Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 27 / 5

In the level-$D$ convention used here, the Selberg upper-bound sieve states that a nonnegative sequence with sieve distribution $A_d=Xg(d)+r(d)$ satisfiesIt follows by minimizing the quadratic form with the Selberg sieve weights and using the Selberg least-common-multiple weights bound $|{\lambda }_d|\le 3^{\omega (d)}$.

Sieve the interval $x<n\le x+z$ by two and by the primes congruent to three modulo four, up to $L=\sqrt{D}$. The interval divisor counts giveThus $X=z$, $g(p)=1/p$, and $r(d)=O(1)$ uniformly in the location of the interval. Take $D=z^{1/2}$ and $L=z^{1/4}$.

To bound $J$, let $y=L^{1/4}$ and form the Euler product over the forbidden primes at most $y$. The given reciprocal-prime sum in residue class one modulo four, subtracted from the Mertens second theorem, yieldsMoreover, the Mertens first theorem bounds the logarithmic mean by ${\sum }_{p\le y}(\log p)/p=\log y+O(1)\le \frac{1}{2}\log L$ for large $L$. The truncated Euler-product lower bound gives $J\ge W/2\gg \sqrt{\log z}$.

The full-range summatory bound for three to the prime omega bounds the error by $O(D{\log }^{2}D)=O(\sqrt{z}{\log }^{2}z)$, which is $O(z/\sqrt{\log z})$. Every integer whose prime factors are all congruent to one modulo four survives this finite sieve. Therefore the half-dimensional interval sieve givesThe constants are independent of $x$. Enlarging them covers the bounded range of $z$ before the asymptotic product estimates apply, so the result holds throughout $1000\le z\le x$.

Finite rearrangement and the Von Mangoldt divisor identity giveFor $m=\lfloor x\rfloor$, integral comparison of the increasing logarithm gives $\log (m!)=m\log m-m+O(\log (m+1))$. Moving $m$ to real $x$ changes the main term by $O(\log (x+1))$. Thus the integrated Chebyshev sum hasFor $x\ge 2$, chooseThese are logarithmic Möbius function weights; they need not be upper-bound sieve weights. For $n\le x$, the Möbius inversion identities $\mu \ast 1=ϵ$ and $\mu \ast \log =\Lambda$ giveHence the expression suggested in the question isWe will use the elementary Möbius harmonic logarithmic moments, $A_j(x)={\sum }_{d\le x}\mu (d){\log }^j(x/d)/d$, for whichHere is a proof of the needed estimates. From ${\sum }_d\mu (d)\lfloor x/d\rfloor =1$, replacing floors by $x/d+O(1)$ gives $A_0=O(1)$. Also ${\sum }_d\mu (d)H_{\lfloor x/d\rfloor }/d=1$, because $\mu \ast 1=ϵ$. Substituting the permitted harmonic number estimate $H_{\lfloor t\rfloor }=\log t+\gamma +O(1/t)$ gives $A_1+\gamma A_0=1+O(1)$, so $A_1=O(1)$.

For $A_2$, the weighted Dirichlet hyperbola method gives the harmonic divisor sum expansionExplicitly, for $r=\lfloor \sqrt{t}\rfloor$, $B(t)=2{\sum }_{a\le r}{H}_{\lfloor t/a\rfloor }/a-H_r^2$. Substitute the harmonic estimate and ${\sum }_{a\le r}(\log a)/a=\frac{1}{2}{\log }^{2}r+c\prime +O(\log (2r)/r)$, obtained by unit-interval integral comparison. This proves the expansion without a prime-distribution theorem. Since $\tau =1\ast 1$, we have $\mu \ast \tau =1$, and thereforeThe summed error is bounded by $x^{-1/2}{\sum }_{d\le x}{d}^{-1/2}\log (2x/d)=O(1)$, by integral comparison. Thus $A_2=2\log x+O(1)$.

Finally substitute the asymptotic for $T$ into $F$. The main term isThe error sum is $O(x)$: partition the integers $d$ into $x/2^{j+1}<d\le x/2^j$; each interval contributes $O(x(j+1{)}^2/2^j)$, and the resulting series converges. Using the moment estimates proves the Selberg symmetry formula:In particular, the factor two comes from the quadratic logarithmic moment, rather than from an assumption of the Prime number theorem.