A Direct Proof of the Prime Number Theorem

Home , Prime Number Theorem, Prime zeta function

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

Direct PNT – p. 1/28 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)

Direct PNT – p. 2/28 New Improving kissing sphere bounds. Applied Math Counting cycles in graphs. Deformation of half-spaces. Integer-valued logistic equation. Micro-pore diffusion in a ﬁnite volume. Fitting data to predator-prey. Froth Flotation. Representing reals using bounded Deformation of spectacle lenses. continued fractions. Pure Math Analysis of “Dreidel”. Fractals in regular graphs. Integral approximations to π. The Hamiltonian cycle problem.

Research Projects

Numerical Analysis Accurately ﬁnding multiple roots. Euler-Maclaurin-like summation for Simp- son’s rule. Placing a circle pack.

Direct PNT – p. 3/28 New Improving kissing sphere bounds. Counting cycles in graphs. Integer-valued logistic equation. Fitting data to predator-prey. Representing reals using bounded continued fractions. Pure Math Analysis of “Dreidel”. Fractals in regular graphs. Integral approximations to π. The Hamiltonian cycle problem.

Research Projects

Numerical Analysis Accurately ﬁnding multiple roots. Euler-Maclaurin-like summation for Simp- son’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.

Research Projects

Direct PNT – p. 3/28 Research Projects

Numerical Analysis Accurately ﬁnding multiple roots. Euler-Maclaurin-like summation for Simp- son’s rule. New Placing a circle pack. Improving kissing sphere bounds. Applied Math Counting cycles in graphs. Deformation of half-spaces. Integer-valued logistic equation. Micro-pore diffusion in a ﬁnite volume. Fitting data to predator-prey. Froth Flotation. Representing reals using bounded Deformation of spectacle lenses. continued fractions. Pure Math Analysis of “Dreidel”. Fractals in regular graphs. Integral approximations to π. The Hamiltonian cycle problem.

Direct PNT – p. 3/28 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) π(x) ∼ or lim = 1. ln x x→∞ x/ ln x

Another form states that x dt π(x) ∼ − (= li(x)) ln t Z0

The Prime Number Theorem

A prime number is any integer ≥ 2 with no divisors except itself and one.

Direct PNT – p. 4/28 The prime number theorem states that x π(x) π(x) ∼ or lim = 1. ln x x→∞ x/ ln x

Another form states that x dt π(x) ∼ − (= li(x)) ln t Z0

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 → ∞)

Direct PNT – p. 4/28 Another form states that x dt π(x) ∼ − (= li(x)) ln t Z0

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) π(x) ∼ or lim = 1. ln x x→∞ x/ ln x

Direct PNT – p. 4/28 The Prime Number Theorem

x π(x) π(x) ∼ or lim = 1. ln x x→∞ x/ ln x

Another form states that

x dt π(x) ∼ − (= li(x)) ln t Z0

Direct PNT – p. 4/28 Plot of π(x), x/ ln x, li(x)

1014 100

-1 1012 x/ln x 10

10-2 1010 s

e -3 10 r m o i r

r 8 10 r P E f

-4 e o 10 v i r t e 6 a b 10 l e

m -5 R

u 10

N li(x) 104 10-6

2 10 10-7

100 10-8 101 103 105 107 109 1011 1013 1015 x

Direct PNT – p. 5/28 Legendre (1808) used numerical evidence to claim x π(x) = , where lim A(x) = 1.08366 . . .. ln x + A(x) x→∞ Gauss (1849, as early as 1792) used numerical evidence to conjecture that x dt π(x) ∼ − . ln t Z0

Dirichlet (1837) introduced Dirichlet series:

∞ f(n) f(s) = . ns n=1 X b

History of the PNT

x Legendre (1798) conjectured π(x) = . A ln x + B

Direct PNT – p. 6/28 Gauss (1849, as early as 1792) used numerical evidence to conjecture that x dt π(x) ∼ − . ln t Z0

Dirichlet (1837) introduced Dirichlet series:

