Complete multiplicativity and absolute convergence give the <Euler product>
$$
D_f(s)=\prod_p\left(1-\frac{f(p)}{p^s}\right)^{-1}
\qquad(\Re s>1).
$$
Set $\sigma=1+1/\log x$. Taking logarithms of absolute values and expanding the local factors gives, uniformly in real $t$,
$$
\log|D_f(\sigma+it)|
=\sum_p\frac{\Re(f(p)p^{-it})}{p^\sigma}+O(1)
=\sum_{p\leq x}\frac{\Re(f(p)p^{-it})}{p}+O(1).
$$
The prime powers with exponent at least two contribute $O(1)$; changing $p^{-\sigma}$ to $p^{-1}$ below $x$ and estimating the tail above $x$ also cost $O(1)$. By <Mertens theorem>,
$$
\sum_{p\leq x}\frac{\Re(f(p)p^{-it})}{p}
=\log\log x-\mathbb D(f,n^{it};x)^2+O(1).
$$
Exponentiating yields
$$
\left|D_f\left(1+\frac1{\log x}+it\right)\right|
\asymp(\log x)\exp\left(-\mathbb D(f,n^{it};x)^2\right).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-150/3/c/solution