Using Vinogradov's theorem for finding prime solutions to a linear equation (an exercise from Vaughan's book) I'm trying to solve an exercise from Vaughan's book, "The Hardy-Littlewood Method" (ex. 3 in chapter 3: Goldbach's problems, p.36), because I want to use the result stated in it. It is a variation of Vinogradov's theorem on every large enough odd integer being the sum of 3 primes.
Here I need to show that there are "many" triples of prime numbers less than or equal to $N$ that solve the equation with integer coefficients:
$b_1p_1+b_2p_2+b_3p_3-b_4=0$.
More specifically, I need to show that if 
$R(N)=\sum_{(p_1,p_2,p_3)\mid p_i\leq n, b_1p_1+b_2p_2+b_3p_3-b_4=0} (\log p_1)(\log p_2)(\log p_3)$
 then $R(N)=J(N)\mathfrak S +O(N^2/\log ^AN)$  where $J(N)$ is the number of integer solutions to $b_1m_1+b_2m_2+b_3m_3-b_4=0$ satisfying $m_i\leq N$.
Following the Proof as given in Vaughan and Nathanzon, I have defined $F_i(x)=\sum _{p\leq N}\log p\cdot e(b_ipx)$ so that $R(N)=\int  _0 ^1 F_1(x)F_2(x)F_3(x)e(-b_4x)dx$.
When integrating over the major arcs $\mathcal M$, I guess I should approximate $F_i(x)$ by $G_i(x)=\frac{c_q(b_i)}{\phi (q)}u_i \left(x-\frac{a}{q}\right)$ where $u_i(y)=\sum _{m\leq N}e(mb_iy)$ and evaluate the integral $\int _\mathcal M G_1(x)G_2(x)G_3(x)e(-b_4x)dx$. 
So, I start by integrating $G_1(x)G_2(x)G_3(x)e(-b_4x)$ over a singel major arc $(\frac {a}{q}-\frac {Q}{N}, \frac {a}{q}+\frac {Q}{N})$. After a change of variables I end up with the integral $\int _\frac {-Q}{N}^\frac {Q}{N}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$, and I want to bound the difference between the last integral to: $J(N)=\int _\frac {-1}{2}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$. I guess this is how I get the $J(N)$ in the main term of the desired expression for $R(N)$.
If so, I want to bound $\int _\frac {-1}{2}^\frac {-Q}{N}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$ and $\int _\frac {Q}{N}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$. In the proof of Vinogradov's Theorem, this difference is $O(Q^2/N^2)$ where $Q=\log ^BN$. But now, in the case dealt with in the exercise, this difference seems to be huge: $O(N^3)$. Since, when $b_iy$ is an integer $e(b_iy)=1$ which gives $u_i(y)=N$ and $|u_1(y)u_2(y)u_3(y)|=O(N^3)$.
To summarize, my questions are:
How do I bound $\int _\frac {Q}{N}^\frac {1}{2}u_1(y)u_2(y)u_3(y)e(-b_4y)dy$? Does anyone know of a place where this claim is proved?
Thanks!
 A: It's a long time since this was posted. I don't know why you didn't ask me directly. There is an oversight in exercise 2. The main term on the RHS in question 2 should be $\gcd(a_1,a_2,a_3)^{-1}J(n)\frak S$. One has to use intervals of length $1/\gcd(a_1,a_2,a_3)$.
A: First, we may assume $(b_1,b_2,b_3)=1$.  That gives that for each $y\in (0,1)$ at most two of $\{b_1y,b_2y,b_3y \}$  can be integers.
Now, take $$J^{(i)}_k=\left[\frac {k}{b_i}-\frac {1}{b_iN^\frac{1}{3}},\frac {k}{b_i}+\frac {1}{b_iN^\frac{1}{3}}\right]$$ for all $1\leq k\leq b_i-1$. Assume $N$ to be large enough such that no two of these intervals (for all $i=1,2,3$ and all $k$) intersect each other, unless when they are both centered around the same point $y_0$, i.e both $b_iy_0$ and $b_jy_0$ are integers. If so, at most two of these intervals can intersect each other.
Also, define: $$I^{(i)}_k=\left[\frac {k}{b_i}-\frac {1}{b_iN^\frac{1}{2}},\frac {k}{b_i}+\frac {1}{b_iN^\frac{1}{2}}\right]\subset J^{(i)}_k.$$
On $I^{(i)}_k$ we have that $|u_i(y)|\sim N$, so if $y\in I^{(i)}_{k_1}\cap I^{(j)}_{k_2}$ then $|u_i(y)u_j(y)|\sim N^2$.But, for $l\neq i,j$ we have that $|u_l(y)|\ll N^\frac {1}{3}$ as $y\notin J^{(l)}_k$ for all $k$. Over a certain $I^{(i)}_k$ we can therefore bound $|u_1(y)u_2(y)u_3(y)| $ by $$|u_1(y)u_2(y)u_3(y)|\ll N^2N^\frac {1}{3}.$$ As $|I^{(i)}_k|=\frac{2}{b_iN^\frac{1}{2}}$, we have that:
$$\int _{I^{(i)}_k} |u_1(y)u_2(y)u_3(y)|dy \ll N^2N^\frac {1}{3}\frac{2}{b_iN^\frac{1}{2}}=\frac{2}{b_i}N^2N^\frac {-1}{6}$$
There are at most $|b_i|$ such $I^{(i)}_k$'s, so the integral over all of them is bounded by $2N^2N^\frac {-1}{6}$.
