Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 25 / 4 / a

Put $Z=\lfloor N^{2/5}\rfloor$ and $S={\sum }_{N<n\le N+M}{n}^{-it}$. For any positive integer shift $h\le Z^2$, translating the interval changes its sum by at most $2h$. Averaging the shifts $h=xy$ gives the bilinear shift averaging for a logarithmic phase identitySince $xy/n\le N^{-1/5}$, the alternating Taylor expansion of the logarithm has remainder at most $(xy/n{)}^{r+1}/(r+1)$. ThusFor $r=\lfloor 5.01\log t/\log N\rfloor$, we have $r+1>5.01\log t/\log N$, hence the error is at most $t^{-1/500}$. The exponential function on an imaginary argument changes by at most the change in that argument. ThereforeUse $Z^2\asymp N^{4/5}$ and $N<n\le N+M\le 2N$. Taking absolute values provesThe boundary error comes from integer shifts, so this argument also covers intervals shorter than a shift. Here $x,y$ range over positive integers; no zero term is needed.