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The rational approximation below is taken in lowest terms, $(a,q)=1$, with $1\le q\le Q$, and $X>0$. This is necessary for denominator-sensitive exponential sum estimates and for the final Ramanujan sum to equal $\mu (q)$. If $0<X<1$, the defining sum over $a\le X$ is empty and $S_4=0$; hence we may assume $1\le X<\sqrt{N}$.

Set $c_u={\sum }_{a\mid u,a\le X}\mu (a)$, so $|c_u|\le \tau (u)$, and write the given Vaughan identity term as the Type II sumUse a dyadic decomposition into $u\in (U,2U]$, $d\in (D,2D]$, retaining $ud\le N$. Empty blocks may be discarded; every remaining block has $U,D\ge X$, $UD<N$, and there are $O({\log }^2(2N))$ blocks. Denote a block by $T$. The Cauchy-Schwarz inequality givesBy the allowed divisor-square summatory bound, the first factor is $O(U{\log }^3(2N))$. Expand the second factor. The diagonal $d_1=d_2$ contributes $O(UD{\log }^2(2N))$, using $0\le \Lambda (d)\le \log (2N)$. For a pair $d_1\ne d_2$, the allowed $u$ form the intervalThus the exponential geometric sum bound applies to this interval even though the original block has a hyperbolic cutoff. For each difference $h=|d_1-d_2|\le D$, there are $O(D)$ ordered pairs, giving an upper boundfor the expanded second factor, with an absolute constant $C$ accommodating integer endpoints. The supplied reciprocal fractional-part sum near a rational, with $M\asymp D$ and $R\asymp U$, bounds the last sum byOne may also derive this directly from the separated reciprocal-distance sum by splitting $h$ into blocks of length at most $q/2$; the approximation error between two block points is at most $1/(2q)$, and the rational rotations are separated by $1/q$.

Combining the estimates yieldsSince $UD<N$ and $U,D\ge X$, both $U,D\le N/X$. Taking square roots provesSumming the dyadic blocks establishes the rational-phase Type II estimate and, in particular, the requested bound with an explicit logarithmic power:

The Siegel–Walfisz theorem states that for fixed $A,C>0$, uniformly for $q\le (\log N{)}^A$ and $(b,q)=1$,The constant is allowed to be ineffective. Apply this with $C=A+B+2$ to the reduced residue classes. Their contribution to the prime exponential sum issince there are at most $q\le (\log N{)}^A$ residue classes. To evaluate the Ramanujan sum, the Möbius divisor-sum identity gives $1_{(b,q)=1}={\sum }_{d\mid (b,q)}\mu (d)$. Then the finite geometric series givesWhen $(a,q)=1$, only $m=1$ remains, giving $c_q(a)=\mu (q)$. The omitted nonunits contribute only prime powers $p^j$ with $p\mid q$. For each such prime, the sum of their Von Mangoldt function weights is at most $\log N$, so their total is at most $\omega (q)\log N=O_A(\log N\log \log (3N))$, absorbed by the stated error. This proves the major-arc value of the prime exponential sum:The original PDF does not state coprimality in this last request. Without it the asserted main term is genuinely false: $a=q=2$ makes $e(an/q)=1$ and hence ${\hat{\Lambda }}_N(a/q)=\psi (N)\sim N$, whereas $\mu (2)/\phi (2)=-1$. For an arbitrary numerator the correct main term is $N{c}_q(a)/\phi (q)$, with $c_q(a)$ given by the divisor formula above. Reducing the fraction before applying the boxed result is the equivalent repair.