Now, say we're in $J^{(i)}_k\setminus I^{(i)}_k$. Then at least one of the $u_i$'s is bounded by $N^\frac{1}{3}$, and the other two are at bounded by $N^\frac{1}{2}$. The length of each $J^{(i)}_k$ is $\frac{2}{b_iN^\frac{1}{3}}$, so:
$$\int _{J^{(i)}_k\setminus I^{(i)}_k} |u_1(y)u_2(y)u_3(y)|dy \ll N^\frac {1}{2}N^\frac {1}{2}N^\frac {1}{3}\frac{2}{b_iN^\frac{1}{3}}=\frac{2}{b_i}N.$$
Again, the integral over the total of intervals $J^{(i)}_k\setminus I^{(i)}_k$ is bounded by $2N$.
Outside of all the intervals $J^{(i)}_k$, for all $i=1,2,3$ and $1\leq k\leq b_i-1$, we know that $|u_i(y)|\ll N^\frac {1}{3}$ and therefore $|u_1(y)u_2(y)u_3(y)|\ll N$. Everything is inside $(0,1)$, so the total of the integral "outside of the $J^{(i)}_k$'s" is bounded by $N$.
To summarize: $$\int _\frac {Q}{N}^\frac{1}{2}|u_1(y)u_2(y)u_3(y)|dy\ll 2N^2N^\frac {-1}{6}+2N+N=O(N^{1\frac {5}{6}}).$$
But, this is true only when we are far enough from 0. At 0, all three functions $u_i$ obtain their maximum simultaneously, so the main contribution to the integral will be near 0. take $b$ such that  $\frac{1}{bN^{\frac{1}{3}}}\leq\frac{1}{2b_{i}}$  for all $i=1,2,3$. Write:
$$\mathcal S= \left[\frac {Q}{N},\frac{1}{bN^{\frac{1}{3}}} \right].$$
 We estimate the integral on $\mathcal S$. Note that if $y\in \mathcal S$ then $b_{i}y\leq\frac{1}{2}$  and therefore $\left\Vert b_{i}y\right\Vert =\left|b_{i}y\right|$  for all $i=1,2,3$. Thus, $u_{i}\left(y\right)\ll\frac{1}{\left\Vert b_{i}y\right\Vert }=\frac{1}{\left|b_{i}y\right|}$  for $y\in\mathcal S$. Now:
$$\int_{\mathcal S}|u_1(y)u_2(y)u_3(y)|dy \ll \frac{1}{\left|b_{1}b_{2}b_{3}\right|} \int_{\mathcal S}\frac{1}{y^{3}}dy=O\left(\frac{N^{2}}{Q^{2}}\right).$$
We get that The integral of $u_1(y)u_2(y)u_3(y)$ on $\left[\frac{Q}{N},\frac{1}{2}\right]$ is $O\left(\frac{N^{2}}{Q^{2}}\right)$, and this bound is tight.
A: Hello, 
I wanted to suggest some simplifications/improvements.
First of all you can avoid doing case-analysis by using the
elementary estimate 
$v_1v_2v_3 \leq \min(v_1,v_2,v_3) \max(v_1,v_2,v_3)^2$,
which is valid for all positive real numbers $v_1,v_2,v_3$.
In your problem you end up with the bound
$\int_{Q/N}^{1/2} |u_1(y)u_2(y)u_3(y)|dy$
$\leq \left(\sup_{y \in [Q/N,1/2]}\min_{i=1,2,3}|u_i(y)| \right) \int_{Q/N}^{1/2}\max_{i=1,2,3}(|u_i(y)|^2)dy$
$\leq \left(\sup_{y \in [Q/N,1/2]}\min_{i=1,2,3}|u_i(y)|\right) \int_{-1/2}^{1/2}\max_{i=1,2,3}(|u_i(y)|^2)dy.$
And by Parseval's identity
$\leq \left(\sup_{y \in [Q/N,1/2]}\min_{i=1,2,3}|u_i(y)|\right) N$
If $b_1,b_2,b_3$ are coprime (as we have to assume for this estimate), most values of $y$ are at least $(2b_1b_2b_3)^{-1}$ away from a point of the form $a/b_i$ for some 
$i \in \{1,2,3\}$. 
For those $y$ we have the estimate 
$\sup_{y \in [1/(b_1b_2b_3),1/2]}\min_{i=1,2,3}|u_i(y)| \ll b_1b_2b_3$.
For the remaining values of $y$, which are close to $0$, we have to use the weaker estimate
$\sup_{y \in [Q/N,1/(b_1b_2b_3)]}\min_{i=1,2,3}|u_i(y)| \ll N/Q$.
Altogether we obtain
$\int_{Q/N}^{1/2} |u_1(y)u_2(y)u_3(y)|dy \ll N^2/Q$.
I hope I didn't screw things up somewhere...
