Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 124 / 4

Expand the two Dirichlet polynomials, keeping the complex conjugate on $B$. The diagonal $m=n$ contributes exactly $T{\sum }_{n\le X}{a}_n\overline{b_n}/n^{2\sigma }$. If $m\ne n$, the exponential integral iswhose absolute value is at most $2/|\log (m/n)|$. Thus the mean-value formula isThis also covers $T=0$.

For $m>n$, integration of $1/x$ over $[n,m]$ gives $\log (m/n)\ge (m-n)/m$; the analogous inequality holds with $m,n$ exchanged. Hence $|\log (m/n){|}^{-1}\le \max (m,n)/|m-n|\le X$. This gives the requested first bound $O(X{\sum }_{m\ne n}|a_n||b_m|/(n^{\sigma }{m}^{\sigma }))$.

For the weighted bound, put ${\alpha }_n=|a_n|n^{-\sigma }$ and ${\beta }_m=|b_m|m^{-\sigma }$. The inequalityreduces the estimate to a row sum and its symmetric counterpart. For a fixed $n$, the preceding logarithm bounds giveHere $n^2/(m(n-m))=n(1/m+1/(n-m))$, and each resulting harmonic sum is $O(\log (2X))$. The symmetric row sum is bounded by $O(m\log (2X))$. Therefore the second error bound is

Let $D(s)={\sum }_{p\le X}{p}^{-s}$, a Dirichlet polynomial supported on primes. The cosine sum is $(D(\sigma +it)+\overline{D(\sigma +it)})/2$, so its $j$th power is a finite linear combination of $D(\sigma +it{)}^k\overline{D(\sigma +it{)}^{j-k}}$, $0\le k\le j$.

The Dirichlet polynomial $D(s{)}^k$ is supported on integers that are products of exactly $k$ primes, counted with multiplicity. Each coefficient is at most $k!$ by unique prime factorization; for $k=0$ the only coefficient is the one at $1$. As $j$ is odd, $k\ne j-k$, so no integer can appear in both supports. Thus every diagonal term vanishes in the mean value of Dirichlet polynomials from part (a).

Both supports lie in $[1,X^j]$. The first error bound in part (a), with that common length, now gives for each termSumming the finitely many terms proves the odd moment of a prime cosine sum estimate:The argument holds for all real $\sigma$ and all $T\ge 0$; no cancellation estimate involving $T$ is needed once the diagonal is absent.

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