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Let $\alpha$ be a badly approximable number, and sum $B={\sum }_{D<d\le 2D}{\sum }_{W<w\le 2W,dw\le X}{a}_d{b}_{w}e(\alpha dw)$ with $|a_d|\le \tau (d)$ and $|b_w|\le 1$. The Cauchy-Schwarz inequality, divisor-square summatory bound and geometric sums give $|B{|}^2\ll D{\log }^{3}X(DW+W{\sum }_{h\le W}\min (D,\Vert h\alpha {\Vert }^{-1}))$. Separated rotations bound the last sum by $O_{\alpha }(W\log (2W))$, giving the displayed estimate. A hyperbolic cutoff remains valid because the allowed first-variable values after expansion form an interval.

The Riemann zeta function satisfies ${\int }_0^T|\zeta (1/2+it){|}^{4}dt\ll T(\log T{)}^4$ for $T\ge 2$. Squaring a short Dirichlet polynomial produces the divisor function; the divisor-square summatory bound controls the off-diagonal and diagonal terms.

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.

A precise useful form of the principle is the type I–type II inverse principle for Möbius correlation. Normalize $|f(n)|\le 1$, and supposeFor any positive integer cutoffs $U,V$ with $U+V\le \delta X/2$, defineThen $|c_d|,|a_d|\le \tau (d)$, where $\tau$ is the divisor function, and at least one of the following two sums has modulus at least $\delta X/4$:The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a periodic function: this is the “somewhat periodic” branch. The second is a bilinear sum with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.

For clarity, the Vaughan identity for the Möbius function proves this version immediately. Split $\mu ={\mu }_{\le U}+{\mu }_{>U}$ and likewise at $V$. Since $\mu \ast \mu \ast 1=\mu$,Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm{I}}+T_{\mathrm{II}}$. The triangle inequality gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.

We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha =\sqrt{2}$ and $e(u)=e^{2\pi iu}$. The Diophantine bound for the square root of two iswhere $\Vert u\Vert$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\ge 1$, while $h\sqrt{2}+a\le 4h$; dividing proves the bound.

Use the preceding Dirichlet convolution identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the type I sum, the exponential geometric sum bound givesConsequentlyusing ${\sum }_{d\le Y}\tau (d)\ll Y\log (2Y)$ from counting factor pairs.

For the type II sum, perform a dyadic decomposition into $d\in (D,2D]$, $w\in (W,2W]$, keeping $dw\le X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the Cauchy-Schwarz inequality and the divisor-square summatory bound,To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w^{\prime }$ still form an interval, with upper endpoint $\min (2D,X/\max (w,w^{\prime }))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w^{\prime }$ there are $O(W)$ pairs.

The points $0,\alpha ,2\alpha ,\dots ,\lfloor W\rfloor \alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore givesSubstitution yields the bilinear cancellation for badly approximable phases estimateThere are $O({\log }^{2}X)$ blocks, so $T_{\mathrm{II}}\ll X^{19/20}{\log }^{4}X$. The divisor-square bound itself can be obtained elementarily from $\tau (n{)}^2\le {\tau }_4(n)$ prime-power by prime-power and ${\sum }_{n\le Y}{\tau }_4(n)\le Y(1+\log Y{)}^3$; no uncontrolled coefficient bound is being suppressed.

Combining the short, type I and type II terms,Hence

Put $P(t)={\sum }_{n\le \sqrt{t/(2\pi )}}{n}^{-1/2-it}$. The given bound for the Riemann zeta function, together with $(a+b{)}^4\le 8(a^4+b^4)$ for $a,b\ge 0$, givesThe integral over $[0,1]$ is bounded, since the Riemann zeta function has no pole on this compact segment. It remains to estimate the fourth moment of the Riemann zeta function through this moving Dirichlet polynomial cutoff.

Let $L=\lfloor \sqrt{T/(2\pi )}\rfloor$ and $Y=L^2$. In the expansion of $|P(t){|}^4$, a quadruple $n_1,n_2,n_3,n_4\le L$ occurs precisely forIts oscillatory factor is $e^{it\log (n_3{n}_4/(n_1{n}_2))}$, and its weight is $(n_1{n}_2{n}_3{n}_4{)}^{-1/2}$. When $n_1{n}_2=n_3{n}_4$, its integral has length at most $T$. When these products differ, its integral over $[t_0,T]$ has absolute value at most $2/|\log (n_3{n}_4/(n_1{n}_2))|$. Thus the moving cutoff affects the lower endpoint but preserves the off-diagonal bound from part (a).

Write $r_L(m)=\#\{(a,b):a,b\le L,ab=m\}$. The divisor function $d(m)$ bounds $r_L(m)$. Grouping the off-diagonal majorant by the two products and using the weighted row-sum argument in part (a), with coefficient $r_L(m)$ and $\sigma =1/2$, bounds it byThis uses only the given divisor-square summatory bound. For bounded $T$, the desired conclusions follow by boundedness on compact segments, so these estimates may be read for large $T$ with $Y\ge 2$. We have proved the requested diagonal-plus-error estimate:

The diagonal sum equals ${\sum }_{m\le Y}{r}_L(m{)}^2/m\le {\sum }_{m\le Y}d(m{)}^2/m$. If $D(x)={\sum }_{m\le x}d(m{)}^2\ll x{\log }^3(2x)$, then partial summation givesConsequently the full fourth-moment bound is

Suppose $|a_u|\le \tau (u)$, $|b_d|\le \log (2N)$, $u,d>X\ge 1$, $ud\le N$, and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$. A dyadic decomposition reduces $S=\sum a_u{b}_{d}e(\theta ud)$ to blocks $u\asymp U,d\asymp D$. The Cauchy-Schwarz inequality, divisor-square summatory bound and the reciprocal fractional-part sum near a rational giveThe condition $ud\le N$ makes the allowed $u$ values for each pair $d_1,d_2$ an interval, so the exponential geometric sum bound applies despite the hyperbolic cutoff. Since $UD\le N$ and $U,D\gg X$, take square roots and sum the $O({\log }^2(2N))$ blocks to get the displayed estimate.