Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 2

The Mellin transform of $F$ isBecause the support is a compact subset of $(0,\infty )$, the integral defines an entire function of $s$. The Mellin inversion formula says that, for every real $\sigma$ and every $y>0$,

Apply this with $y=n/X$ and initially $\sigma =2$. Absolute convergence of the Dirichlet series for the logarithmic derivative permits interchange of sum and integral, givingTruncate at height $T=(\log X{)}^A$, where $A$ is a sufficiently large fixed constant. The assumed bound on $\tilde{F}$ makes the discarded tails smaller than the required error. The classical Zero-free region of the Riemann zeta function and the standard bound ${\zeta }^{\prime }(s)/\zeta (s)\ll (\log (|t|+3){)}^{O(1)}$ there allow the truncated contour to move toThe only singularity crossed is the simple pole of $-{\zeta }^{\prime }/\zeta$ at $s=1$, whose residue is $1$. Its contribution isOn the new contour, $|X^s|=X^{1-c/\log \log X}$; the zeta bounds, contour length, and exponential decay of $\tilde{F}$ absorb into a slight decrease of $c$. ThereforeThis is the smoothed prime number theorem from a zero-free region.

PutIf $Y<n\le Y+X$ is prime, then $n>D=X^{1/10}$. Since $f$ is supported on $[-1,1]$, the only divisor of $n$ contributing to $W(n)$ is $d=1$, and $W(n)=f(0)=1$. Every summand $W(n{)}^2$ is nonnegative, so

LetSince $g$ is the Fourier transform of $x\mapsto e^{x}f(x)$, the Fourier inversion theorem gives the Fourier representation of a smooth Selberg weightExpanding the square, interchanging the absolutely convergent sums and integrals, and usingwe obtainThe error is $O(D^2)$ because the support of $f$ restricts both divisors to $d_j\le D$. The main factor isand its Euler product is

It remains to estimate the integral using the assumed zeta-factor bound. The transform $g$ is a Schwartz function, since $e^{x}f(x)$ is smooth and compactly supported. We may therefore truncate to $|t_1|,|t_2|\le T$, losing an arbitrarily large negative power of $T$. Uniformly in the needed truncated range, the standard estimates near the pole of the Riemann zeta function giveandAfter multiplication, one net factor $(\log D{)}^{-1}$ remains. The polynomial factors in $t_1,t_2$ are integrable against the rapidly decreasing $g(t_1)g(t_2)$, and choosing $T$ as a sufficiently large power of $\log D$ makes the tails negligible. HenceSince $D=X^{1/10}$ and $D^2=X^{1/5}\ll X/\log X$,This proves the short-interval prime upper bound from a smooth divisor weight.