Proof and method of three prime number theorem (I.)

The purpose of this article is to let oneself learn Goldbach conjecture study of specific methods, specific reference Pan book "Prime Distribution and Goldbach conjecture", in this I will prove the details as far as possible to write more detailed, convenient later reference again. Because it is all self-taught, in this first article, only a few coarser estimates are considered, which is enough to prove the three prime theorems below. Even so, the proof of the theorem is no easy feat.

The three primes theorem each sufficiently large odd number is the sum of three singular primes.

The theorem was first demonstrated by Vinogradov in 1937, using the Hardy-littlewood circle method and his own triangle and estimation methods to prove the above conclusions, which are shown in detail below. It is important to note that the proof here is not practical. That is, we can only get the existence of a constant $c _1$, so that when the odd $n >c_1$, $n $ for the sum of three odd primes, but the method does not specifically calculate the constant $c _1$. Before proving, list some lemma. Note that all occurrences of the symbol $p $, $p _1$, $p _2$, and so on represent prime numbers.

lemma 1.1. $\tau \geq 1$, then for any real number $\alpha$, there are rational numbers $\frac{a}{q}$,$ (a,q) =1$,$1 \leq q \leq \tau$, making $$\left| \alpha-\frac{a}{q} \right| \leq \frac{1}{q \cdot \tau}. $$ further, when $\alpha > 0$, $a \geq 0$.

lemma 1.2. set $q $, $m $ for a positive integer, $ (q,m) =1$, then $$\underset{(h,q) =1}{\sum^{q}_{h=1}} e^{2 \pi i \frac{hm}{q}} = \mu (q), $$ here $ \MU (q) $ is defined as follows: $$\mu (q) =\begin{cases} 1, & \text{if}~q=1,\\ ( -1) ^s, & ~q=p_1 p_2 \cdots p_s,\\ 0, & if ~q~ is The square of a prime number is removed. \end{cases}$$ $\MU (q) $ is called the Möbius function.

We remember that $\pi (X;Q.L) $ represents no more than the number of $x $ in arithmetic progression $l + qm$. The following lemma is very profound, known as the Siegel-walfisz theorem. It is obvious that it can launch the Dirichlet theorem. The proof of Siegel-walfisz theorem is described in chapter 19-20 of Harold Davenport's book multiplicative number theory (Second Edition, GTM74, 1980).

lemma 1.3. set $x \geq 2$, then the arbitrary fixed positive number $A >1$, and any positive integer $q $, $l $, satisfies $$1\leq Q \leq \log^a x, (l,q) =1$$ There is an asymptotic formula $$\pi (x;q,l) =\ Frac{\text{li} (x)}{\varphi (q)} + \mathcal{o} (xe^{-c_2 \sqrt{\log x}) $$ is established, where the constant $c _2$ depends only on $A $, and the $\mathcal{o}$ constant is an absolute constant, $c _2$ cannot be actually calculated.

The third prime theorem is then proved.

Set $N \geq 9$ is odd, with $r (N) $ to represent the number of solutions of the equation $ $N =p_1 + p_2 + p_3, ~~p_1, ~p_2, ~p_3 \leq n$$. Then the weak Goldbach conjecture is that $r (N) >0$ to the $N \geq 9$ are set up, and currently can not testify so strong conclusion. Consider integral $$\int^1_0 e^{2 \pi i \alpha (p_1 + p_2 + p_3-n)} d\alpha. \tag{1}$$ Obviously, if $N =p_1 + p_2 + p_3$, then $ (1) $ type integral equals $1$. If $N \neq p_1 + p_2 + p_3$, then $ (1) $ equals $0$. Easy to know

\begin{align*}
R (N) &= \sum_{p_1 \leq n}\sum_{p_2 \leq n}\sum_{p_3 \leq N} \int^1_0 e^{2 \pi i \alpha (p_1 + p_2 + p_3-n)} D\alpha \\
&= \int^1_0 (\sum_{p_1 \leq N} e^{2 \pi i \alpha p_1} \sum_{p_2 \leq n} e^{2 \pi i \alpha p_2} \sum_{p_3 \leq N} e^{2 \pi i \alpha p_3}) e^{-2 \pi i \alpha N} d\alpha \
&= \int^1_0 (\sum_{p \leq N} e^{2 \pi i \alpha p}) ^3 e^{-2 \pi i \alpha N} d\alpha \ \
&= \int^1_0 s^3 (\alpha) e^{-2 \pi i \alpha N} d\alpha, \tag{2}
\end{align*}

