Honors 1: Undergraduate Math Lab^{∗} The Circle Method and Germain Primes

Peter Sarnak^{†}, Steven Miller^{‡}, Alex Barnett^{§}

November 18th, 2002

Courant Institute of Mathematical Sciences New York University New York, NY

Abstract

Using the Hardy-Littlewood Circle Method (and assuming no main term contribution from the Minor Arcs), we calculate the expected number of Germain primes. Calculations and notes by Steven Miller.

Contents

1 Preliminaries 3 1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Partial Summation . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Siegel-Walfisz . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.4 Germain Integral . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.5 Major and Minor Arcs . . . . . . . . . . . . . . . . . . . . . . 5 1.6 Reformulation of Germain Integral . . . . . . . . . . . . . . . 6 2 f_{N}(x) and u(x) 7

∗Homepage: http://www.math.nyu.edu/∼millerj/ †E-mail: sarnak@math.princeton.edu ‡E-mail: millerj@cims.nyu.edu or sjmiller@math.princeton.edu §E-mail: barnett@nmr.mgh.harvard.edu

2.1 f(\frac{a}{q}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 u(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 f_{N}(α) − \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}u(α − \frac{a}{q}), α ∈ M_{a,q} 8 3.1 Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 3.2 S_{1 ∑;a,q} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3.2.1 First Piece . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.2.2 Second Piece . . . . . . . . . . . . . . . . . . . . . . . 11

3.3 S_{1 ∫;a,q} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3.3.1 First Piece . . . . . . . . . . . . . . . . . . . . . . . . . 12 3.3.2 Second Piece . . . . . . . . . . . . . . . . . . . . . . . 12

4 Integrals of u(x) 13 4.1 Formulations . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.2 ∫− \frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.3 ∫_{− \frac{1}{2}}^{− \frac{Q}{N}} + ∫ \frac{1}{2} ; \frac{Q}{N} u(x)e(−x)dx . . . . . . . . . . . . . . . . . . . . . 14

4.4 Integral over I_{2}, I_{3} . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.5 Integral over I_{1}, I_{4} . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.5.1 0 < |m_{1} − 2m_{2} − 1| ≤ N^{1−ϵ} . . . . . . . . . . . . . . . 16 4.5.2 N^{1−ϵ} < |m_{1} − 2m_{2} − 1| ≤ N . . . . . . . . . . . . . . . 16

4.6 Collecting the Pieces . . . . . . . . . . . . . . . . . . . . . . . 17

5 Determination of the Main Term 18

5.1 Properties of C_{q}(a) and ρ_{q} . . . . . . . . . . . . . . . . . . . . 20 5.1.1 c_{q}(a) is Multiplicative . . . . . . . . . . . . . . . . . . . 20 5.1.2 c_{q}(a) for (a, q) = 1 . . . . . . . . . . . . . . . . . . . . 21 5.1.3 C_{q}(a) is Multiplicative . . . . . . . . . . . . . . . . . . 22 5.1.4 ρ_{q} is Multiplicative . . . . . . . . . . . . . . . . . . . . 23 5.1.5 Calculation of ρ_{q} . . . . . . . . . . . . . . . . . . . . . 25

5.2 Determination of 𝔖_{N} and 𝔖 . . . . . . . . . . . . . . . . . . . 26

5.2.1 𝔖 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 5.2.2 𝔖_{N} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5.3 Number of Germain Primes and Weighted Sums . . . . . . . . 27

1 Preliminaries

1.1 Definitions

Let

e(x) = e^{2πix}  (1)

and

λ(n) = \begin{cases}log p & if n = p is prime \\ 0 & otherwise \end{cases}  (2)

Finally, define

c_{q}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q}).  (3)

1.2 Partial Summation

Lemma 1.1 (Partial Summation: Discrete Version). Let A_{N} = ∑_{n=1}^{N} a_{n}. then

∑_{n=M}^{N} a_{n}b_{n} = A_{N}b_{N} − A_{M−1}b_{M} + ∑_{n=M}^{N−1} A_{n}(b_{n} − b_{n+1})  (4)

Proof. Since A_{n} − A_{n−1} = a_{n},

∑_{n=M}^{N} a_{n}b_{n} = ∑_{n=M}^{N}(A_{n} − A_{n−1})b_{n}
= (A_{N} − A_{N−1})b_{N} + (A_{N−1} − A_{N−2})b_{N−1} + ⋯ + (A_{M} − A_{M−1})b_{M}
= A_{N}b_{N} + (−A_{N−1}b_{N} + A_{N−1}b_{N−1}) + ⋯ + (−A_{M}b_{M+1} + A_{M}b_{M}) − a_{M−1}b_{M}
= A_{N}b_{N} − a_{M_{1}}b_{M} + ∑_{n=M}^{N−1} A_{n}(b_{n} − b_{n+1}).  (5)

Lemma 1.2 (Abel’s Summation Formula - Integral Version). Let h(x) be a continuously differentiable function. Let A(x) = ∑_{n≤x} a_{n}. Then

∑_{n≤x} a_{n}h(n) = A(x)h(x) − ∫_{1}^{x} A(u)h′(u)du  (6)

See, for example, W. Rudin, Principles of Mathematical Analysis, page 70.

1.3 Siegel-Walfisz

Theorem 1.3. [Siegel-Walfisz] Let C, B > 0, and let a and q be relatively prime. Then

∑_{{p≤x \atop p≡a(q)}} log p = \frac{x}{ϕ(q)} + O(\frac{x}{log^{C} x})  (7)

for q ≤ log^{B} x, and the constant above does not depend on x, q or a (ie, it only depends on C and B).

