[proofplan]
We combine Dirichlet approximation with the classical Vinogradov bound for prime exponential sums. The exclusion from the logarithmic major arcs forces every Dirichlet approximant at scale $N/(\log N)^B$ to have denominator larger than $(\log N)^B$. The standard [Vaughan identity](/theorems/7192) estimate then bounds $S_N(\alpha)$ in terms of this denominator, and choosing $B$ sufficiently large in terms of $A$ absorbs all logarithmic losses.
[/proofplan]
[step:Choose logarithmic parameters and record the standard Vinogradov estimate]
Fix $A>0$. We shall use the following classical consequence of Vaughan's identity, dyadic decomposition, Type I estimates, and Type II estimates: there is an absolute constant $C_0>0$ such that, whenever $N\ge 2$, $\alpha\in\mathbb R/\mathbb Z$, and $a,q\in\mathbb Z$ satisfy $q\in\mathbb N$, $\gcd(a,q)=1$, and
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{1}{q^2},
\end{align*}
one has
\begin{align*}
|S_N(\alpha)|\le C_0(\log N)^4\left(\frac{N}{q^{1/2}}+N^{4/5}+N^{1/2}q^{1/2}\right).
\end{align*}
This is the standard Vinogradov-Vaughan estimate for the von Mangoldt exponential sum, obtained by applying Vaughan's identity with balanced parameters and estimating the resulting Type I and Type II bilinear sums.
Choose
\begin{align*}
B:=2A+12.
\end{align*}
After increasing the final threshold $N_0$ if necessary, we may assume throughout that $N\ge 3$ and
\begin{align*}
N^{-1/5}(\log N)^{A+4}\le 1.
\end{align*}
Equivalently, for such $N$,
\begin{align*}
N^{4/5}(\log N)^4\le \frac{N}{(\log N)^A}.
\end{align*}
[guided]
Fix $A>0$. The main technical input is the classical estimate for exponential sums over primes due to Vinogradov, in Vaughan's identity form. It says that if $\alpha$ has a reduced rational approximant $a/q$ with approximation quality at least $q^{-2}$, then the von Mangoldt sum is controlled by three terms depending on $q$:
\begin{align*}
|S_N(\alpha)|\le C_0(\log N)^4\left(\frac{N}{q^{1/2}}+N^{4/5}+N^{1/2}q^{1/2}\right).
\end{align*}
Here $C_0>0$ is an absolute constant. The estimate is not a formal consequence of the theorem being proved: it comes from Vaughan's identity, dyadic subdivision of the resulting convolutions, Type I bounds for sums with one short variable, and Type II bounds obtained by Cauchy-Schwarz and estimates for sums involving $\|\alpha h\|_{\mathbb R/\mathbb Z}$. Since this prerequisite is not yet available as a wiki theorem, we state it explicitly as a classical input.
The role of the rest of the proof is to put the denominator $q$ in the correct range. The minor arc hypothesis will imply that $q$ is not small, while Dirichlet approximation will give an approximant with $q$ not too large. We choose
\begin{align*}
B:=2A+12.
\end{align*}
This value is deliberately larger than needed by a fixed margin, because the Vinogradov estimate carries a factor $(\log N)^4$ and the endpoint terms will cost square roots of $(\log N)^B$.
Finally, we choose the threshold $N_0$ large enough that for every $N\ge N_0$ the power-saving term is also logarithmically small:
\begin{align*}
N^{4/5}(\log N)^4\le \frac{N}{(\log N)^A}.
\end{align*}
This is possible because
\begin{align*}
\frac{N^{4/5}(\log N)^4}{N/(\log N)^A}=N^{-1/5}(\log N)^{A+4}\to 0
\end{align*}
as $N\to\infty$.
[/guided]
[/step]
[step:Use Dirichlet approximation at the minor arc scale]
Let $N\ge N_0$, and let
\begin{align*}
\alpha\in \mathbb R/\mathbb Z\setminus \mathfrak M_N((\log N)^B,(\log N)^B).