Here $S (\alpha) =\sum_{p \leq N} e^{2 \pi i \alpha p}. $

To estimate $r (N) $, the first round method is used. The following constant set $N $ for full size odd.

Since $ (2) $ in the integrand to $1$ for the period, so the arbitrary $\tau \geq 1$,$ (2) $ can be changed to write $ $r (N) = \int^{1-\frac{1}{\tau}}_{-\frac{1}{\tau}} s^3 (\alpha) E^{-2 \pi i \alpha N} d\alpha. \tag{3}$$ according to lemma 1.1 ,  arbitrary $\alpha \in \left[-\frac{1}{\tau}, 1-\frac{1}{\tau } \right) $, with $$\alpha = \frac{a}{q} + \beta, to \leq q \leq \tau, ~ (a,q) =1, ~|\beta| \leq \frac{1}{q \cdot \tau} \tag{4},$$ where $ \leq a \leq q-1$. This is because, if $\alpha \in \left[-\frac{1}{\tau}, 0\right]$, there is $a =0$, $q =1$, making $|\alpha-\frac{a}{q}| \leq \frac{1}{q \cdot \tau}$. If $\alpha \in \left (0,1-\frac{1}{\tau} \right) $, this time $\tau>1$. Therefore there is lemma 1.1 knowledge exists $a, ~q:~a \geq 0, to \leq q \leq \tau, ~ (a,q) =1$, making $$\left| \alpha-\frac{a}{q} \right| \leq \frac{1}{q \cdot \tau}. $$ $|\alpha-\beta| \leq |\alpha| + |\beta| < 1-\frac{1}{\tau}+\frac{1}{\tau}= 1$, thus $ \leq \frac{a}{q} < 1$, pushed to \leq a \leq q-1$.

Now take $\tau = N (\log N) ^{-20}$, kee $$\mathscr{a} = \left\{\frac{a}{q}: ~ 0 \leq A \leq q-1, ~ (a,q) =1, ~q \leq \log^{15}n \ri ght\},$$ when $N $ full large, meet $2q \leq 2 \log^{15} N < \tau < n$. For any $\frac{a}{q} \in \mathscr{a}$, Kee $\mathfrak{m} (a,q) = \left[\frac{a}{q}-\frac{1}{q \cdot \tau}, \frac{a}{q}+\frac{1 }{q \cdot \tau} \right]$. Obviously, arbitrary $\alpha \in \mathscr{a}$, there are $$\left| \alpha-\frac{a}{q} \right| \leq \frac{1}{q \cdot \tau}. $$

We want to prove that for any $\frac{a_1}{q_1}, \frac{a_2}{q_2} \in \mathscr{a}$ when $ (a_1-a_2) ^2 + (q_1-q_2) ^2 \neq 0$, $\mathfrak{m} (A_ 1,q_1) $ with $\mathfrak{m} (a_2,q_2) $ is disjoint.

Because $q _1 \leq \log^{15} n$, $q _2 \leq \log^{15} n$, the distance between $\frac{a_1}{q_1}$ and $\frac{a_2}{q_2}$ can not be too small, that is, the following formula: $$\left| \frac{a_1}{q_1}-\frac{a_2}{q_2} \right| = \left| \frac{a_1q_2-a_2q_1}{q_1q_2} \right| \geq \frac{1}{q_1q_2}. $$ on the other hand, apparently $$\frac{1}{q_1 \cdot \tau} + \frac{1}{q_2 \cdot \tau} = \frac{q_1 + Q_2}{q_1q_2\tau} < \frac{\tau}{q_1q_2\ta U} = \frac{1}{q_1q_2}. If the $$ intersect, the $\left| \frac{a_1}{q_1}-\frac{a_2}{q_2} \right| \leq \frac{1}{q_1 \cdot \tau} + \frac{1}{q_2 \cdot \tau} < \frac{1}{q_1q_2}$, contradiction.