1.4 Germain Integral

Define

f_{1N}(x) = ∑_{p_{1}≤N} log p_{1} · e(p_{1}x)
f_{2N}(x) = ∑_{p_{2}≤N} log p_{2} · e(−2p_{2}x)
f_{N}(x) = ∑_{p_{1}≤N} ∑_{p_{2}≤N} log p_{1} log p_{2} · e((p_{1} − 2p_{2})x).  (8)

Consider

∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx = ∑_{p_{1}≤N} ∑_{p_{2}≤N} log p_{1} log p_{2} ∫− \frac{1}{2}^{\frac{1}{2}} e((p_{1} − 2p_{2} − 1)x)dx. (9)

Note

∫− \frac{1}{2}^{\frac{1}{2}} e((p_{1} − 2p_{2} − 1)x)dx = \begin{cases}1 & if p_{1} − 2p_{2} − 1 = 0 \\ 0 & if p_{1} − 2p_{2} − 1 ≠ 0 \end{cases}  (10)

Thus, we get a contribution of log p_{1} log p_{2} if p_{1} and p_{2} = \frac{p_{1}−1}{2} are both primes. Thus,

∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx = ∑_{{p_{1}≤N \atop p_{2}= \frac{p_{1}−1}{2} prime}} log p_{1} log p_{2}.  (11)

The above is a weighted counting of Germain primes.

1.5 Major and Minor Arcs

Let B be a positive integer, Q = log^{B} N, and define the Major Arc M_{a,q}

M_{a,q} = \{x ∈ [0, 1) : |x − \frac{a}{q}| < \frac{Q}{N} \}.  (12)

We also add in one interval centered at either 0 or 1, ie, the ”interval” (or wrapped-around interval)

[0, \frac{Q}{N} ] ∪ [1 − \frac{Q}{N} , 1].  (13)

For convenience, we often use the interval [−\frac{1}{2}, \frac{1}{2}] instead of [0, 1], in which case we would have

[ − \frac{1}{2}, −\frac{1}{2} + \frac{Q}{N} ] ⋃ [\frac{1}{2} − \frac{Q}{N} , \frac{1}{2}].  (14)

For functions that are periodic of period one, we could instead consider

[\frac{1}{2} − \frac{Q}{N} , \frac{1}{2} + \frac{Q}{N} ].  (15)

The Major Arcs are defined by

M = ⋃_{q≤Q} ⋃_{{a=1 \atop (a,q)=1}}^{q} M_{a,q}.  (16)

The Minor Arcs, m, are whatever is not in the Major Arcs. Then

∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx = ∫_{M} f_{N}(x)e(−x)dx + ∫_{m} f_{N}(x)e(−x)dx. (17)

We will assume that there is no net contribution over the minor arcs. Thus, in the sequel we investigate

∫_{M} f_{N}(x)e(−x)dx.  (18)

1.6 Reformulation of Germain Integral

f_{1N}(x) = ∑_{m_{1}≤N} λ(m_{1}) · e(m_{1}x)
f_{2N}(x) = ∑_{m_{2}≤N} λ(m_{2}) · e(−2m_{2}x)
f_{N}(x) = ∑_{m_{1}≤N} ∑_{m_{2}≤N} λ(m_{1})λ(m_{2}) · e((m_{1} − 2m_{2})x).  (19)

We investigate

∫_{M} f_{N}(x)e(−x)dx.  (20)

We will show the Major Arcs contribute, up to lower order terms, T_{2}N, where T_{2} is a constant independent of N. The length of the Major Arc M_{a,q} is \frac{Q}{N}. We sum over (a, q) = 1 and q ≤ Q. Thus, the total length is bounded by

∑_{q≤Q} q · \frac{Q}{N} ≪ \frac{Q^{3}}{N} ≪ \frac{log^{B}}{N} .  (21)

By choosing B sufficiently large, we will be able to make all the errors from the Major Arc calculations less than the main term from the Major Arcs. Of course, we have absolutely no control over what happens on the minor arcs, and we will simply assume there is no contribution from the minor arcs.

Thus, on the Major Arc M_{a,q}, success will be in finding a function of size N^{2} such that the error from this function to f_{N}(x) on M_{a,q} is much smaller than N^{2}, say N^{2} divided by a large power of log N.

Similarly, when we integrate over the Major Arcs, we will find the main terms will be of size N; again, success will be in showing the errors in the approximations are much smaller than N, say N divided by a large power of log N.

We are able to do this because of the Siegel-Walfisz Theorem (Theorem 1.3). Given any B > 0, we can find a C > 0 such that, if q ≤ log^{B} N, then

∑_{{p≤N \atop p≡r(q)}} log p = \frac{N}{ϕ(q)} + O(\frac{N}{log^{C} N} ),  (22)

(r, q) = 1. Thus, we can take C enormous, large enough so that even when we multiply by the length of the Major Arcs (of size \frac{log^{3B} N}{N}, we still have something small.

2 f_{N}(x) and u(x)

2.1 f(\frac{a}{q})

We now calculate f_{N}(\frac{a}{q}) for q ≤ log^{B} N.

Up to lower order terms,

f_{N}(\frac{a}{q}) = ∑_{p_{1}≤N} log p_{1} · e(p_{1} \frac{a}{q}) ∑_{p_{2}≤N} log p_{2} · e( − 2p_{2} \frac{a}{q})

= ∑_{r_{1}=1}^{q} ∑_{{p_{1}≤N \atop p_{1}≡r_{1}(q)}} log p_{1} · e(p_{1} \frac{a}{q}) ∑_{r_{2}=1}^{q} ∑_{{p_{2}≤N \atop p_{2}≡r_{1}(q)}} log p_{2} · e( − 2p_{2} \frac{a}{q})

= ∑_{r_{1}=1}^{q} e(r_{1} \frac{a}{q}) ∑_{r_{2}=1}^{q} e(r_{2} \frac{−2a}{q}) ∑_{{p_{1}≤N \atop p_{1}≡r_{1}(q)}} log p_{1} ∑_{{p_{2}≤N \atop p_{2}≡r_{2}(q)}} log p_{2}

= \frac{N^{2}}{ϕ^{2}(q)} ∑_{{r_{1}=1 \atop (r_{1},q)=1}}^{q} e(r_{1} \frac{a}{q}) ∑_{{r_{2}=1 \atop (r_{2},q)=1}}^{q} e(r_{2} \frac{−2a}{q})

= \frac{N^{2}}{ϕ^{2}(q)}c_{q}(a)c_{q}(−2a),  (23)

where the second to last line follows from the Siegel-Walfisz Theorem (Theorem 1.3). We restrict to (r_{i}, q) = 1 because if (r_{i}, q) > 1, there is at most one prime p_{i} ≡ r_{i} mod q.