∞ f(n) f(s) = . ns n=1 X b

History of the PNT

x Legendre (1798) conjectured π(x) = . A ln x + B Legendre (1808) used numerical evidence to claim x π(x) = , where lim A(x) = 1.08366 . . .. ln x + A(x) x→∞

Direct PNT – p. 6/28 Dirichlet (1837) introduced Dirichlet series:

∞ f(n) f(s) = . ns n=1 X b

History of the PNT

x Legendre (1798) conjectured π(x) = . A ln x + B Legendre (1808) used numerical evidence to claim x π(x) = , where lim A(x) = 1.08366 . . .. ln x + A(x) x→∞ Gauss (1849, as early as 1792) used numerical evidence to conjecture that x dt π(x) ∼ − . ln t Z0

Direct PNT – p. 6/28 History of the PNT

∞ f(n) f(s) = . ns n=1 X b

Direct PNT – p. 6/28 Riemann (1860) introduced the Riemann zeta function:

∞ 1 1 −1 ζ(s) = = 1 − , <(s) > 1. ns ps n=1 p X Y

History of the PNT (cont’d)

Chebyshev (1851) introduced

θ(x) = ln p, ψ(x) = ln p, ≤ m≤ pXx pXx and showed that the PNT is equivalent to

θ(x) ψ(x) lim = 1, lim = 1. x→∞ x x→∞ x

Direct PNT – p. 7/28 History of the PNT (cont’d)

Chebyshev (1851) introduced

θ(x) = ln p, ψ(x) = ln p, ≤ m≤ pXx pXx and showed that the PNT is equivalent to

θ(x) ψ(x) lim = 1, lim = 1. x→∞ x x→∞ x

Riemann (1860) introduced the Riemann zeta function:

∞ 1 1 −1 ζ(s) = = 1 − , <(s) > 1. ns ps n=1 p X Y

Direct PNT – p. 7/28 1 Only singularity at s = 1, ζ(s) = + γ + γ (s − 1) + · · ·. s − 1 1 Showed using residue calculus that xρ ζ0(0) 1 ψ(x) = x − − − ln 1 − x−2 , ρ ζ(0) 2 ρ X where ρ are the non-trivial zeros of ζ(s), so the PNT is equivalent to 1 xρ lim = 0. x→∞ x ρ ρ X

1 Riemann hypothesis: <(ρ) = . 2

Riemann Zeta Function

Integrals extend to whole plane by analytic continuation.

Direct PNT – p. 8/28 Showed using residue calculus that xρ ζ0(0) 1 ψ(x) = x − − − ln 1 − x−2 , ρ ζ(0) 2 ρ X where ρ are the non-trivial zeros of ζ(s), so the PNT is equivalent to 1 xρ lim = 0. x→∞ x ρ ρ X

1 Riemann hypothesis: <(ρ) = . 2

Riemann Zeta Function

Integrals extend to whole plane by analytic continuation. 1 Only singularity at s = 1, ζ(s) = + γ + γ (s − 1) + · · ·. s − 1 1

Direct PNT – p. 8/28 1 Riemann hypothesis: <(ρ) = . 2

Riemann Zeta Function

Integrals extend to whole plane by analytic continuation. 1 Only singularity at s = 1, ζ(s) = + γ + γ (s − 1) + · · ·. s − 1 1 Showed using residue calculus that

xρ ζ0(0) 1 ψ(x) = x − − − ln 1 − x−2 , ρ ζ(0) 2 ρ X where ρ are the non-trivial zeros of ζ(s), so the PNT is equivalent to 1 xρ lim = 0. x→∞ x ρ ρ X

Direct PNT – p. 8/28 Riemann Zeta Function

Integrals extend to whole plane by analytic continuation. 1 Only singularity at s = 1, ζ(s) = + γ + γ (s − 1) + · · ·. s − 1 1 Showed using residue calculus that

xρ ζ0(0) 1 ψ(x) = x − − − ln 1 − x−2 , ρ ζ(0) 2 ρ X where ρ are the non-trivial zeros of ζ(s), so the PNT is equivalent to 1 xρ lim = 0. x→∞ x ρ ρ X 1 Riemann hypothesis: <(ρ) = . 2

Direct PNT – p. 8/28 First Proof of the PNT

Hadamard & de la Valèe Poussin (1896, independently) showed ∃a, t0 1 such that ζ(σ + it) 6= 0 if σ ≥ 1 − , |t| ≥ t , so a log |t| 0

1/14 ψ(x) = x + O xe−c(log x) .