\end{align*}
Define the auxiliary integer approximation scale
\begin{align*}
K_N:=\left\lfloor \frac{N}{(\log N)^B}\right\rfloor.
\end{align*}
Increasing $N_0$ if necessary, assume $K_N\ge 1$.
By Dirichlet's approximation theorem on $\mathbb R/\mathbb Z$, applied with integer denominator bound $K_N$, there exist $a,q\in\mathbb Z$ with $1\le q\le K_N$, $\gcd(a,q)=1$, and
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{1}{q(K_N+1)}.
\end{align*}
Since $K_N+1\ge N/(\log N)^B$, this gives
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{(\log N)^B}{qN}.
\end{align*}
If $q\le(\log N)^B$, then the defining conditions for $\mathfrak M_N((\log N)^B,(\log N)^B)$ are satisfied, contradicting the choice of $\alpha$. Hence
\begin{align*}
q>(\log N)^B.
\end{align*}
Moreover, from the construction,
\begin{align*}
q\le K_N\le \frac{N}{(\log N)^B}.
\end{align*}
[guided]
We now choose the rational approximant in a way that matches the major-arc radius exactly. Let
\begin{align*}
K_N:=\left\lfloor \frac{N}{(\log N)^B}\right\rfloor.
\end{align*}
After increasing $N_0$ if necessary, $K_N\ge 1$ for all $N\ge N_0$. Dirichlet's approximation theorem on $\mathbb R/\mathbb Z$, with integer denominator bound $K_N$, gives integers $a,q\in\mathbb Z$ such that $1\le q\le K_N$, $\gcd(a,q)=1$, and
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{1}{q(K_N+1)}.
\end{align*}
Because $K_N+1\ge N/(\log N)^B$, the approximation quality becomes
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{(\log N)^B}{qN}.
\end{align*}
This is exactly the width used in the definition of $\mathfrak M_N((\log N)^B,(\log N)^B)$. Therefore, if $q\le (\log N)^B$, then the integers $a,q$ satisfy every defining condition for membership of $\alpha$ in that major arc set. This contradicts
\begin{align*}
\alpha\in \mathbb R/\mathbb Z\setminus \mathfrak M_N((\log N)^B,(\log N)^B).
\end{align*}
Hence
\begin{align*}
q>(\log N)^B.
\end{align*}
The same Dirichlet construction also gives the upper bound
\begin{align*}
q\le K_N\le \frac{N}{(\log N)^B}.
\end{align*}
[/guided]
[/step]
[step:Verify the rational approximant satisfies the Vinogradov hypothesis]
The estimate from the first step requires the approximation quality $q^{-2}$. Since
\begin{align*}
q\le \frac{N}{(\log N)^B},
\end{align*}
we have
\begin{align*}
\frac{(\log N)^B}{qN}\le \frac{1}{q^2}.
\end{align*}
Therefore
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{1}{q^2}.
\end{align*}
The hypotheses of the Vinogradov-Vaughan estimate are satisfied for this reduced fraction $a/q$.
[guided]
The Vinogradov-Vaughan estimate from the first step requires a reduced rational approximant $a/q$ satisfying the stronger approximation condition
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{1}{q^2}.
\end{align*}
The previous step gave
\begin{align*}
\left\|\alpha-\frac{a}{q}\right\|_{\mathbb R/\mathbb Z}\le \frac{(\log N)^B}{qN}
\end{align*}
and also
\begin{align*}
q\le \frac{N}{(\log N)^B}.
\end{align*}
Multiplying the last inequality by $(\log N)^B$ gives $q(\log N)^B\le N$. Dividing by $q^2N$, which is positive, gives
\begin{align*}
\frac{(\log N)^B}{qN}\le \frac{1}{q^2}.
\end{align*}
Therefore the selected reduced fraction $a/q$ satisfies the approximation hypothesis required by the Vinogradov-Vaughan estimate.
