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!



Using Vinogradov's theorem for finding prime solutions to a linear equation (an exercise from Vaughan's book)
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.