It is easy to know that these neighborhoods are contained within the interval $\left[-\frac{1}{\tau}, 1-\frac{1}{\tau}\right) $. This is because of the arbitrary $\mathfrak{m} (A,Q) \in \mathscr{a}$ and any $\alpha \in \mathfrak{m} (A,Q) $, with $\frac{a}{q} \leq 1-\frac{1}{q} < 1-\frac{2}{\tau}$, where $\alpha = \frac{a}{q} + \beta$ and satisfies $ (4) $, thus $$|\alpha| = |\alpha-\frac{a}{q} + \frac{a}{q}| \leq \frac{1}{q \cdot \tau} + \frac{a}{q} < \frac{1}{\tau} +1-\frac{2}{\tau} = 1-\frac{1}{\tau}. $$

It may therefore be recalled that the whole of these communities is $\mathfrak{m}$, i.e. $\mathfrak{m} = \bigcup \left\{\mathfrak{m} (A,Q): \frac{a}{q} \in \mathscr{a} \right\} $ In the interval $\left[-\frac{1}{\tau}, 1-\frac{1}{\tau} \right) The remainder of the set $\mathfrak{m}$ after the drop of the collection is recorded as $\mathfrak{e}$, there is $ $r (n) =r_1 (n) + r_ 2 (n), \tag{5}$$ here $ $r _1 (n) = \int_{\mathfrak{m}} s^{3} (\alpha) e^{-2 \pi i \alpha N} d\alpha,\tag{6}$$ $ $r _2 (n) = \int_{\m Athfrak{e}} s^{3} (\alpha) e^{-2 \pi i \alpha N} d\alpha,\tag{7}$$ Our aim is to prove $r _1 (n) $ is the main part of $r (N) $, while $r _2 (n) $ is a minor part, which can be rolled out When $N $ for full large odd numbers, there is a constant $ $r (n) \geq r_1 (N)-|r_2 (n) | > 0. $$

Hardy and Littlewood said the above method is "round method". Because when \leq \alpha < 1$ \leq 2 \pi \alpha < 2\pi$, and $e ^{2 \pi i \alpha}$ is the unit circle on the complex plane, so $[0,1) $ one by one corresponds to the unit circle. Thus the division on the unit circle is divided on the half-closed half-open interval $\left[-\frac{1}{\tau}, 1-\frac{1}{\tau} \right) of length $1$. $\mathfrak{m}$ is called the split-arc (Major Arcs), and the split $\mathfrak{e}$ is known as counterclockwise (Minor Arcs).

The main parts are estimated first. We have the following theorem 2.1:

theorem 2.1. set $N $ for a sufficiently large odd number, then the following progressive formula is established: $ $r _1 (n) = \frac{1}{2} \mathscr{o} (n) \frac{n^2}{\log^3 (n)} + \mathcal{o}\left (\frac{n^2}{\log^4 (N)} \right), $$ where $$\mathscr{o} (n) = \underset{p}{\prod} \left (1 + \frac{1}{(p-1) ^3} \right) \underse t{p| N}{\prod} \left (1-\frac{1}{p^2-3p +3} \right) > 1 ~. $$ for proof theorem 2.1, we need the following lemma:

lemma 2.2. Set $\alpha = \frac{a}{q} + \beta$,$ (a,q) =1$, $q \leq \log^{15} n$,$|\beta| \leq \frac{1}{q \cdot \tau}$, there is $ $S (\alpha) = \FRAC{\MU (q)}{\varphi (q)} \sum^{n}_{n=3} \frac{e^{2 \pi i \beta n}}{\log N} + \mathcal{o}\left (ne^{-c_4 \sqrt{\log N}} \right), $$ where $c _4$ is a positive absolute constant.