Direct PNT – p. 9/28 Ikehara (1930) & Wiener (1932) used Tauberian theorems to prove Ikehara’s theorem: ∞ a Let f(s) = n , {a } real and nonnegative. ns i n=1 X If f(sb) converges for <(s) > 1, and ∃A > 0 s.t. for all t ∈ R, A fˆ(s) − → ﬁnite limit as s → 1+ + it, then a ∼ Ax. b (s − 1) n≤x n This can be used to show lim ψ(x)/x = 1 directlyP. x→∞

Improvements to the PNT Proof

Mertens (1898) gave a short proof that ζ(s) 6= 0, <(s) = 1.

Direct PNT – p. 10/28 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: ∞ a Let f(s) = n , {a } real and nonnegative. ns i n=1 X If f(sb) converges for <(s) > 1, and ∃A > 0 s.t. for all t ∈ R, A fˆ(s) − → ﬁnite limit as s → 1+ + it, then a ∼ Ax. b (s − 1) n≤x n This can be used to show lim ψ(x)/x = 1 directlyP. x→∞

Direct PNT – p. 10/28 Newman (1980) showed lim ψ(x)/x = 1 using straightforward x→∞ contour integration.

Lucas, Martin & Lever – current work, looks at π(x) directly.

Improvements (cont’d)

Erdös & Selberg (1949) produced an “elementary” proof – no complex analysis.

Direct PNT – p. 11/28 Lucas, Martin & Lever – current work, looks at π(x) directly.

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.

Direct PNT – p. 11/28 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.

Direct PNT – p. 11/28 1 +i∞ L−1{f¯(s)} = H(x)f(x) = f¯(s)esx ds. 2πi Z−i∞ H is the unit step function. to the right of any singularities in f¯(s). assume f¯(s) → 0 as |s| → ∞.

Choose g¯(s) such that f¯(s) − g¯(s) is analytic at the rightmost singularity in f¯(s). If g¯(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)

Laplace Transforms

∞ L{f(x)} = f¯(s) = e−sxf(x) dx. Z0

Direct PNT – p. 12/28 Choose g¯(s) such that f¯(s) − g¯(s) is analytic at the rightmost singularity in f¯(s). If g¯(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)

Laplace Transforms

∞ L{f(x)} = f¯(s) = e−sxf(x) dx. Z0 1 +i∞ L−1{f¯(s)} = H(x)f(x) = f¯(s)esx ds. 2πi Z−i∞ H is the unit step function. to the right of any singularities in f¯(s). assume f¯(s) → 0 as |s| → ∞.

Direct PNT – p. 12/28 If g¯(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)

Laplace Transforms

Choose g¯(s) such that f¯(s) − g¯(s) is analytic at the rightmost singularity in f¯(s).

Direct PNT – p. 12/28 Laplace Transforms

Direct PNT – p. 12/28 ∞ f(n) Its Dirichlet series is f(s) = , ns n=1 X − convergent for <(s) >bc if f(n) = O(nc 1) as n → ∞. Given that 1 +i∞ xs 0, x < 1, ds = > 0, 2πi −i∞ s  1, x > 1, Z  then  ∞ 1 +i∞ ds 0, x < n, xsf(s) = f(n) × 2πi s  −i∞ n=1 1, x > n, Z X  b = f(n). ≤ ≤ 1 Xn x

Arithmetic Functions and Dirichlet Series

An arithmetic function is a map f : N → C.

Direct PNT – p. 13/28 Given that 1 +i∞ xs 0, x < 1, ds = > 0, 2πi −i∞ s  1, x > 1, Z  then  ∞ 1 +i∞ ds 0, x < n, xsf(s) = f(n) × 2πi s  −i∞ n=1 1, x > n, Z X  b = f(n). ≤ ≤ 1 Xn x

Arithmetic Functions and Dirichlet Series

An arithmetic function is a map f : N → C. ∞ f(n) Its Dirichlet series is f(s) = , ns n=1 X − convergent for <(s) >bc if f(n) = O(nc 1) as n → ∞.