2.2 u(x)

Let

u(x) = ∑_{m_{1}≤N} ∑_{m_{2}≤N} e((m_{1} − 2m_{2})x).  (24)

We will often look at

\frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}u(x).  (25)

Note

u(0) = N^{2}.  (26)

3 f_{N}(α) − \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}u(α − \frac{a}{q}), α ∈ M_{a,q}

Let

C_{q}(a) = \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}.  (27)

We write α as β + \frac{a}{q}, β ∈ [ − \frac{Q}{N} , \frac{Q}{N} ], Q = log^{B} N. As always, we ignore lower order terms.

Note f_{N}(x) is approximately C_{q}(a)N^{2} for x near \frac{a}{q}. We now expand and

show f_{N}(α) is C_{q}(a)u(α − \frac{a}{q}) plus errors of size \frac{N^{2}}{log^{C−2B} N} for α ∈ M_{a,q}.

3.1 Setup

S_{a,q}(α) = f_{N}(α) − C_{q}(a)u(α − \frac{a}{q})
= ∑_{m_{1},m_{2}≤N} λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})α) − C_{q}(a) ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2})β)

= ∑_{m_{1},m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e((m_{1} − 2m_{2})β)

= ∑_{m_{1}≤N} [ ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(−2m_{2}β)]e(m_{1}β)

(28)

We now apply Partial Summation multiple times. First, we apply Partial Summation to the m_{2}-sum:

S_{2;a,q} = ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(−2m_{2}β)
= ∑_{m_{2}≤N} a_{m_{2}}b_{m_{2}}

= A_{2}(N)e(−2Nβ) + 4πiβ ∫_{0}^{N} ∑_{m_{2}≤u} a_{m_{2}}e(−uβ)du.  (29)

We hit the above with e(m_{1}β), and sum from m_{1} = 1 to N. We get two pieces:

S_{1 ∑;a,q} = ∑_{m_{1}≤N} A_{2}(N)e(−2Nβ) · e(m_{1}β)

S_{1 ∫;a,q} = ∑_{m_{1}≤N} 4πiβ ∫_{0}^{N} ∑_{m_{2}≤u} a_{m_{2}}e(−uβ)du · e(m_{1}β)
S_{a,q} = S_{1 ∑;a,q} + S_{1 ∫;a,q}.  (30)

3.2 S_{1 ∑;a,q}

S_{1 ∑;a,q} = ∑_{m_{1}≤N} A_{2}(N)e(−2Nβ) · e(m_{1}β)
= e(−2Nβ) ∑_{m_{1}≤N} A_{2}(N)e(m_{1}β)
= e(−2Nβ) ∑_{m_{1}≤N} ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(m_{1}β)

= e(−2Nβ)[A_{1}(N)e(Nβ)

−2πiβ ∫_{0}^{N} ∑_{m_{1}≤t} ∑_{m_{2}≤N} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(tβ)dt.

(31)

3.2.1 First Piece

The first piece, the A_{1}(N)e(Nβ) term, is small for q ≤ Q. Why? We have (up to lower order terms)

A_{1}(N)e(Nβ) = ∑_{m_{1},m_{2}≤N} λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − ∑_{m_{1},m_{2}≤N} C_{q}(a)
= C_{q}(a)N^{2} − N^{2}C_{q}(a) = 0.  (32)

Thus, because of our choice of functions, the leading terms vanish, and the remaining term is small.

3.2.2 Second Piece

We now study the second piece. Note |β| ≤ \frac{Q}{N} = \frac{log^{2} B}{N}, and C_{q}(a) = \frac{c_{q}(a)}{ϕ^{2}(q)} \frac{c_{q}(−2a)}{ϕ^{2}(q)}.

Up to lower order terms, the m_{2}-sum will leave us with

β \frac{c_{q}(−2a)N}{ϕ(q)} ∫_{0}^{N} ∑_{m_{1}≤t} [λ(m_{1})e(m_{1} \frac{a}{q}) − \frac{c_{q}(a)}{ϕ(q)}]e(tβ)dt.  (33)

Note f_{N}(x) is a multiple of N^{2} for x near \frac{a}{q}. Thus, we want to make sure the above is well dominated by N^{2}.

For t ≤ \sqrt{N}, this is immediate. For t ≥ \sqrt{N}, using Siegel-Walfisz (Theorem 1.3), we can make the bracketed quantity in the integrant dominated by \frac{N}{log^{C} N} for any C when q ≤ log^{B} N. Thus, we integrate a quantity that is at most \frac{N}{log^{C} N} over an interval of length N, we multiply by Nβ ≪ Q = log^{B} N.

Thus, choosing C appropriately, the integral contributes \frac{N^{2}}{log^{C−B} N}, and hence is negligible.

Remark 3.1. Note, of course, that the contribution is only negligible while |β| ≤ \frac{Q}{N}.

Lemma 3.2. S_{1 ∑;a,q} is a lower order correction.