Now proveslemma 2.2。 \begin{align*} s (\alpha) &= s (\frac{a}{q} + \beta) = \sum_{p \leq N} e^{2 \pi i \frac{a}{q} p} e^{2 \pi i \beta p} \ \ &= \sum_{\sqrt{n} < P \leq N} e^{2 \pi i \frac{a}{q} p} e^{2 \pi i \beta p} + \mathcal{o}\left (\sqrt{N} \right) \ \ &= \sum^{q}_{l=1} \underset{p \equiv L (\text{mod}q)}{\sum_{\sqrt{n} < P \leq N}} e^{2 \pi i \frac{a}{q} P} e^{2 \pi i \beta p} + \mathcal{o} \left (\sqrt{n} \right) \ &= \underset{(l,q) =1}{\sum^{q}_{l=1}} e^{2 \ Pi i \frac{a}{q}l} \underset{p \equiv L (\text{mod}q)}{\sum_{\sqrt{n} < P \leq N}} e^{2 \pi i \beta p} + \mathcal{ O}\left (\sqrt{n} \right), \end{align*} where, when $N $ full, there $q \leq \log^{15} N < \sqrt{n} <p$. Also $p $ is a prime number, thereby $ (p,q) =1$, also has $ (l,q) = 1$. This gets \begin{align*} S (\alpha) &= s\left (\frac{a}{q} + \beta \right) \ &= \underset{(l,q) =1}{\sum^{q}_{l=1}} e^{2 \pi i \frac{a}{q}l} \underset{p \equiv l \left (\text{mod} q \right)}{\sum_{\sqrt{n} < P \leq N}} e^{2 \pi I \beta p} + \MATHCAL{o}\left (\sqrt{n} \right). \TAG{8} \end{align*}  First consider $ $T (l) = \underset{p \equiv l \left (\text{mod}q \right)}{\sum_{\sqrt{n} < P \leq N}} e^{ 2 \pi i \beta p}. \tag{9}$$ The following formula is clearly established: $$\PI (n;q,l)-\pi (n-1;q,l) =\begin{cases} 1, ~~& n ~ is prime and ~n \equiv l \left (\text{mod}q \right);  \\  0, ~~& other situations. \ \ \end{cases}$$ will be substituting $ (9) $ with  \begin{align*}t (l) &= \underset{p \equiv l \left (\text{mod}q \right)}{\SUM_{\SQR T{n} < P \leq N}} e^{2 \pi i \beta p} \ &= \sum_{\sqrt{n} < N \leq n} (\pi (n;q,l)-\pi (n-1;q,l)) e^{2 \pi I \beta n} \ &=  \sum_{\sqrt{n} < n \leq N-1} \pi (n;q,l) e^{2 \pi i \beta n} +\sum_{\sqrt{n} < n \leq n} \pi (n-1;q,l) e^{2 \pi i \beta n} + \pi (n;q,l) e^{2 \pi i \beta n} \ \ &=  \sum_{\sqrt{n} < n \leq N-1} \pi (n;q,l) E^{2 \pi i \beta n} + \sum_{\sqrt{n}-1 < n \leq N-1} \pi (n;q,l) e^{2 \pi i \beta (n+1)} + \pi (n;q,l) e^{2 \pi i \beta n } \ \ &= \sum_{\sqrt{n} < N \leq N-1} \pi (n;q,l) \left (e^{2 \piI \beta n}-e^{2 \pi i \beta (n+1)} \right) +  \pi (n;q,l) e^{2 \pi i \beta n} + \mathcal{o}\left (\sqrt{n} \right) \ TAG{10}, \end{align*} where, $ (10) $ is due to  $\left| \pi (\lfloor \sqrt{n} \rfloor; q, l) e^{2 \pi i \beta \lfloor \sqrt{n} \rfloor} \right| \leq \sqrt{n}$, and  $\sqrt{n}-1 < \lfloor \sqrt{n} \rfloor \leq \sqrt{n}$. By lemma 1.3 when $N $ full size  \begin{align*} (&= \sum_{\sqrt{n} < N \leq N-1} \left (\frac{\text{li} (N)}{\varphi (q)} + \mathcal{o} (NE ^{-c_2 \sqrt{\log n}) \right) \left (e^{2 \pi i \beta n}-e^{2 \pi i \beta (n+1)} \right) + \ \ &+ ~ ~\left (\frac{\t Ext{li} (N)}{\varphi (q)} + \mathcal{o}\left (n e^{-c_2 \sqrt{\log n}} \right) \right) e^{2 \pi i \beta N} + \mathcal{o}\lef T (\sqrt{n} \right) \ &= \sum_{\sqrt{n} < N \leq N-1} \frac{\text{li} (n)}{\varphi (q)}\left (e^{2 \pi i \beta N}- E^{2 \pi i \beta (n+1)} \right) + \ \ &+ \sum_{\sqrt{n} < n \leq N-1} \mathcal{o}\left (ne^{-c_2 \sqrt{\log n}} \rig HT) \left (e^{2 \pi i \beta n}-e^{2 \pi i \beta (n+1)} \right) + \ \ &+ \frac{\text{li} (n)}{\varphi (q)} e^{2 \pi I \beta n} + \mathcal{o}\left (n e^{-c_2 \sqrt{\log n}} \right) + \mathcal{o}\left (\sqrt{n} \right) \tag{11} \end{a lign*} derivation is known, when the $x $ full large, $f _1 (x) = \frac{x}{e^{c_2 \sqrt{\log x}}}$ monotonically increment. In fact, $ $f ' _1 (x) = \frac{1-\frac{c_2}{2} (\log x) ^{-1/2}}{e^{c_2 \sqrt{\log x}}}。 and by \left|. E^{2 \pi i \beta n-e^{2 \pi i \beta (n+1)}} \right| = \left| E^{2 \pi I \beta}-1 \right|$$, and when $N $ full large, $|\beta| \leq \frac{1}{\tau} = \frac{\log^{20} n}{n}$ can be sufficiently small, and at this time there are $e ^{2 \pi i \beta}-1 = \mathcal{o} \left (|\beta| \right) $. Thus when $\sqrt{n} < n \leq n-1$ there is  $$\mathcal{o}\left (ne^{-c_2 \sqrt{\log n}} \right) \left (e^{2 \pi i \beta N}-e^ {2 \pi i \beta (n+1)} \right) = \mathcal{o}\left (ne^{-c_2 \sqrt{\log n}} \right) \mathcal{o}\left (|\beta| \right). $$ therefore, \begin{align*} &~~~~\sum_{\sqrt{N}< N \leq N-1} \mathcal{o} (ne^{-c_2 \sqrt{\log N}) \left (e^{2 \pi i \beta n}-e^{2 \pi i \beta (n+1)} \right) \ &= \sum_{\sqrt{n}< n \leq N-1} \mathcal{o}\left (|\beta| ne^{-c_2 \sqrt{\log N}} \right) \ &=  \mathcal{o}\left (|\beta| n^2 e^{-c_2 \sqrt{\log N}} \right)  \\ &= \mathcal{o}\left (\frac{n^2 e^{-c_2 \sqrt{\log N}}}{\tau} \right) \ \ &A mp;= \mathcal{o}\left (ne^{-c_2 \sqrt{\log n}} \log^{20} n \right). \end{align*}  Kee $f _2 (x) = x \log^{20}x e^{-c_2 \sqrt{\log x}}$, $f _3 (x) = x e^{-\frac{c_2}{2} \sqrt{\log x}}$. Then $f _4 (x) = \frac{f_2 (x)}{f_3 (x)} = \log^{20}x