Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-23/6/b/solution

Assume Riemann hypothesis. First fix and work on . We prove the subpower zeta bound to the right of the critical line, then move back to the line by the functional equation and Phragmén–Lindelöf principle.
Put for large positive and integrate the supplied smoothed logarithmic derivative identity horizontally from to two. There are no zeros on this path under Riemann hypothesis, so the Euler-product logarithm at continues along it. Each prime-power term contributes at most , and the smoothing weights are at most one. Hence the integrated prime terms are bounded by
All zeros have . The local zero-count estimate from Question 3, with its reflected version for negative ordinates, gives uniformly for
To see the uniformity, sum the zeros in successive unit ordinate intervals against ; the distant dyadic tails are summable. The zero-term numerator has modulus at most . Its integrated contribution is therefore at most . The integrated supplied remainder is , also . Since is bounded, we obtain
Thus for every fixed and , . Negative follow by complex conjugation.
The zeta functional equation and the gamma ratio give . Zeta is holomorphic throughout this strip, since its pole at one is outside it, and Euler summation supplies polynomial vertical growth. The strip convexity conclusion of Phragmén–Lindelöf principle therefore gives at the midpoint
For a prescribed , choose and with ; the bounded range is harmless. This proves
The explicit-formula estimate is deliberately first made a fixed distance to the right of the critical line. No divergent zero bound at is used.

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