3.3 S_{1 ∫;a,q}

We must evaluate

S_{1 ∫;a,q} = ∑_{m_{1}≤N} 4πiβ ∫_{0}^{N} ∑_{m_{2}≤u} a_{m_{2}}e(−uβ)du · e(m_{1}β),  (34)

where

a_{m_{2}} = [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)].  (35)

We bring the sum over m_{1} inside the integral and again use Partial Summation.

We will ignore the integration and β for now, as these will contribute βN ≪ Q = log^{B} N times the maximum value of the integrand. We will leave the e(−uβ)du with this integration.

When u ≤ \sqrt{N}, we can immediately show the above is a lower order correction. Thus, below we always assume u ≥ \sqrt{N}.

3.3.1 First Piece

We have

S_{1 ∫ ∑;a,q} = ∑_{{m_{1}≤N \atop m_{2}≤u}} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]e(Nβ)

= e(Nβ)[ ∑_{{m_{1}≤N \atop m_{2}≤u}} λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a) ∑_{{m_{1}≤N \atop m_{2}≤u}} 1].

= e(Nβ)[C_{q}(a)uN − C_{q}(a)uN + Lower Order Terms], (36)

where by the Siegel-Walfisz Theorem (Theorem 1.3), the error in the bracketed quantity is of size \frac{uN}{log^{C} N}.

We then integrate from u = \sqrt{N} to N and multiply by β, giving a contribution bounded by

βN · \frac{N^{2}}{log^{C} N} ≪ \frac{log^{B}}{N} \frac{N^{3}}{log^{C} N} ≪ \frac{N^{2}}{log^{C−B} N} ,  (37)

again getting a lower order correction to f_{N}(x) for x near \frac{a}{q} (remember f_{N}(x) is of size N^{2}).

3.3.2 Second Piece

Again, u ≥ \sqrt{N}, and we have

2πiβ ∫_{0}^{N} ∑_{m_{1}≤t} [ ∑_{m_{2}≤u} [λ(m_{1})λ(m_{2})e((m_{1} − 2m_{2})\frac{a}{q}) − C_{q}(a)]]e(tβ)dt. (38)

Again, for t ≤ \sqrt{N}, the contribution will be a lower order correction. For t, u ≥ \sqrt{N},

Again, executing the sum over m_{1} and m_{2} will give us

C_{q}(a)ut − C_{q}(a)ut + Lower Order Terms,  (39)

with the lower order terms of size \frac{ut}{log^{C} N}. Integrating over t (from \sqrt{N} to N), then integrating over u (from \sqrt{N} to

N) and then multiplying by β^{2} gives an error bounded by

β^{2}N^{2} · \frac{N^{2}}{log^{C} N} ≪ \frac{log^{2B} N}{N^{2}} \frac{N^{4}}{log^{C} N} ≪ \frac{N^{2}}{log^{C−2B} N} ,  (40)

again a lower order correction.

4 Integrals of u(x)

4.1 Formulations

Remember

u(x) = ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2})x).  (41)

We need to study ∫− \frac{1}{2}^{\frac{1}{2}} f_{N}(x)e(−x)dx. We have shown that

f_{N}(α) = C_{q}(a)u(α − \frac{a}{q}) + O(\frac{N^{2}}{log^{C−2B} N} ), α ∈ M_{a,q}.  (42)

Thus, we must evaluate

∫_{M_{a,q}} u(α − \frac{a}{q}) · e(−α)dα = ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} u(α − \frac{a}{q}) · e(−α)dα
= ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β) · e( − \frac{q}{q} − β)dβ
= e( − \frac{a}{q}) ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β)e(−β)dβ. (43)

4.2 ∫−\frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx

∫− \frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx = ∫− \frac{1}{2}^{\frac{1}{2}} ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2})x) · e(−x)dx

= ∑_{m_{1},m_{2}≤N} ∫− \frac{1}{2}^{\frac{1}{2}} e((m_{1} − 2m_{2} − 1)x)dx.  (44)

If m_{1} − 2m_{2} − 1 = 0, the integral gives 1. There are approximately \frac{N}{2} ways to choose m_{1}, m_{2} ≤ N such that m_{1} − 2m_{2} − 1 = 0.

Assume now m_{1} − 2m_{2} − 1 ≠ 0. Then the integral vanishes.

Hence,

Lemma 4.1.

∫− \frac{1}{2}^{\frac{1}{2}} u(x)e(−x)dx = \frac{N}{2} + O(1).  (45)

4.3 ∫_{−\frac{1}{2}}^{−\frac{Q}{N}} + ∫ \frac{1}{2} ; \frac{Q}{N} u(x)e(−x)dx

Define

I_{1} = [ − \frac{1}{2}, −\frac{1}{2} + \frac{Q}{N} ]
I_{2} = [ − \frac{1}{2} + \frac{Q}{N} , −\frac{Q}{N} ]
I_{3} = [\frac{Q}{N} , \frac{1}{2} − \frac{Q}{N} ]
I_{4} = [\frac{1}{2} − \frac{Q}{N} , \frac{1}{2}]
I = I_{1} ∪ I_{2} ∪ I_{3} ∪ I_{4}.  (46)

4.4 Integral over I_{2}, I_{3}

We have

∫_{I_{i}} u(x)e(−x)dx = ∫_{I_{i}} ∑_{m_{1},m_{2}≤N} e((m_{1} − 2m_{2} − 1)x)dx

= ∫_{I_{i}} ∑_{m_{1}≤N} e(m_{1}x) ∑_{m_{2}≤N} e(−2m_{2}x) · e(−x)dx

= ∫_{I_{i}} \frac{e(x) − e((N + 1)x)}{1 − e(x)} \frac{e(−2x) − e(−2(N + 1)x)}{1 − e(−2x)}e(−x)dx.