Direct PNT – p. 13/28 Arithmetic Functions and Dirichlet Series

An arithmetic function is a map f : N → C. ∞ f(n) Its Dirichlet series is f(s) = , ns n=1 X − convergent for <(s) >bc if f(n) = O(nc 1) as n → ∞. Given that 1 +i∞ xs 0, x < 1, ds = > 0, 2πi −i∞ s  1, x > 1, Z  then  ∞ 1 +i∞ ds 0, x < n, xsf(s) = f(n) × 2πi s  −i∞ n=1 1, x > n, Z X  b = f(n). ≤ ≤ 1 Xn x

Direct PNT – p. 13/28 Some useful transform pairs are:

M f ←→ M(f) 1 1 γ + ln(ln x) log s s 1 c xali(xc) log , <(s) > c > 0 s − a s − a − c 1 Γ(k)−1xc(ln x)k−1 , <(s) > c; k > 0 (s − c)k

The Transform

We can recognize this as the inversion of a Laplace-like transform (x = et): ∞ (Mf)(t) = f(x)x−s−1 dx, Z1 1 +i∞ H(x − 1)f(x) = (Mf)(t)xt dt. 2πi Z−i∞

Direct PNT – p. 14/28 The Transform

We can recognize this as the inversion of a Laplace-like transform (x = et): ∞ (Mf)(t) = f(x)x−s−1 dx, Z1 1 +i∞ H(x − 1)f(x) = (Mf)(t)xt dt. 2πi Z−i∞ Some useful transform pairs are:

M f ←→ M(f) 1 1 γ + ln(ln x) log s s 1 c xali(xc) log , <(s) > c > 0 s − a s − a − c 1 Γ(k)−1xc(ln x)k−1 , <(s) > c; k > 0 (s − c)k

Direct PNT – p. 14/28 1 +i∞ Then xf(x) = f(s)xs−1 ds, 2πi Z−i∞ e M f(s) b M f(s + 1) and xf(x) ←→ and f(x) ←→ . s s + 1 b b So, to ﬁnde the asymptotic forme of an average order: ﬁnd its Dirichlet series f(s) in closed form, f(s) identify the position andbform of the singularities in , s sum the inverse transforms of the singular parts. b

Average Order Asymptotics

1 The average order of an arithmetic function is f(x) = f(n). x ≤ ≤ 1 Xn x e

Direct PNT – p. 15/28 M f(s) M f(s + 1) and xf(x) ←→ and f(x) ←→ . s s + 1 b b So, to ﬁnde the asymptotic forme of an average order: ﬁnd its Dirichlet series f(s) in closed form, f(s) identify the position andbform of the singularities in , s sum the inverse transforms of the singular parts. b

Average Order Asymptotics

1 The average order of an arithmetic function is f(x) = f(n). x ≤ ≤ 1 Xn x e 1 +i∞ Then xf(x) = f(s)xs−1 ds, 2πi Z−i∞ e b

Direct PNT – p. 15/28 So, to ﬁnd the asymptotic form of an average order:

ﬁnd its Dirichlet series f(s) in closed form, f(s) identify the position andbform of the singularities in , s sum the inverse transforms of the singular parts. b

Average Order Asymptotics

1 The average order of an arithmetic function is f(x) = f(n). x ≤ ≤ 1 Xn x e 1 +i∞ Then xf(x) = f(s)xs−1 ds, 2πi Z−i∞ e M f(s) b M f(s + 1) and xf(x) ←→ and f(x) ←→ . s s + 1 b b e e

Direct PNT – p. 15/28 Average Order Asymptotics

1 The average order of an arithmetic function is f(x) = f(n). x ≤ ≤ 1 Xn x e 1 +i∞ Then xf(x) = f(s)xs−1 ds, 2πi Z−i∞ e M f(s) b M f(s + 1) and xf(x) ←→ and f(x) ←→ . s s + 1 b b So, to ﬁnde the asymptotic forme of an average order: ﬁnd its Dirichlet series f(s) in closed form,

f(s) identify the position andbform of the singularities in , s sum the inverse transforms of the singular parts. b

