Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2008 / A (Analytic Number Theory)

A multiplicative arithmetic function has $f(1)=1$ and satisfies $f(mn)=f(m)f(n)$ whenever $m,n$ are coprime. For a prime power $p^e$, the divisors are exactly $1,p,\dots ,p^e$, so there are $e+1$. Multiplicativity therefore givesFix $\epsilon >0$. If $p\ge 2^{1/\epsilon }$, then $(e+1)p^{-\epsilon e}\le (e+1)2^{-e}\le 1$ for every integer $e\ge 0$. There are only finitely many primes below the threshold. For each such prime the sequence $(e+1)p^{-\epsilon e}$ tends to zero, so its supremum $C_p$ is finite and at least one. ThusThe product is over a fixed finite set, independent of $n$, proving $\boxed{d(n)\le C(\epsilon )n^{\epsilon }}$.

The change of divisor $k=m/d$ gives $\sigma (m)/m={\sum }_{k\mid m}1/k$. List the divisors as $k_1<\cdots <k_{d(m)}$. They are distinct positive integers, so $k_j\ge j$ and $\sigma (m)/m\le {\sum }_{j\le d(m)}1/j$.

For an integer $K\ge 1$, monotonicity of $1/x$ gives ${\sum }_{k=2}^{K}1/k\le {\int }_1^{K}dx/x=\log K$; the same bound for real $K\ge 1$ follows by taking its integer part. ConsequentlyHence the requested bound holds with $\boxed{K(\epsilon )=1+\log C(\epsilon )}$, including $m=1$.

For $\mathrm{Re}s>1$, the Euler product theorem isThe local series converge absolutely, and the product is absolutely convergent in the sense that the sum over primes of the absolute deviations of the local factors from one is finite. Indeed this sum is bounded by ${\sum }_{n\ge 2}|f(n)|n^{-\mathrm{Re}s}$. To prove the identity, expand the product over primes at most $y$. Absolute convergence permits all those local expansions, and multiplicativity identifies the coefficients with $f(n)$ for integers whose prime factors are at most $y$. Every omitted integer exceeds $y$, so the difference from the full Dirichlet series is bounded by its absolute tail beyond $y$, which tends to zero. The same tail bound on a closed right half-plane inside $\mathrm{Re}s>1$ proves local uniform convergence. With signed coefficients a local factor can be zero; the identity does not require claiming every such general Euler product is nonzero.

For $d(n^2)$, multiplicativity gives local coefficients $d(p^{2e})=2e+1$. With $z=p^{-s}$,The divisor bound from (1), applied with a sufficiently small exponent, proves absolute convergence when $\mathrm{Re}s>1$; divergence at one follows from $d(n^2)\ge 1$. Multiplying the local identities therefore givesFor $n\le x$, the inequality $n^{\delta }\le x^{\delta }$ gives $d(n^2)/n\le x^{\delta }d(n^2)/n^{1+\delta }$. Sum those nonnegative terms and enlarge to the whole infinite series to get the required comparison.

For real $1<s<2$, integral comparison gives $\zeta (s)\le 1+(s-1{)}^{-1}\le 2/(s-1)$, while $\zeta (2s)\ge 1$. Hence $\zeta (s{)}^3/\zeta (2s)=O((s-1{)}^{-3})$ uniformly in that interval. Choosing $\delta =1/\log x$ for $x\ge 3$ makes $0<\delta <1$ and $x^{\delta }=e$. Thus

Each finite-interval integral $f_n(s)$ is entire in $s$: the integrand is $(x-n)e^{-(s+1)\log x}$ and its complex derivatives may be integrated on the compact interval. On a compact subset of $\mathrm{Re}s>0$, choose $\delta >0$ with $\mathrm{Re}s\ge \delta$. Then $|f_n(s)|\le n^{-1-\delta }$. The Weierstrass uniform-convergence theorem gives convergence and holomorphy of ${\sum }_n{f}_n(s)$ throughout that half-plane.

Integration by parts givesFor $\mathrm{Re}s>1$, sum absolutely and use the integral over $[1,\infty )$ to obtain $s{\sum }_n{f}_n(s)=1-\zeta (s)+1/(s-1)$. Rearranging gives the first requested identity.

On $[n,n+1]$, the magnitude of $(x-n)(x-n-1)$ is at most $1/4$. On a compact subset of $\mathrm{Re}s>-1$, its integrals are bounded by a convergent constant multiple of $n^{-1-\delta }$ for some $\delta >0$. Thus ${\sum }_n{g}_n(s)$ is holomorphic there. The prefactor $s(s+1)$ is entire, leaving only the explicit simple pole at one in $\tilde{\zeta }$, with residue one.

Differentiate $(x-n)(x-n-1)x^{-s-1}$ and integrate. Its endpoint values both vanish, givingAt $s=0$ the last quotient has its continuous value $-\log ((n+1)/n)$. For $\mathrm{Re}s>1$, telescoping over $n$ gives $(s+1){\sum }_n{g}_n-2{\sum }_n{f}_n=-1/s$. Substitution into $\tilde{\zeta }$ yieldsThis supplies the analytic continuation to $\mathrm{Re}s>-1$ apart from the pole. Evaluating its holomorphic expression at zero makes the series term vanish and gives $\boxed{\zeta (0)=-1/2}$.

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