(47)

On I_{2} and I_{3}, the integral is

≪ ∫_{I_{i}} \frac{2}{x} \frac{2}{x}dx ≪ \frac{N}{Q} = \frac{N}{log^{B} N} ,  (48)

see, for example, Nathanson (Additive Number Theory: The Classical Bases, Chapter 8).

4.5 Integral over I_{1}, I_{4}

Each of these intervals has length \frac{Q}{N} = \frac{log^{B} N}{N}. There are \frac{N}{2} + O(1) pairs such that m_{1} − 2m_{2} − 1 = 0. Each of these pairs will contribute (bound the integrand by 1) \frac{Q}{N}. As there are at most \frac{N}{2} pairs, these contribute at most \frac{N}{2} \frac{Q}{N} ≪ log^{B} N.

Henceforth we assume m_{1} − 2m_{2} − 1 ≠ 0. We write

I_{1} ∪ I_{4} = [\frac{1}{2} − \frac{Q}{N} , \frac{1}{2} + \frac{Q}{N} ] = I′.  (49)

We have

∑_{{m_{1},m_{2}≤N \atop m_{1}−2m_{2}−1≠0}} ∫_{I′} e((m_{1} − 2m_{2} − 1)x)dx

= e( − \frac{1}{2}) ∑_{{m_{1},m_{2}≤N \atop m_{1}−2m_{2}−1≠0}}(−1)^{m_{1}} ∫− \frac{Q}{N}^{\frac{Q}{N}} e((m_{1} − 2m_{2} − 1)x)dx

= e( − \frac{1}{2})\frac{1}{2πi} ∑_{{m_{1},m_{2}≤N \atop m_{1}−2m_{2}−1≠0}}(−1)^{m_{1}} \frac{2 sin ((m_{1} − 2m_{2} − 1)\frac{Q}{N} )}{m_{1} − 2m_{2} − 1}, (50)

because, changing variables by sending x to (x − \frac{1}{2}) + \frac{1}{2} gives factors of

e((m_{1} − 2m_{2} − 1)\frac{1}{2}) = e(−\frac{1}{2})e(\frac{m_{1}}{2} )e(−m_{2}), and e(\frac{m_{1}}{2} ) = (−1)^{m_{1}}.

4.5.1 0 < |m_{1} − 2m_{2} − 1| ≤ N^{1−ϵ}

Let w = m_{1} − 2m_{2} − 1. We will do the case 0 < w ≤ N^{1−ϵ}, the case with −N^{1−ϵ} > w > 0 being handled similarly.

For each w, there are at most N pairs of m_{1}, m_{2} giving rise to such a w. For such w, \frac{sin(w \frac{Q}{N})}{w} ≪ \frac{Q}{N} (because we are taking the sin of a quantity very close to zero).

Thus, these pairs contribute at most

≪ N · \frac{Q}{N} ≪ Q = log^{B} N.  (51)

Inserting absolute values in Equation 50 gives a contribution of at most log^{B} N for such w, 0 < w < N^{1−ϵ}.

4.5.2 N^{1−ϵ} < |m_{1} − 2m_{2} − 1| ≤ N

Again, let w = m_{1} − 2m_{2} − 1 and assume N^{1−ϵ} < |w| ≤ N. We will only consider w > 0; w < 0 is handled similarly.

The cancellation is due to the presence of the factor (−1)^{m_{1}}; note that for the pair (m_{1}, m_{2}) we only care about the parity of m_{1}.

Consider w and w − 1.

For m_{1} − 2m_{2} − 1 = w, the solutions are

m_{1} = w + 3, m_{2} = 1
m_{1} = w + 5, m_{2} = 2
m_{1} = w + 7, m_{2} = 3  (52)

and so on; thus there are about \frac{N−w}{2} pairs, all with parity −(−1)^{w}.

For m_{1} − 2m_{2} − 1 = w − 1, we again have about \frac{N−w}{2} pairs, but now the parity is (−1)^{w}. Thus, each of the \frac{N−w}{2} pairs with m_{1} − 2m_{2} − 1 = w is matched with one of the \frac{N−w}{2} pairs with m_{1} − 2m_{2} − 1 = w − 1, and we are off by at most O(1) pairs, which will contribute

≪ ∑_{w=N^{1−ϵ}}^{N} \frac{1}{w} ≪ log N.  (53)

For the remaining terms, we subtract in pairs, using the first order Taylor Expansion of sin(x). We have

∑_{w=N^{1−ϵ}}^{N} [\frac{sin (w\frac{Q}{N} )}{w} − \frac{sin (w\frac{Q}{N} − \frac{Q}{N} )}{w − 1}].  (54)

The Main Term of the Taylor Expansion gives ≪ \frac{1}{w^{2}} , which when summed over w gives \frac{1}{N^{1−ϵ}}. As we have about \frac{N−w}{2} ≪ N pairs for each w, this contributes at most N · \frac{1}{N^{1−ϵ}} ≪ N^{ϵ}.

We also have the first order term from the Taylor Expansion:

sin (w\frac{Q}{N} − \frac{Q}{N} ) = sin (w\frac{Q}{N} ) + O(\frac{Q}{N} ).  (55)

This error leads to (remembering there are \frac{N−w}{2} ≪ N pairs for each w)

≪ N ∑_{w=N^{1−ϵ}}^{N} \frac{\frac{Q}{N}}{w − 1} ≪ Q log N^{ϵ} ≪ log^{B+1} N.  (56)

4.6 Collecting the Pieces

We have shown

∫_{[− \frac{1}{2} , \frac{1}{2} ]} u(x)e(−x)dx = \frac{N}{2} + O(1)
∫_{[− \frac{1}{2} , \frac{1}{2} ]−[− \frac{Q}{N} , \frac{Q}{N}]} u(x)e(−x)dx = O(\frac{N}{log^{B} N} ).  (57)

Therefore Lemma 4.2.

