Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2008 / A / 4

For $\mathrm{Re}s>1$, logarithmic differentiation of the absolutely convergent Euler product gives $-{\zeta }^{\prime }(s)/\zeta (s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$, where $\Lambda (p^k)=\log p$ and it is zero otherwise. On the line $\mathrm{Re}s=c>1$, this series is absolutely bounded by ${\sum }_n(\log n)n^{-c}<\infty$, and $1/[s(s+1)]$ is integrable in the vertical variable. Hence absolute convergence justifies interchanging sum and contour integral. The given first Mellin kernel, with $y=x/n$, contributes $x(1-n/x)=x-n$ if $n\le x$ and zero otherwise. ThusSplitting off $1/(s-1)$ gives the desired decomposition with $f$ for $x\ge 1$. There is a range misprint in the second supplied kernel: for $0<y<1$ its integral is zero, not $(y-1{)}^2/2$. The globally correct value is $(y-1{)}_{+}^2/2$. For $y\ge 1$ it is exactly the expression used here, so the subsequent large-$x$ argument is unchanged.

Near one, write $h=s-1$ and $\zeta (s)=h^{-1}+\gamma +O(h)$. Then $-{\zeta }^{\prime }/\zeta =h^{-1}-\gamma +O(h)$, so subtraction cancels the pole and $f$ has the removable value $\boxed{f(1)=-\gamma /2}$.

For a continuously differentiable $F$ with the given decay, both $F$ and $F^{\prime }$ are integrable and $F(t)\to 0$ at either end. Integration by parts on finite intervals and passage to the limit gives, for $x>1$,Its absolute value is at most $\Vert F^{\prime }{\Vert }_1/\log x$, proving the estimate.

We also need uniform control of $f(c+it)$ on bounded $t$ as $c\downarrow 1$. Here is the three-four-one zero-free-region argument supplying it. For $\sigma >1$, the Euler product and $3+4\cos u+\cos 2u=2(1+\cos u{)}^2\ge 0$ implyIf $\zeta (1+it)$ vanished for a nonzero real $t$, holomorphy from (3) would make the left side $O((\sigma -1{)}^{-3+4})\to 0$, since $\zeta (\sigma +2it)$ remains bounded, a contradiction. Thus there are no zeros on that line, and there are none to its right by the Euler product. After removal of the singularity at one, $f$ and $f^{\prime }$ are bounded on the compact rectangle $1\le \mathrm{Re}s\le 2$, $|\mathrm{Im}s|\le 3$.

For $|t|\ge 3$, put $H=-{\zeta }^{\prime }/\zeta -1/(s-1)$. The supplied estimates give $H=O((\log |t|{)}^9)$ anduniformly in $1<c<2$. Since $f=H/[s(s+1)]$, it follows that $f=O(t^{-2}(\log |t|{)}^9)$ and $f^{\prime }=O(t^{-2}(\log |t|{)}^{18})$, both $O(|t{|}^{-3/2})$. Also $df(c+it)/dt=i{f}^{\prime }(c+it)$. Together with the compact bound this proves both required decay bounds with constants independent of $c$.

Parametrizing the contour, the remainder in the decomposition is $x^{c+1}/(2\pi )$ times the oscillatory integral just bounded. Choose $c=1+1/\log x$ for $x>e$, so $x^{c+1}=e{x}^2$. Hence the remainder is $O(x^2/\log x)$. Finally $(x-1{)}^2/2=x^2/2+O(x)$, and $O(x)$ is absorbed, proving