E^{-\frac{c_2}{2}\sqrt{\log x}}$. Make $x =e^{y^2}$, then $f _4 (y) = y^{40} e^{-\frac{c_2}{2} y}$. therefore $f _4 (x) = \mathcal{o} (1) $, thereby $$\mathcal{o}\left (ne^{-c_2 \sqrt{\log n}} \log^{20} n \right) = \mathcal{o}\left (n e^{-\ FRAC{C_2}{2} \sqrt{\log N}} \right). $$ therefore \begin{align*} (one) &= \sum_{\sqrt{n} < n \leq N-1} \frac{\text{li} (n)}{\varphi (q)} \left (e^{2 \pi i \beta N }-e^{2 \pi i \beta (n+1)} \right) + \mathcal{o}\left (n e^{-\frac{c_2}{2} \sqrt{\log n}} \right) + \frac{\text{li} (n)}{\ Varphi (q)} e^{2 \pi i \beta n} + \\ &+ \mathcal{o}\left (N e^{-c_2 \sqrt{\log n}} \right) + \mathcal{O}\left (\sqrt{n} \right) \ &= \sum_{\sqrt{n} < n \leq N-1} \frac{\text{li} (n)}{\varphi (q)} \left (e^{2 \pi i \beta N}- E^{2 \pi i \beta (n+1)} \right) + \frac{\text{li} (n)}{\varphi (q)} e^{2 \pi i \beta n} \ \ &+ \mathcal{o}\left (n e^{-\f rac{c_2}{2} \sqrt{\log N}} \right), \end{align*} where, when $N $ full, there are $\sqrt{n} < n e^{-\frac{c_2}{2} \sqrt{\log n}}$ and $N e^{-c_2 \sqrt{\log N}} < n e^{-\frac{c_2}{2} \sqrt{\log n}}$. You may wish to remember that the right-hand side is $ (12) $, then  \begin{align*} (&=) \frac{1}{\varphi (q)} \sum_{\sqrt{n} < N \leq n} \left (\text{li} (N) -\text{li} (n-1) \right)  e^{2 \pi i \beta n} + \ \ &+ \frac{\text{li} (\lfloor \sqrt{n} \rfloor)}{\varphi (q)} &NB Sp;e^{2 \pi i \beta (\lfloor \sqrt{n} \rfloor + 1)} + \mathcal{o}\left (N e^{-\frac{c_2}{2} \sqrt{\log N}} \right). \TAG{13} \end{align*} obviously $\varphi (q) \geq 1$, due to $\text{li} (\lfloor \sqrt{n} \rfloor) = \text{li} (3)  + \int^{\lfloor \ Sqrt{n} \rfloor}_{3} \frac{1}{\log t} DT \leq \text{li} (3) + \int^{\sqrt{n}}_{3} \frac{1}{\log t} dt \leq \text{Li} (3 ) + \sqrt{n}$, so $\frac{\text{li} (\lfloor \sqrt{n} \rfloor)}{\varphi (q)} e^{2 \pi i \beta (\lfloor \sqrt{n} \rfloor +1)} = \mathcal{o}\left (\sqrt{n} \right) $. thereby $ (13) $ \begin{align*}  (&=) \frac{1}{\varPhi (Q)} \sum_{\sqrt{n} < n \leq n} \left (\text{li} (N)-\text{li} (n-1) \right)  e^{2 \pi i \beta N} + \ \ &+ \ Mathcal{o}\left (n e^{-\frac{c_2}{2} \sqrt{\log n}} \right) \ &= \frac{1}{\varphi (q)} \sum_{\sqrt{n} < n \leq n} \ Left (\int^{n}_{n-1} \frac{1}{\log t} dt \right) e^{2 \pi i \beta n} + \mathcal{o}\left (n e^{-\frac{c_2}{2} \sqrt{\log N} } \right), \tag{14} \end{align*} 