∫− \frac{Q}{N}^{\frac{Q}{N}} u(x)e(−x)dx = \frac{N}{2} + O(\frac{N}{log^{B} N} ).  (58)

Remembering that we had

∫_{M_{a,q}} u(α − \frac{a}{q}) · e(−α)dα = ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} u(α − \frac{a}{q}) · e(−α)dα
= ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β) · e( − \frac{q}{q} − β)dβ
= e( − \frac{a}{q}) ∫− \frac{Q}{N}^{\frac{Q}{N}} u(β)e(−β)dβ, (59)

we see that Lemma 4.3.

∫_{M_{a,q}} u(α − \frac{a}{q}) · e(−α)dα = e( − \frac{a}{q})\frac{N}{2} .  (60)

5 Determination of the Main Term

We now calculate the contribution from the Major Arcs. Up to lower order terms,

∫_{M} f_{N}(x)e(−x)dx = ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} f_{N}(α)e(−α)dα
= ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} ∫_{\frac{a}{q} − \frac{Q}{N}}^{\frac{a}{q} + \frac{Q}{N}} C_{q}(a)u(α − \frac{a}{q})e(−α)dα
= ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} e( − \frac{a}{q}) ∫− \frac{Q}{N}^{\frac{Q}{N}} C_{q}(a)u(β)e(−β)
= ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q})\frac{N}{2}
= \frac{N}{2} ∑_{q≤Q} ∑_{{a=1 \atop (a,q)=1}}^{q} \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)} · e( − \frac{a}{q})
= \frac{N}{2} ∑_{q=1}^{Q} [ ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q})]
= \frac{N}{2} ∑_{q=1}^{Q} ρ_{q}
= 𝔖_{N} \frac{N}{2} ,  (61)

where we have defined

c_{q}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q})

C_{q}(a) = \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)}
ρ_{q} = ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q})

𝔖_{N} = ∑_{{a=1 \atop (a,q)=1}}^{q} ρ_{q}.  (62)

5.1 Properties of C_{q}(a) and ρ_{q}

We will follow the presentation of Nathanson (Additive Number Theory: The Classical Bases, Chapter 8 and Appendix A).

5.1.1 c_{q}(a) is Multiplicative

We follow Nathanson, Pages 320 − 321, Theorem A.23. Note that we are labeling by r what he labels a, and we are labeling by a what he labels n.

Lemma 5.1. c_{q}(a) is multiplicative; ie, if (q, q′) = 1, then c_{qq′}(a) = c_{q}(a)c_{q′}(a).

Proof: We have

∑_{{r̃=1 \atop (r̃,qq′)=1}}^{qq′} e(r̃\frac{a}{qq′} ).  (63)

Exercise 5.2. Show that we can write the r̃s above as r̃ ≡ rq′ + r′q mod qq′, where 1 ≤ r ≤ q, 1 ≤ r′ ≤ q′, and (r, q) = (r′, q′) = 1.

Thus

c_{q}(a)c_{q′}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q}) ∑_{{r′=1 \atop (r′,q′)=1}}^{q′} e(r′ \frac{a}{q′} )
= ∑_{{r=1 \atop (r,q)=1}}^{q} ∑_{{r′=1 \atop (r′,q′)=1}}^{q′} e(\frac{(rq′ + r′q)a}{qq′})
= ∑_{{r̃=1 \atop (r̃,qq′)=1}}^{qq′} e(r̃\frac{a}{q}) = c_{qq′}(a).  (64)

5.1.2 c_{q}(a) for (a, q) = 1 Exercise 5.3. Show that

h_{d}(a) = ∑_{r=1}^{d} e(r\frac{a}{d}) = \begin{cases}d & if d|a \\ 0 & otherwise \end{cases}  (65)

Recall the moebius function:

μ(d) = \begin{cases}(−1)^{r} & if d is the product of r distinct primes \\ 0 & otherwise \end{cases}  (66)

Exercise 5.4. Prove

∑_{d|(r,q)} μ(d) = \begin{cases}1 & if (r, q) = 1 \\ 0 & otherwise \end{cases}  (67)

Then

c_{q}(a) = ∑_{{r=1 \atop (r,q)=1}}^{q} e(r\frac{a}{q})

= ∑_{r=1}^{q} e(r\frac{a}{q}) ∑_{d|(r,q)} μ(d)

= ∑_{d|q} μ(d) ∑_{{r=1 \atop d|r}}^{q} e(r\frac{a}{q})

= ∑_{d|q} μ(d) ∑_{l=1}^{\frac{q}{d}} e(l \frac{a}{\frac{q}{d}})
= ∑_{d|q} μ(d)h_{\frac{q}{d}} (a)
= ∑_{d|q} μ(\frac{q}{d})h_{d}(a)
= ∑_{{d|q \atop d|a}} μ(\frac{q}{d}) · d

= ∑_{d|(a,q)} μ(\frac{q}{d})d.  (68)

Note that if (a, q) = 1, then there is only one term above, namely d = 1, which yields

c_{q}(a) = μ(q) if (a, q) = 1.  (69)

Corollary 5.5. If q = p^{k}, k ≥ 2 and (a, q) = 1, then c_{q}(a) = 0.

5.1.3 C_{q}(a) is Multiplicative

We have shown c_{qq′}(a) = c_{q}(a)c_{q′}(a) if (q, q′) = 1. Recall the Euler phifunction, ϕ(q), is the number of numbers less than q which are relatively prime to q.

Exercise 5.6. Prove that ϕ(q) is multiplicative; ie, if (q, q′) = 1, then ϕ(qq′) = ϕ(q)ϕ(q′).

We now have Lemma 5.7. C_{q}(a) is multiplicative.