Direct PNT – p. 15/28 1 ζ(s) only singularity ∼ 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, 1 f(n) = : P (s) = .  ps 0, otherwise, p  X 

Riemann and Prime Zeta-Functions

The Riemann zeta-function is the Dirichlet series of f(n) = 1:

∞ 1 1 ζ(s) = = . ns 1 − p−s n=1 p X Y

Direct PNT – p. 16/28 The prime zeta-function is the Dirichlet series of

1, n prime, 1 f(n) = : P (s) = .  ps 0, otherwise, p  X 

Riemann and Prime Zeta-Functions

The Riemann zeta-function is the Dirichlet series of f(n) = 1:

∞ 1 1 ζ(s) = = . ns 1 − p−s n=1 p X Y 1 ζ(s) only singularity ∼ 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:

∞ 1 1 ζ(s) = = . ns 1 − p−s n=1 p X Y 1 ζ(s) only singularity ∼ 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, 1 f(n) = : P (s) = .  ps 0, otherwise, p  X 

Direct PNT – p. 16/28 Using the Möbius inversion formula,

∞ µ(n) P (s) = log ζ(ns), n n=1 X 0, n contains a square factor, where µ(n) =  (−1)m, n a product of m distinct primes.  

The Prime Zeta-Function

Now ∞ 1 1 log ζ(s) = log = 1 − p−s mpms p p m=1 X X X ∞ 1 = P (ms). m m=1 X

Direct PNT – p. 17/28 The Prime Zeta-Function

Now ∞ 1 1 log ζ(s) = log = 1 − p−s mpms p p m=1 X X X ∞ 1 = P (ms). m m=1 X Using the Möbius inversion formula,

∞ µ(n) P (s) = log ζ(ns), n n=1 X 0, n contains a square factor, where µ(n) =  (−1)m, n a product of m distinct primes.  

Direct PNT – p. 17/28 Then f(s) = P (s) and xf(x) = π(x). b e So M P (s) π(x) ←→ . s

P (s) 1 1 The rightmost singularity of is log at s − 1, whose inverse s s s − 1 transform is li(x). So π(x) ∼ li(x).

The Prime Number Theorem

Let 1, n prime, f(n) =  0, otherwise,  

Direct PNT – p. 18/28 So M P (s) π(x) ←→ . s

P (s) 1 1 The rightmost singularity of is log at s − 1, whose inverse s s s − 1 transform is li(x). So π(x) ∼ li(x).

The Prime Number Theorem

Let 1, n prime, f(n) =  0, otherwise,  Then  f(s) = P (s) and xf(x) = π(x).

b e

Direct PNT – p. 18/28 P (s) 1 1 The rightmost singularity of is log at s − 1, whose inverse s s s − 1 transform is li(x). So π(x) ∼ li(x).

The Prime Number Theorem

Let 1, n prime, f(n) =  0, otherwise,  Then  f(s) = P (s) and xf(x) = π(x).

So b e M P (s) π(x) ←→ . s

Direct PNT – p. 18/28 So π(x) ∼ li(x).

The Prime Number Theorem

Let 1, n prime, f(n) =  0, otherwise,  Then  f(s) = P (s) and xf(x) = π(x).

So b e M P (s) π(x) ←→ . s P (s) 1 1 The rightmost singularity of is log at s − 1, whose inverse s s s − 1 transform is li(x).

Direct PNT – p. 18/28 The Prime Number Theorem

Let 1, n prime, f(n) =  0, otherwise,  Then  f(s) = P (s) and xf(x) = π(x).

So b e M P (s) π(x) ←→ . s P (s) 1 1 The rightmost singularity of is log at s − 1, whose inverse s s s − 1 transform is li(x). So π(x) ∼ li(x).

