1
1
1
log +
ln ζ(m) +
log + O(1).
s
s s m=2 m
s+1
s
Inversion gives
∞
X
li(x)
µ(m)
ln ζ(m) + (γ − 1)
.
ω
e (x) ∼ ln(ln x) + γ +
m
x
m=2
Direct PNT – p. 22/28
Asymptotics for ω
We can show that
ω
b (s) = ζ(s)P (s),
M
and ω
e (x) ←→
ζ(s + 1)
P (s + 1).
(s + 1)
Expanding around the rightmost singularity at the origin,
∞
1 1 X µ(m)
γ−1
1
1
log +
ln ζ(m) +
log + O(1).
s
s s m=2 m
s+1
s
Inversion gives
∞
X
li(x)
µ(m)
ln ζ(m) + (γ − 1)
.
ω
e (x) ∼ ln(ln x) + γ +
m
x
m=2
Previously ω
e (n) ∼ ln(ln n) + γ +
X
p
ln 1 −
1
p
+
1
.
p
Direct PNT – p. 22/28
Asymptotics for Ω
Similarly,
b
Ω(s)
−ω
b (s) = ζ(s)
∞
X
P (ks),
k=2
Direct PNT – p. 23/28
Asymptotics for Ω
Similarly,
b
Ω(s)
−ω
b (s) = ζ(s)
∞
X
P (ks),
k=2
∞
X
M ζ(s + 1)
e
P (k(s + 1)),
Ω(x) − ω
e (x) ←→
s+1
k=2
Direct PNT – p. 23/28
Asymptotics for Ω
Similarly,
b
Ω(s)
−ω
b (s) = ζ(s)
∞
X
P (ks),
k=2
∞
X
M ζ(s + 1)
e
P (k(s + 1)),
Ω(x) − ω
e (x) ←→
s+1
k=2
e
Ω(x)
∼ω
e (x) +
∞
X
P (k).
k=2
Direct PNT – p. 23/28
Asymptotics for Ω
Similarly,
b
Ω(s)
−ω
b (s) = ζ(s)
∞
X
P (ks),
k=2
∞
X
M ζ(s + 1)
e
P (k(s + 1)),
Ω(x) − ω
e (x) ←→
s+1
k=2
e
Ω(x)
∼ω
e (x) +
e
Old Ω(n)
=ω
e (n) +
∞
X
P (k).
k=2
X
p
1
p(p − 1)
!
.
Direct PNT – p. 23/28
Results for ω and Ω
4
(a)
3.5
~
ω
Average Orders
3
2.5
2
~
Ω
1.5
1
0.5
0
0
250
500
750
n
1000
4
(b)
~
ω
Average Orders
3.5
3
~
Ω
2.5
2
0.0x10+00
Direct PNT – p. 24/28
2.5x10+06
5.0x10+06
n
7.5x10+06
1.0x10+07
Asymptotic Errors for ω and Ω
-1
10
ω
Ω
Errors
10-2
10-3
10-4
-5
10
104
105
n
106
107
Direct PNT – p. 25/28
Conclusion
We have developed a generalization of Ikehara’s theorem that relates the
asymptotic behavior of the average order of an arithmetic function to the
singularities in its Dirichlet series.
We have used this technique to prove the prime number theorem directly,
without recourse to ψ(x), derived both the Riemann and “exact” forms of
π(x), and found a correction to the classical ω, Ω average order
asymptotics.
Direct PNT – p. 26/28
Further work:
Apply technique to other arithmetic functions whose Dirichlet
functions are known in closed form. Will the results improve the
classical results?
Reformulation of the twin prime conjecture:
π2 (x) is the number of twin primes between 1 and x, and is x times
the average order of t(n), where t(n) = p(n)p(n − 2)
Can we use results for p to find the Dirichlet series for t, find the
form of the rightmost singularity of b
t, and find a result relating this
2Cx
, where
to the conjectured asymptotic π2 (x) ∼
2
(log x)
Y p(p − 2)
?
C=
(p − 1)2
p≥3
Direct PNT – p. 27/28
An Integral Proof That π < 355/113
Z
0
1
22
22
x4 (1 − x)4
− π, which shows π <
(Dalzell 1971, Mahler).
dx =
1 + x2
7
7
Direct PNT – p. 28/28
An Integral Proof That π < 355/113
Z
1
22
22
x4 (1 − x)4
− π, which shows π <
(Dalzell 1971, Mahler).
dx =
2
1
+
x
7
7
0
Z 1 m
x (1 − x)n
dx = a + bπ + c ln(2), (Backhouse 1995),
In fact Im,n =
2
1+x
0
ab < 0, and if 2m − n(mod 4) ≡ 0, then c = 0.
Direct PNT – p. 28/28
An Integral Proof That π < 355/113
Z
1
22
22
x4 (1 − x)4
− π, which shows π <
(Dalzell 1971, Mahler).
dx =
2
1
+
x
7
7
0
Z 1 m
x (1 − x)n
dx = a + bπ + c ln(2), (Backhouse 1995),
In fact Im,n =
2
1+x
0
ab < 0, and if 2m − n(mod 4) ≡ 0, then c = 0.
Unfortunately, no integers m, n lead to Im,n involving other continued
fractions for π.
Direct PNT – p. 28/28
An Integral Proof That π < 355/113
Z
1
22
22
x4 (1 − x)4
− π, which shows π <
(Dalzell 1971, Mahler).
dx =
2
1
+
x
7
7
0
Z 1 m
x (1 − x)n
dx = a + bπ + c ln(2), (Backhouse 1995),
In fact Im,n =
2
1+x
0
ab < 0, and if 2m − n(mod 4) ≡ 0, then c = 0.
Unfortunately, no integers m, n lead to Im,n involving other continued
fractions for π.
However,
Z
1
0
355
x8 (1 − x)8 (25 + 816x2 )
dx
=
− π,
2
3164(1 + x )
113
which proves π < 355/113.
Direct PNT – p. 28/28