Because $\frac{1}{\log t} = \frac{1}{\log n} + \mathcal{o}\left (\frac{1}{n \log^2 n} \right) $,$ (N-1 < T <n) $, and $\sum^{\ infty}_{n=3} \frac{1}{n \log^2 N} = \mathcal{o} (1) $, with \begin{align*} (+) &= \frac{1}{\varphi (q)} \sum_{\sqrt{n} &lt ; n \leq N} \left (\frac{e^{2 \pi i \beta n}}{\log n} + \mathcal{o}\left (\frac{1}{n \log^2 n} \right) \right) + \mathcal{O} \left (n e^{-\frac{c_2}{2} \sqrt{\log n}} \right) \ &= \frac{1}{\varphi (q)} \sum_{3 \leq n \leq N} \left (\frac{ E^{2 \pi i \beta n}}{\log n} + \mathcal{o}\left (\frac{1}{n \log^2 n} \right) \right)-\frac{1}{\varphi (q)} \sum_{3 \leq n \leq \sqrt{n}} \left (\frac{e^{2 \pi i \beta n}}{\log N} + \ \ + \mathcal{o}\left (\frac{1}{n \log^2 N} \right) \right) & nbsp;+ \mathcal{o}\left (n e^{-\frac{c_2}{2} \sqrt{\log N}} \right)  \end{align*} may wish to remember that the rightmost is $ (15) $, Kee $g (n) =-\frac{1 }{\varphi (q)} \sum_{3 \leq n \leq \sqrt{n}} \left (\frac{e^{2 \pi i \beta n}}{\log n}  + \mathcal{o}\left (\frac{1}{n \log^2 n} \right) \rigHT) $, then \begin{align*} |g| &\leq \sum_{3 \leq n \leq \sqrt{n}}  \frac{1}{\log n} + \mathcal{o}\left (\sum_{3 \leq n \leq \sqrt{N}} \frac{1}{n \log^2 n} \right) \ &\leq \sqrt{n} + \mathcal{o} (1). \end{align*} is thus easy to know, $g (N) = \mathcal{o}\left (\sqrt{n} \right) $. So  \begin{align*} = \frac{1}{\varphi (q)} \sum_{3 \leq n \leq n} \frac{e^{2 \pi I \beta n}}{\log n} + \mathcal{O} \left (n e^{-\frac{c_2}{2} \sqrt{\log n} \right). \TAG{16} \end{align*} will $ (16) $ into $ (8) $ and combinelemma 1.2and $ (a,q) =1$ get \begin{align*} S (\alpha) &= \underset{(l,q) =1}{\sum^{q}_{l=1}} e^{2 \pi I \frac{a}{q} l} \frac{1}{\var Phi (Q)} \sum_{3 \leq n \leq n} \frac{e^{2 \pi I \beta n}}{\log n} + \mathcal{o}\left (q n e^{-\frac{c_2}{2} \sqrt{\log n}} \right) \ &= \frac{\mu (q)}{\varphi (q)} \sum_{3 \leq n \leq n} \frac{e^{2 \pi I \beta n}}{\log n} + \mathcal{o}\left ( N e^{-\frac{c_2}{4} \sqrt{\log n}} \right), \end{align*} wherein, $q \leq \log^{15}n$, so $N $ full, $qNe ^{-\frac{c_2}{2} \sqrt{\lo G N}} < Ne^{-\frac{c_2}{4} \sqrt{\log n}}$.lemma 2.2The certificate is completed.