Direct PNT – p. 18/28 P (s) 1 Singularity in ζ(s) at s = 1 means singularities in at s = , µ(n) 6= 0. s n µ(n) 1 At these points, singularities take the form log , whose inverse ns ns − 1 li(x1/n) transforms are µ(n) . n So ∞ µ(n) π(x) ∼ li(x) + li x1/n = R(x). n n=2 X

Riemann’s Form of π(x)

∞ µ(n) Recall P (s) = log ζ(ns). n n=1 X

Direct PNT – p. 19/28 µ(n) 1 At these points, singularities take the form log , whose inverse ns ns − 1 li(x1/n) transforms are µ(n) . n So ∞ µ(n) π(x) ∼ li(x) + li x1/n = R(x). n n=2 X

Riemann’s Form of π(x)

∞ µ(n) Recall P (s) = log ζ(ns). n n=1 X P (s) 1 Singularity in ζ(s) at s = 1 means singularities in at s = , µ(n) 6= 0. s n

Direct PNT – p. 19/28 So ∞ µ(n) π(x) ∼ li(x) + li x1/n = R(x). n n=2 X

Riemann’s Form of π(x)

∞ µ(n) Recall P (s) = log ζ(ns). n n=1 X P (s) 1 Singularity in ζ(s) at s = 1 means singularities in at s = , µ(n) 6= 0. s n µ(n) 1 At these points, singularities take the form log , whose inverse ns ns − 1 li(x1/n) transforms are µ(n) . n

Direct PNT – p. 19/28 Riemann’s Form of π(x)

Direct PNT – p. 19/28 Zero of ζ(s) at s = ρm means a singularity in P (s)/s at s = ρm of the form 1 s log − 1 , whose inverse is −li(xρm ). s ρ m As for Riemann’s form, singularity in ζ(s) at s = ρm also means singularities in P (s) at s = ρm/n, µ(n) 6= 0. These singularities are of the µ(n) ns m form log − 1 , whose inverses are −µ(n)li xρ /n n. ns ρ m ρm The contributions of all singularities related to s = ρm contribute −R(x ) to π(x), and so k ρm π(x) = lim Rk(x) where Rk(x) = R(x) − R(x ). k→∞ − mX= k

“Exact” Form of π(x)

∞ µ(n) Recall P (s) = log ζ(ns). n n=1 X

Direct PNT – p. 20/28 As for Riemann’s form, singularity in ζ(s) at s = ρm also means singularities in P (s) at s = ρm/n, µ(n) 6= 0. These singularities are of the µ(n) ns m form log − 1 , whose inverses are −µ(n)li xρ /n n. ns ρ m ρm The contributions of all singularities related to s = ρm contribute −R(x ) to π(x), and so k ρm π(x) = lim Rk(x) where Rk(x) = R(x) − R(x ). k→∞ − mX= k

“Exact” Form of π(x)

∞ µ(n) Recall P (s) = log ζ(ns). n n=1 X Zero of ζ(s) at s = ρm means a singularity in P (s)/s at s = ρm of the form 1 s log − 1 , whose inverse is −li(xρm ). s ρ m

Direct PNT – p. 20/28 ρm The contributions of all singularities related to s = ρm contribute −R(x ) to π(x), and so k ρm π(x) = lim Rk(x) where Rk(x) = R(x) − R(x ). k→∞ − mX= k

“Exact” Form of π(x)

singularities in P (s) at s = ρm/n, µ(n) 6= 0. These singularities are of the µ(n) ns m form log − 1 , whose inverses are −µ(n)li xρ /n n. ns ρ m

Direct PNT – p. 20/28 “Exact” Form of π(x)

∞ µ(n) Recall P (s) = log ζ(ns). n n=1 X Zero of ζ(s) at s = ρm means a singularity in P (s)/s at s = ρm of the form 1 s log − 1 , whose inverse is −li(xρm ). s ρ m As for Riemann’s form, singularity in ζ(s) at s = ρm also means singularities in P (s) at s = ρm/n, µ(n) 6= 0. These singularities are of the µ(n) ns m form log − 1 , whose inverses are −µ(n)li xρ /n n. ns ρ m ρm The contributions of all singularities related to s = ρm contribute −R(x ) to π(x), and so k ρm π(x) = lim Rk(x) where Rk(x) = R(x) − R(x ). k→∞ − mX= k

Direct PNT – p. 20/28 α1 α2 αk If n = p1 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(n) = ln(ln n) + B1 + o(1), 1 1 where B = γ + ln 1 − + = 0.26149 72128 47642 . . ., 1 p p p e X 1 and Ω(n) = ω(n) + = ω(n) + B − B , p(p − 1) 2 1 p X wheree B2 = e1.03465 38818 97438 . . .e.

