For $\Re s>1$, absolute convergence permits a shift of the complex variable:
$$
\sum_{n=1}^{\infty}\frac{\Lambda(n)n^{iu}}{n^s}
=\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^{s-iu}}
=-\frac{\zeta'(s-iu)}{\zeta(s-iu)}.
$$

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 150 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-150/2/c/solution