Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / A

The Dirichlet convolution is $(f_1\ast f_2)(n)={\sum }_{d\mid n}{f}_1(d)f_2(n/d)$. Fix $s$ with $\sigma =\mathrm{Re}s>1$. Absolute convergence of the two Dirichlet series givesThus the double series may be regrouped by $n=ab$, provingThe same estimate proves absolute convergence of the resulting Dirichlet series.

For a finite set of primes $p\le y$, multiply the absolutely convergent local series. Unique prime factorization and multiplicativity, with $g(1)=1$, giveThe omitted integers all exceed $y$, so the absolute difference from $G(s)$ is at most ${\sum }_{n>y}|g(n)|n^{-\sigma }$, which tends to zero. Hence the finite products converge to $G(s)$:This proves the Euler product directly from absolute convergence, rather than treating an infinite formal multiplication as automatically valid.

The Möbius function satisfies $\mu (p^e{)}^2=1$ for $e=0,1$ and is zero for $e\ge 2$. ThereforeThe inverse local factor is $(1+p^{-s}{)}^{-1}={\sum }_{e\ge 0}(-1{)}^e{p}^{-es}$. Consequently the inverse under Dirichlet convolution is the Liouville functionwhere $\Omega (n)$ counts prime factors with multiplicity. Both functions are multiplicative functions. At $p^e$ with $e\ge 1$, their convolution is $(-1{)}^e+(-1{)}^{e-1}=0$, while its value at one is one. Multiplicativity then proves $h_1\ast h_2=I$ on every positive integer. Its Dirichlet series is $H_2(s)=\zeta (2s)/\zeta (s)$ for $\mathrm{Re}s>1$.

For $n\le x$, write $F(n)=F(x)-{\int }_n^x{F}^{\prime }(t)dt$, multiply by $f(n)$ and sum. Interchanging a finite sum with the integral gives the Abel summation formulaApply it with $F(t)=t^{-s}$ and $S(t)={\sum }_{n\le t}{a}_n$. If $\sigma >\alpha$, the boundary term is $O(x^{\alpha -\sigma })\to 0$ and ${\int }_1^{\infty }|S(t)t^{-s-1}|dt$ converges. Therefore the Dirichlet series converges andChoose $C$ with $|S(t)|\le C{t}^{\alpha }$ for $t\ge 1$. For $\sigma >\alpha +\delta$,The Weierstrass M-test proves uniform convergence of $\sum I_m$ on this entire open half-plane. For fixed $s$, the same estimate with its actual $\sigma$ gives $I_m(s)=O(m^{\alpha -1-\sigma })$. Each $I_m$ is an entire function, since its integral has bounded positive $t$ range and can be differentiated in $s$ under an integrable bound on each compact set. Thus the locally uniform sum is holomorphic, and so is its product with $s$. This gives holomorphy from polynomial growth of arithmetic partial sums and proves holomorphy of $A$ for $\mathrm{Re}s>\alpha$. Multiplication by the unbounded factor $s$ does not assert uniform convergence of $A$ on a whole unbounded half-plane; local uniform convergence is what the holomorphy argument needs.

With $a_n=\mu (n)$ and the hypothetical bound, $A_{\mu }(s)$ is holomorphic for $\mathrm{Re}s>3/4$. On $\mathrm{Re}s>1$, the Euler product gives $\zeta (s)A_{\mu }(s)=1$. The identity theorem extends this as a meromorphic identity throughout the connected half-plane $\mathrm{Re}s>3/4$. Hence $\zeta$ has no zeros in that half-plane: at any zero, the product with the holomorphic $A_{\mu }$ would be zero. The pole at one is not a zero; $A_{\mu }$ has a compensating zero there.

The double-angle formula givesFor $\sigma >1$, the logarithm of the absolutely convergent Euler product givesExponentiation provesSuppose $\zeta (1+it)=0$ for $t\ne 0$, of order $r\ge 1$. As $\sigma \downarrow 1$, its middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$ because the pole at one is simple, and the final factor is bounded because $1+2it\ne 1$. The product is $O((\sigma -1{)}^{4r-3})\to 0$, contradicting the lower bound one. ThereforeThis is the three-four-one zero-free-region argument specialized to the line with real part one.

Complete multiplicativity and $f(p)=\pm 1$ give, for $\mathrm{Re}s>1$,The product is nonzero there; its absolutely convergent logarithm has terms $f(p{)}^m{p}^{-ms}/m$. ThusEach numerator is one of $3\pm 4\cos \theta +\cos 2\theta$, so it is nonnegative. Exponentiation gives the required lower bound one.

If $F(1+it)=0$ with $t\ne 0$, its assumed holomorphy makes $|F(\sigma +it){|}^4=O((\sigma -1{)}^4)$, or smaller for a higher-order zero. The zeta factor at real $\sigma$ grows at most as $(\sigma -1{)}^{-3}$, while the zeta factor near $1+2it$ stays bounded. The product tends to zero, a contradiction. Hence

For $x>0$, $(e^x-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nx}$. If $\sigma =\mathrm{Re}s>1$, the integral of the sum of absolute values is $\Gamma (\sigma )\sum n^{-\sigma }<\infty$. Therefore interchange of sum and integral is justified, and the substitution $u=nx$ givesOn the two unbounded rays of the specified Hankel contour, $x\ge \pi$ and $1/(e^x-1)$ decays exponentially, dominating every fixed power $x^{\sigma -1}$. The branch multiplier on the lower ray is a fixed factor $e^{2\pi i(s-1)}$. On the circular part, the chosen branch is continuous along the parametrized arc, and $e^z-1$ never vanishes because its nonzero zeros have modulus at least $2\pi$. Thus $I(s)$ exists for every $s$. The estimates are uniform when $s$ ranges over a compact set, including after differentiation in $s$, so $I$ is an entire function.

Let $C(s)=(e^{2\pi is}-1)\Gamma (s)$. The given identity shows that $I(s)/C(s)$ agrees with $\zeta (s)$ for $\mathrm{Re}s>1$ wherever the quotient is directly defined. It supplies the meromorphic continuation elsewhere wherever $C$ is finite and nonzero, with removable points filled by limits. At a nonpositive integer $-k$, the simple zero of the exponential factor cancels the supplied gamma pole, givingAt $s=-k$, the integrals on the two rays cancel because the integrand is single-valued. The remaining positively oriented circle has only the pole at zero inside it, soThe coefficient at $k=0$ is $-1/2$, hence $\boxed{\zeta (0)=-1/2}$. For positive even $k$, the relevant odd coefficient has degree at least three and is zero, giving $\boxed{\zeta (-2m)=0}$ for $m\ge 1$. For $k=2m-1$, the given generating function gives the coefficient $(-1{)}^{m-1}{B}_m/(2m)!$. Division by $C(-(2m-1))=-2\pi i/(2m-1)!$ yieldsThe symbols $B_m$ here follow the printed generating function: they are the positive magnitudes of the usual even-index Bernoulli numbers, so $B_1=1/6$, $B_2=1/30$. This convention accounts for the alternating sign in the final expression.