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!