For $\Re s>1$, the <Dirichlet series> multiplication rule and $-\zeta'/\zeta(s)=\sum_{n\geq1}\Lambda(n)n^{-s}$ give
$$
\left(\frac{\zeta'(s)}{\zeta(s)}\right)^2
=\sum_{n=1}^{\infty}\frac{(\Lambda*\Lambda)(n)}{n^s}.
$$
Apply an effective <Perron formula> on the line $\kappa=1+1/\log x$ and truncate at
$$
T=\exp(\sqrt{\log x}).
$$
The bound from part (c) controls the truncation error.

Use the classical <Zero-free region of the Riemann zeta function>
$$
\zeta(s)\ne0
\quad\text{when}\quad
\Re s\geq1-\frac{c_0}{\log(|\Im s|+3)},
$$
together with $\zeta'/\zeta(s)\ll\log^2(|\Im s|+3)$ there. <Contour shifting> moves the Perron contour to $\Re s=1-c_1/\log T$. The only crossed singularity is the double pole at $s=1$, whose <residue> is $x\log x+Ax$ by part (b). On the new contour,
$$
|x^s|
\leq x\exp\left(-\frac{c_1\log x}{\log T}\right)
=x\exp(-c_1\sqrt{\log x}),
$$
and the logarithmic-derivative bounds contribute only powers of $\log x$, which can be absorbed by reducing the positive constant in the exponential. The horizontal integrals and Perron truncation error are $O(x\exp(-c_2\sqrt{\log x}))$ as well. Therefore, for some $c>0$,
$$
\sum_{n\leq x}(\Lambda*\Lambda)(n)
=x\log x+Ax+O\bigl(x\exp(-c\sqrt{\log x})\bigr).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 3 / d / Solution

Contour shifting deforms a complex integration path through a region where the integrand is <meromorphic function>[meromorphic]. The <residue theorem> says that the change equals $2\pi i$ times the sum of residues of the poles crossed.

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