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.

Direct PNT – p. 21/28 It is well known (e.g. Hardy & Wright) that ω(n) = ln(ln n) + B1 + o(1), 1 1 where B = γ + ln 1 − + = 0.26149 72128 47642 . . ., 1 p p p e X 1 and Ω(n) = ω(n) + = ω(n) + B − B , p(p − 1) 2 1 p X wheree B2 = e1.03465 38818 97438 . . .e.

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.

α1 α2 αk If n = p1 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.

α1 α2 αk If n = p1 p2 . . . pk (pi’s some primes), then ω(n) = k and Ω(n) = α1 + α2 + · · · + αk.

Direct PNT – p. 21/28 Expanding around the rightmost singularity at the origin,

∞ 1 1 1 µ(m) γ − 1 1 log + ln ζ(m) + log + O(1). s s s m s + 1 s m=2 X Inversion gives

∞ µ(m) li(x) ω(x) ∼ ln(ln x) + γ + ln ζ(m) + (γ − 1) . m x m=2 X e 1 1 Previously ω(n) ∼ ln(ln n) + γ + ln 1 − + . p p p X e

Asymptotics for ω

We can show that

M ζ(s + 1) ω(s) = ζ(s)P (s), and ω(x) ←→ P (s + 1). (s + 1)

b e

Direct PNT – p. 22/28 Inversion gives

∞ µ(m) li(x) ω(x) ∼ ln(ln x) + γ + ln ζ(m) + (γ − 1) . m x m=2 X e 1 1 Previously ω(n) ∼ ln(ln n) + γ + ln 1 − + . p p p X e

Asymptotics for ω

We can show that

M ζ(s + 1) ω(s) = ζ(s)P (s), and ω(x) ←→ P (s + 1). (s + 1)

Expandingbaround the rightmost singulare ity at the origin,

∞ 1 1 1 µ(m) γ − 1 1 log + ln ζ(m) + log + O(1). s s s m s + 1 s m=2 X

Direct PNT – p. 22/28 1 1 Previously ω(n) ∼ ln(ln n) + γ + ln 1 − + . p p p X e

Asymptotics for ω

We can show that

M ζ(s + 1) ω(s) = ζ(s)P (s), and ω(x) ←→ P (s + 1). (s + 1)

Expandingbaround the rightmost singulare ity at the origin,

∞ 1 1 1 µ(m) γ − 1 1 log + ln ζ(m) + log + O(1). s s s m s + 1 s m=2 X Inversion gives

∞ µ(m) li(x) ω(x) ∼ ln(ln x) + γ + ln ζ(m) + (γ − 1) . m x m=2 X e

Direct PNT – p. 22/28 Asymptotics for ω

We can show that

M ζ(s + 1) ω(s) = ζ(s)P (s), and ω(x) ←→ P (s + 1). (s + 1)

Expandingbaround the rightmost singulare ity at the origin,

∞ 1 1 1 µ(m) γ − 1 1 log + ln ζ(m) + log + O(1). s s s m s + 1 s m=2 X Inversion gives

∞ µ(m) li(x) ω(x) ∼ ln(ln x) + γ + ln ζ(m) + (γ − 1) . m x m=2 X e 1 1 Previously ω(n) ∼ ln(ln n) + γ + ln 1 − + . p p p X e Direct PNT – p. 22/28 ∞ M ζ(s + 1) Ω(x) − ω(x) ←→ P (k(s + 1)), s + 1 kX=2 e e ∞ Ω(x) ∼ ω(x) + P (k). kX=2 e e 1 Old Ω(n) = ω(n) + . p(p − 1) p ! X e e