Proof: Assume (q, q′) = 1. We have

C_{qq′}(a) = \frac{c_{qq′}(a)c_{qq′}(−2a)}{ϕ^{2}(qq′)}
= \frac{c_{q}(a)c_{q′}(a)c_{q}(−2a)c_{q′}(−2a)}{ϕ^{2}(q)ϕ^{2}(q′)}
= \frac{c_{q}(a)c_{q}(−2a)}{ϕ^{2}(q)} · \frac{c_{q′}(a)c_{q′}(−2a)}{ϕ^{2}(q′)}
= C_{q}(a)C_{q′}(a).  (70)

5.1.4 ρ_{q} is Multiplicative We first prove a needed lemma. Lemma 5.8. Consider C_{q_{1}}(a_{1}q_{2}). Then

C_{q_{1}}(a_{1}q_{2}) = C_{q_{1}}(a_{1})  (71)

if (q_{1}, q_{2}) = 1.

Proof:

C_{q_{1}}(a_{1}q_{2}) = ∑_{{r_{1}=1 \atop (r_{1},q_{1})=1}}^{q_{1}} e(r_{1} \frac{a_{1}q_{2}}{q_{1}} )
= ∑_{{r_{1}=1 \atop (r_{1},q_{1})=1}}^{q_{1}} e(r_{1}q_{2} \frac{a_{1}}{q_{1}} )
= ∑_{{r=1 \atop (r,q_{1})=1}}^{q_{1}} e(r\frac{a_{1}}{q_{1}} ) = C_{q_{1}}(a),  (72)

because (q_{1}, q_{2}) = 1 implies that as r_{1} goes through all residue classes that are relatively prime to q_{1}, so too does r = r_{1}q_{2}. □

Lemma 5.9. ρ_{q} is multiplicative.

Recall

ρ_{q} = ∑_{{a=1 \atop (a,q)=1}}^{q} C_{q}(a)e( − \frac{a}{q}).  (73)

Assume (q_{1}, q_{2}) = 1. Then we can write the congruence classes mod q_{1}q_{2} as a_{1}q_{2} + a_{2}q_{1}, with 1 ≤ a_{1} ≤ q_{1}, 1 ≤ a_{2} ≤ q_{2} and (a_{1}, q_{1}) = (a_{2}, q_{2}) = 1.

ρ_{q_{1}q_{2}} = ∑_{{a=1 \atop (a,q_{1}q_{2})=1}}^{q_{1}q_{2}} C_{q_{1}q_{2}}(a)e( − \frac{a}{q_{1}q_{2}} )

= ∑_{{a=1 \atop (a,q_{1}q_{2})=1}}^{q_{1}q_{2}} C_{q_{1}}(a)C_{q_{2}}(a)e( − \frac{a}{q_{1}q_{2}} )

= ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{1}}(a_{1}q_{2} + a_{2}q_{1})C_{q_{2}}(a_{1}q_{2} + a_{2}q_{1})e( − \frac{a_{1}q_{2} + a_{2}q_{1}}{q_{1}q_{2}} ).

(74)

Exercise 5.10. With a_{1}, a_{2}, q_{1}, q_{2} as above,

C_{q_{1}}(a_{1}q_{2} + a_{2}q_{1}) = C_{q_{1}}(a_{1}q_{2}) and C_{q_{2}}(a_{1}q_{2} + a_{2}q_{1}) = C_{q_{2}}(a_{2}q_{1}). (75)

Thus, we have

ρ_{q_{1}q_{2}} = ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{1}}(a_{1}q_{2})C_{q_{2}}(a_{2}q_{1})e( − \frac{a_{1}q_{2} + a_{2}q_{1}}{q_{1}q_{2}} )

= ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} C_{q_{1}}(a_{1}q_{2})e( − \frac{a_{1}}{q_{1}} ) ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{2}}(a_{2}q_{1})e( − \frac{a_{2}}{q_{2}} )

= ∑_{{a_{1}=1 \atop (a_{1},q_{1})=1}}^{q_{1}} C_{q_{1}}(a_{1})e( − \frac{a_{1}}{q_{1}} ) ∑_{{a_{2}=1 \atop (a_{2},q_{2})=1}}^{q_{2}} C_{q_{2}}(a_{2})e( − \frac{a_{2}}{q_{2}} )
= ρ_{q_{1}} · ρ_{q_{2}}.  (76)

Thus, ρ_{q} is multiplicative. □

5.1.5 Calculation of ρ_{q} Lemma 5.11. ρ_{p^{k}} = 0 if k ≥ 2 and p is a prime.

Proof: This follows immediately from C_{p^{k}}(a) = 0. □ Lemma 5.12. If p > 2 is prime, ρ_{p} = −\frac{1}{(p−1)^{2}} .

Proof:

ρ_{p} = ∑_{{a=1 \atop (a,p)=1}}^{p} C_{p}(a)e( − \frac{a}{p})
= ∑_{a=1}^{p−1} \frac{c_{p}(a)c_{p}(−2a)}{ϕ^{2}(p)}e( − \frac{a}{p}).  (77)

But as p > 2, c_{p}(a) = c_{p}(−2a) = μ(p) as (a, p) = 1. As μ^{2}(p) = 1 and ϕ(p) = p − 1 we have

ρ_{p} = ∑_{a=1}^{p−1} \frac{1}{(p − 1)^{2}} e( − \frac{a}{p})
= \frac{1}{(p − 1)^{2}} [ − e( − \frac{0}{p}) + ∑_{a=0}^{p−1} e( − \frac{a}{p})]
= −\frac{1}{(p − 1)^{2}} .  (78)

Lemma 5.13. If p = 2, then ρ_{2} = 1.

Proof:

ρ_{2} = ∑_{{a=1 \atop (a,2)=1}}^{2} C_{2}(a)e( − \frac{a}{2})
= C_{2}(1)e( − \frac{1}{2})
= \frac{c_{2}(1)c_{2}(−2)}{ϕ^{2}(2)} · e^{−πi}
= \frac{e^{πi}e^{−2πi}}{1^{2}} · e^{−πi} = 1,  (79)

where we have used c_{2}(1) = e^{πi} and c_{2}(−2) = e^{−2πi}.