By lemma 2.2 can be launched \begin{align*} s^3 \left (\alpha \right) &= s^3 \left (\frac{a}{q} + \beta \right) \ \ &= \frac{\mu^3 (q) }{\varphi^3 (q)} \left (\sum_{3 \leq n \leq n} \frac{e^{2 \pi I \beta n}}{\log n} \right) ^3 + \mathcal{o}\left (N^3 e^{-\fr AC{C_2}{4} \sqrt{\log N}} \right) \ &= \frac{\mu^3 (q)}{\varphi^3 (q)} m^3 (\beta) + \mathcal{o}\left (N^3 e^{-\frac{c_ 2}{4} \sqrt{\log N}} \right). \TAG{17} \end{align*} here $ $M (\beta) = \sum_{3 \leq n \leq n} \frac{e^{2 \pi I \beta n}}{\log n}. $$
In fact, \begin{align*} s^3 \left (\alpha \right) &= \left (\FRAC{\MU (q)}{\varphi (q)} \sum_{3 \leq n \leq n} \frac{e^{2 \pi I \beta n}}{\log n} + \mathcal{o}\left (n e^{-\frac{c_2}{4} \sqrt{\log n}} \right) \right) ^3 \ &\triangleq \left (A_ 1 + b_1 \right) ^3. \end{align*} easy to $A _1 = \mathcal{o}\left (N \right) $, thereby \begin{align*} s^3 (\alpha) &= a_1^3 + 3a_1^2b_1 + 3A_1B_1^2 + B _1^3 \ &= a_1^3 + \mathcal{o}\left (n^3 e^{-\frac{c_2}{4} \sqrt{\log N}} \right) \end{align*} by $ (6) $ and $ (17) $ can be obtained: \begin{align*} r_1 (n) &= \int_{\mathfrak{m}} s^3 (\alpha) e^{-2 \pi \alpha N} d\alpha \ &= \sum_{\frac{a}{q} \in \mathscr{a}} \int_{\mathfrak{m} (A,q)} s^3 (\alpha) e^{-2 \pi i \alpha N} d\alpha \ &= \sum_{q \leq \log^{15} N} \under set{(A,Q) =1}{\sum^{q}_{a=1}} \int^{\frac{1}{q \cdot \tau}}_{-\frac{1}{q \cdot \tau}} s^3\left (\frac{a}{q} + \beta \ right) e^{-2 \pi i \left (\frac{a}{q} + \beta \right) n} d\beta \ &= \sum_{q \leq \log^{15} N} \underset{(a,q) =1}{\su M^{q}_{a=1}} \int^{\frac{1}{q \cdot \tau}}_{-\frac{1}{q \cdot \tau}} \left (\frac{\mu^3 (q)}{\varphi^3 (q)} M^3 (\beta) + \mathcal{O}\ Left (n^3 e^{-\frac{c_2}{4} \sqrt{\log N}} \right) \right) e^{-2 \pi i \left (\frac{a}{q} + \beta \right) N} d\beta \ &am p;= \sum_{q \leq \log^{15}n} \frac{\mu^3 (q)}{\varphi^3 (q)} \underset{(A,q) =1}{\sum^{q}_{a=1}} e^{-2 \pi i \frac{a}{q}N} \int^{\frac{1}{q \cdot \tau}}_{-\frac{1}{q \cdot \tau}}m^3 (\beta) e^{-2\pi i \beta N} D\beta + \ \ &+ \mathcal{O}\left ( N^3 e^{-\frac{c_2}{4} \sqrt{\log N}} \sum_{q \leq \log^{15} N} q \times \frac{2}{q \cdot \tau} \right) \ \ &= \sum_{q \leq \log^{15}n} \frac{\mu^3 (q)}{\varphi^3 (q)} \underset{(A,q) =1}{\sum^{q}_{a=1}} e^{-2 \pi i \frac{a}{q}n} \int^{\ Frac{1}{q \cdot \tau}}_{-\frac{1}{q \cdot \tau}}m^3 (\beta) e^{-2\pi i \beta N} D\beta + \ \ &+ \mathcal{O}\left (N^2 e^ {-\frac{c_2}{8} \sqrt{\log N}} \right), \tag{18} \end{align*} where the order above $ (18) $ is due to $$\sum_{q \leq \log^{15}n} \frac{2}{\tau } = \frac{2\log^{35}n}{n},$$ thereby when $NAfter full size, $N ^2 \log^{35}n e^{-\frac{c_2}{4} \sqrt{\log N}} < n^2 e^{-\frac{c_2}{8} \sqrt{\log n}}$. Now, the problem turns into research points $$\int^{\frac{1}{q \cdot \tau}}_{-\frac{1}{q \cdot \tau}} m^3 (\beta) e^{-2 \pi i \beta N} d\beta. $$ for this we first prove the following lemma:

lemma 2.3. Set $ $M _0 (\beta) = \frac{1}{\log n} \sum_{3 \leq N \leq n} e^{2 \pi i \beta n},$$ there are $$\int^{\frac{1}{q \CDO T \tau}}_{-\frac{1}{q \cdot \tau}} | M^3 (\beta)-m_0^3 (\beta) | D\beta = \mathcal{o}\left (\frac{n^2}{\log^4 N} \right). $$

So far, the proof is far from over (half of the proofs have not yet arrived), and the full proof of lemma 2.3 and the three prime theorem is put on hold for the time being.

Proof and method of three prime number theorem (I.)