Asymptotics for Ω

Similarly, ∞ Ω(s) − ω(s) = ζ(s) P (ks), Xk=2 b b

Direct PNT – p. 23/28 ∞ Ω(x) ∼ ω(x) + P (k). kX=2 e e 1 Old Ω(n) = ω(n) + . p(p − 1) p ! X e e

Asymptotics for Ω

Similarly, ∞ Ω(s) − ω(s) = ζ(s) P (ks), Xk=2 b b ∞ M ζ(s + 1) Ω(x) − ω(x) ←→ P (k(s + 1)), s + 1 kX=2 e e

Direct PNT – p. 23/28 1 Old Ω(n) = ω(n) + . p(p − 1) p ! X e e

Asymptotics for Ω

Similarly, ∞ Ω(s) − ω(s) = ζ(s) P (ks), Xk=2 b b ∞ M ζ(s + 1) Ω(x) − ω(x) ←→ P (k(s + 1)), s + 1 kX=2 e e ∞ Ω(x) ∼ ω(x) + P (k). kX=2 e e

Direct PNT – p. 23/28 Asymptotics for Ω

Similarly, ∞ Ω(s) − ω(s) = ζ(s) P (ks), Xk=2 b b ∞ M ζ(s + 1) Ω(x) − ω(x) ←→ P (k(s + 1)), s + 1 kX=2 e e ∞ Ω(x) ∼ ω(x) + P (k). kX=2 e e 1 Old Ω(n) = ω(n) + . p(p − 1) p ! X e e

Direct PNT – p. 23/28 Results for ω and Ω

4 (a) 3.5

ω~ 3 s r 2.5 e d r O

e 2

g ~

a Ω r e v 1.5 A

0.5

0 0 250 500 750 1000 n

4 (b)

ω~

3.5 s r e d r O

e 3 g ~ a

r Ω e v A

2.5

2 Direct PNT – p. 24/28 0.0x10+00 2.5x10+06 5.0x10+06 7.5x10+06 1.0x10+07 n Asymptotic Errors for ω and Ω

10-1

10-2 ω Ω s r -3 o

r 10 r E

10-4

10-5 104 105 106 107 n

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 t, and find a result relating this 2Cx to the conjectured asymptotic π2(x) ∼ , where b (log x)2 p(p − 2) C = ? (p − 1)2 ≥ pY3

Direct PNT – p. 27/28 1 xm(1 − x)n In fact , (Backhouse 1995), Im,n = 2 dx = a + bπ + c ln(2) 0 1 + x ab < 0, and if 2Zm − n(mod 4) ≡ 0, then c = 0.

Unfortunately, no integers m, n lead to Im,n involving other continued fractions for π. However, 1 x8(1 − x)8(25 + 816x2) 355 dx = − π, 3164(1 + x2) 113 Z0 which proves π < 355/113.

An Integral Proof That π < 355/113

1 x4(1 − x)4 22 22 dx = − π, which shows π < (Dalzell 1971, Mahler). 1 + x2 7 7 Z0

Direct PNT – p. 28/28 Unfortunately, no integers m, n lead to Im,n involving other continued fractions for π. However, 1 x8(1 − x)8(25 + 816x2) 355 dx = − π, 3164(1 + x2) 113 Z0 which proves π < 355/113.

An Integral Proof That π < 355/113

1 x4(1 − x)4 22 22 dx = − π, which shows π < (Dalzell 1971, Mahler). 1 + x2 7 7 Z0 1 xm(1 − x)n In fact , (Backhouse 1995), Im,n = 2 dx = a + bπ + c ln(2) 0 1 + x ab < 0, and if 2Zm − n(mod 4) ≡ 0, then c = 0.

Direct PNT – p. 28/28 However, 1 x8(1 − x)8(25 + 816x2) 355 dx = − π, 3164(1 + x2) 113 Z0 which proves π < 355/113.

An Integral Proof That π < 355/113

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

Unfortunately, no integers m, n lead to Im,n involving other continued fractions for π. However, 1 x8(1 − x)8(25 + 816x2) 355 dx = − π, 3164(1 + x2) 113 Z0 which proves π < 355/113.

Direct PNT – p. 28/28