Exercise 5.14. Prove c_{2}(1) = e^{πi} and c_{2}(−2) = e^{−2πi}.

5.2 Determination of 𝔖_{N} and 𝔖

Recall

𝔖_{N} = ∑_{q≤Q} ρ_{q}.  (80)

We define

𝔖 = ∑_{q} ρ_{q}.  (81)

Exercise 5.15. Let h_{q} be any multiplicative sequence (with whatever growth conditions are necessary to ensure the convergence of all sums below). Then

∑_{q} h_{q} = ∏_{p prime} (1 + ∑_{k=1}^{∞} h_{p^{k}}).  (82)

5.2.1 𝔖

We have

𝔖 = ∑_{q} ρ_{q}

= ∏_{p prime} (1 + ∑_{k=1}^{∞} ρ_{p^{k}})
= ∏_{p} (1 + ρ_{p})  (83)

because ρ_{p^{k}} = 0 for k ≥ 2 and p prime by Lemma 5.11. We have previously shown (see Lemmas 5.12 and 5.13) that ρ_{2} = 1 and ρ_{p} = −\frac{1}{(p−1)} for p > 2 prime. Therefore

𝔖 = ∏_{p} (1 + ρ_{p})
= (1 + ρ_{2}) ∏_{p>2}(1 + ρ_{p})
= 2 ∏_{p>2} [1 − \frac{1}{(p − 1)^{2}} ]
= 2T_{2},  (84)

where Definition 5.16 (Twin Prime Constant).

T_{2} = ∏_{p>2} [1 − \frac{1}{(p − 1)^{2}} ] ≈ .6601618158  (85)

is the twin prime constant.

5.2.2 𝔖_{N}

We need to estimate |𝔖 − 𝔖_{N}|. As ρ_{q} is multiplicative and zero if q = p^{k} (k ≥ 2), we see we need only look at sums of ρ_{p}. As ρ_{p} = −\frac{1}{(p−1)^{2}} , one can show that the difference between 𝔖 and 𝔖_{N} tends to zero as N → ∞.

Thus, Lemma 5.17.

𝔖 = 2T_{2}.  (86)

5.3 Number of Germain Primes and Weighted Sums

Combining the above arguments, we have shown that, up to lower order terms,

∑_{{p≤N \atop p, \frac{p−1}{2} prime}} log(p) · log (\frac{p − 1}{2}) = 𝔖\frac{N}{2}
= 2T_{2} \frac{N}{2}
= T_{2}N.  (87)

Note that we are counting Germain prime pairs by (\frac{p−1}{2}, p) and not (p, 2p + 1). Such a difference in counting will introduce a factor of 2.

We can pass from this weighted sum to a count of the number of Germain prime pairs (\frac{p−1}{2}, p) with p ≤ N.

Again we follow Nathanson, Chapter 8. Define

π_{G}(N) = ∑_{{p≤N \atop p, \frac{p−1}{2} prime}} 1

G(N) = ∑_{{p≤N \atop p, \frac{p−1}{2} prime}} log(p) · log (\frac{p − 1}{2}).  (88)

Clearly

G(N) ≤ log^{2} N · π_{G}(N).  (89)

Therefore,

Lemma 5.18. Up to lower order terms,

π_{G}(N) ≥ \frac{G(N)}{log^{2} N} = \frac{T_{2}N}{log^{2} N} .  (90)

We now provide a bound in the opposite direction.

π_{G}(N^{1−δ}) = ∑_{{p≤N^{1−δ} \atop p, \frac{p−1}{2} prime}} 1 ≪ \frac{N^{1−δ}}{log N} .  (91)

Then

G(N) ≥ ∑_{{p≥N^{1−δ} \atop p, \frac{p−1}{2} prime}} log p · log (\frac{p − 1}{2})
= (1 − δ)^{2} log^{2} N ∑_{{p≥N^{1−δ} \atop p, \frac{p−1}{2} prime}} 1
= (1 − δ)^{2} log^{2} N(π_{G}(N) − π_{G}(N^{1−δ}))
≥ (1 − δ)^{2} log^{2} Nπ_{G}(N) + O((1 − δ)^{2} log^{2} N · \frac{N^{1−δ}}{log N} ). (92)

Therefore

log^{2} N · π_{G}(N) ≤ (1 − δ)^{−2} · G(N) + O( log^{2} N · \frac{N^{1−δ}}{log N} )
0 ≤ log^{2} N · π_{G}(N) − G(N) ≤ [(1 − δ)^{−2} − 1]G(N) + O( log N · N^{1−δ}). (93)

If 0 < δ < \frac{1}{2}, then (1 − δ)^{−2} − 1 ≪ δ. We thus have

0 ≤ log^{2} N · π_{G}(N) − G(N) ≪ N[δ + O(\frac{log N}{N^{δ}} )].  (94)

Choose δ = \frac{2 log log N}{log N}. Then we get

0 ≤ log^{2} N · π_{G}(N) − G(N) ≤ O(N \frac{log log N}{log N} ).  (95)

Recalling G(N) ≈ T_{2}N gives Lemma 5.19.

π_{G}(N) ≤ \frac{T_{2}N}{log^{2} N} .  (96)

Combining with the other bound we have finally shown Theorem 5.20. Assuming there is no contribution to the main term from the Minor Arcs, up to lower order terms we have

π_{G}(N) = \frac{T_{2}N}{log^{2} N} ,  (97)

where T_{2} is the twin prime constant

T_{2} = ∏_{p>2} [1 − \frac{1}{(p − 1)^{2}} ] ≈ .6601618158.  (98)