[/guided]
[/step]
[step:Apply the Vinogradov estimate and bound each denominator term]
Applying the Vinogradov-Vaughan estimate gives
\begin{align*}
|S_N(\alpha)|\le C_0(\log N)^4\left(\frac{N}{q^{1/2}}+N^{4/5}+N^{1/2}q^{1/2}\right).
\end{align*}
The lower bound $q>(\log N)^B$ gives
\begin{align*}
\frac{N}{q^{1/2}}\le \frac{N}{(\log N)^{B/2}}.
\end{align*}
The upper bound $q\le N/(\log N)^B$ gives
\begin{align*}
N^{1/2}q^{1/2}\le \frac{N}{(\log N)^{B/2}}.
\end{align*}
Thus
\begin{align*}
|S_N(\alpha)|\le C_0(\log N)^4\left(\frac{2N}{(\log N)^{B/2}}+N^{4/5}\right).
\end{align*}
By the choice $B=2A+12$,
\begin{align*}
(\log N)^4\frac{N}{(\log N)^{B/2}}=\frac{N}{(\log N)^{A+2}}.
\end{align*}
Therefore, for all $N\ge N_0$,
\begin{align*}
|S_N(\alpha)|\le 2C_0\frac{N}{(\log N)^{A+2}}+C_0\frac{N}{(\log N)^A}.
\end{align*}
Since $(\log N)^{-A-2}\le(\log N)^{-A}$ for $N\ge 3$, we obtain
\begin{align*}
|S_N(\alpha)|\le 3C_0\frac{N}{(\log N)^A}.
\end{align*}
[guided]
We now apply the Vinogradov-Vaughan estimate to the reduced approximant $a/q$ verified in the previous step. It gives
\begin{align*}
|S_N(\alpha)|\le C_0(\log N)^4\left(\frac{N}{q^{1/2}}+N^{4/5}+N^{1/2}q^{1/2}\right).
\end{align*}
The point of the minor-arc argument is that $q$ is neither too small nor too large. From $q>(\log N)^B$, we get
\begin{align*}
\frac{N}{q^{1/2}}\le \frac{N}{(\log N)^{B/2}}.
\end{align*}
From $q\le N/(\log N)^B$, we get
\begin{align*}
N^{1/2}q^{1/2}\le \frac{N}{(\log N)^{B/2}}.
\end{align*}
Substituting these two bounds into the Vinogradov-Vaughan estimate yields
\begin{align*}
|S_N(\alpha)|\le C_0(\log N)^4\left(\frac{2N}{(\log N)^{B/2}}+N^{4/5}\right).
\end{align*}
Since $B=2A+12$, we have $B/2=A+6$, and hence
\begin{align*}
(\log N)^4\frac{N}{(\log N)^{B/2}}=\frac{N}{(\log N)^{A+2}}.
\end{align*}
The threshold $N_0$ was chosen so that
\begin{align*}
N^{4/5}(\log N)^4\le \frac{N}{(\log N)^A}.
\end{align*}
Therefore
\begin{align*}
|S_N(\alpha)|\le 2C_0\frac{N}{(\log N)^{A+2}}+C_0\frac{N}{(\log N)^A}.
\end{align*}
Finally, $N\ge 3$ implies $(\log N)^{-A-2}\le (\log N)^{-A}$, so
\begin{align*}
|S_N(\alpha)|\le 3C_0\frac{N}{(\log N)^A}.
\end{align*}
[/guided]
[/step]
[step:Choose the final constants and conclude the minor arc estimate]
Define
\begin{align*}
C_A:=3C_0.
\end{align*}
The preceding step proves that, for every $N\ge N_0$ and every
\begin{align*}
\alpha\in \mathbb R/\mathbb Z\setminus \mathfrak M_N((\log N)^B,(\log N)^B),
\end{align*}
one has
\begin{align*}
|S_N(\alpha)|\le C_A\frac{N}{(\log N)^A}.
\end{align*}
This is the asserted Vinogradov minor arc estimate for the von Mangoldt exponential sum.
[/step]
