Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2010 / A (Analytic number theory)

For $N=\lfloor x\rfloor$, interchange a finite sum and an integral:This proves the Abel's summation formula, including the term at $n=1$.

Apply it with $f(n)=1$, $F(t)=1/t$. Writing $\{t\}=t-\lfloor t\rfloor$ givesThe last two terms are $O(x^{-1})$, proving the first expansion. For the refinement let $B_1(t)=\{t\}-1/2$. It has mean zero on each unit interval, so its primitive $P(t)={\int }_0^t{B}_1(u)du$ is bounded and periodic. Integration by parts givesThus for integer $x$, where $\{x\}=0$,For the zeta representation, integrating each defining integral by parts yieldsSumming from $1$ to $N-1$ and substituting into the definition of $\zeta$ givesSince $0\le x-n\le 1$,This is uniform on compact subsets of $\mathrm{Re}s>0$, so the series is holomorphic there and limits at $s=1$ are legitimate after subtracting the pole. Expanding $N^{1-s}=1-(s-1)\log N+O((s-1{)}^2)$ givesThe final tail is $O(N^{-1})$, and the harmonic expansion sends the remaining difference to $\gamma$. Therefore

Use the convention $f(1)=1$ in the definition of a multiplicative arithmetic function. For a finite set of primes, multiplying the absolutely convergent local series gives the Dirichlet sum over integers supported on those primes, by unique prime factorization. If all primes up to $Y$ are included, any missing integer is larger than $Y$, so the difference from the full series is bounded by ${\sum }_{n>Y}|f(n)|n^{-\sigma }$, which tends to zero. This proves the Euler product in its absolute-convergence half-plane.

For coprime $m,n$, every divisor of $mn$ is uniquely a product of a divisor of $m$ and a divisor of $n$. Hence ${\sigma }_{\alpha }(mn)={\sigma }_{\alpha }(m){\sigma }_{\alpha }(n)$. The squared modulus is also multiplicative when $\alpha =i\theta$.

At a prime put $z=p^{-s}$ and $a=p^{i\theta }$, so $|a|=1$. Summing by the two divisor exponents givesFor example, expand the square as ${\sum }_{j,l\le k}{a}^{j-l}$ and sum $z^k$ from $k=\max (j,l)$ to obtain this rational expression. It remains valid for $a=1$. Absolute convergence for $\sigma >1$ follows from $|{\sigma }_{i\theta }(p^k)|\le k+1$ and convergence of ${\sum }_p{p}^{-\sigma }$. Multiplying the factors therefore provesSuppose now that $\zeta (1+i\theta )=0$. Necessarily $\theta \ne 0$, since $s=1$ is a pole. Complex conjugation gives a zero also at $1-i\theta$. In the numerator, the double pole at $s=1$ is canceled by those two zeros in the shifted factors. At $s=1\pm i\theta$, the possible simple pole of a shifted factor is canceled by the double zero of $\zeta (s{)}^2$. Thus on $\mathrm{Re}s>0$ the only possible remaining poles come from zeros of $\zeta (2s)$. Since the Euler product excludes zeros for $\mathrm{Re}(2s)>1$, these possible poles lie in $0<\mathrm{Re}s\le 1/2$. In particular $f$ is regular for $\mathrm{Re}s>1/2$.

The assumed nonnegative-series principle now makes its Dirichlet series converge and equal $f$ there. Its first coefficient is one and all coefficients are nonnegative, so $f(\sigma )\ge 1$ for real $\sigma >1/2$. But as $\sigma \downarrow 1/2$, the numerator remains finite while $\zeta (2\sigma )\to +\infty$, forcing $f(\sigma )\to 0$. This contradiction provesAt the real point $s=1$ the statement is understood as a pole rather than a finite nonzero value. The pole-location argument did not assume the zero-free boundary before proving it.

The vertical integral is absolutely convergent because the denominator is quadratic in the imaginary part. If $x<1$, move its line to the right: no poles are crossed, and the integral on $\mathrm{Re}s=R$ is $O(x^R/R)$, tending to zero. If $x>1$, move left across the simple poles at $0,-1$. Their residues are $1$ and $-x^{-1}$, and the far-left integral tends to zero. Horizontal segments vanish by their $O(|\mathrm{Im}s{|}^{-2})$ decay for a fixed pair of vertical lines. At $x=1$, dominated convergence on the original line gives the limiting value zero. Hence the kernel is $0$ below one and $1-x^{-1}$ at or above one.

On a line $c>2$, absolute convergence permits interchanging the totient Dirichlet series and this integral. The $n$th term is $x\varphi (n)$ times the kernel at $x/n$, giving exactly $\varphi (n)(x-n)$ for $n\le x$. ThusWe may take $c=5/2$ and shift to $\mathrm{Re}s=3/2$. The denominator has no zeros in this strip; the only pole crossed is $s=2$, where $\zeta (s-1)$ has residue one. Its residue in the integrand isOn the new line, the supplied estimates bound the integrand by $O(x^{5/2}|t{|}^{-3/2}{\log }^8|t|)$ for $|t|\ge 2$, an integrable function. The compact portion is bounded as well. The same estimates, uniformly across the strip, make the horizontal integrals tend to zero. ConsequentlyFor the Tauberian step let $A(x)={\sum }_{n\le x}\varphi (n)$. It is nondecreasing because the summands are nonnegative, and $S(x)={\int }_0^{x}A(t)dt$. For fixed $0<ϵ<1$,After division by $x^2$ and passage to lower and upper limits, the asymptotic for $S$ gives bounds $(3-3ϵ+{ϵ}^2)/{\pi }^2$ and $(3+3ϵ+{ϵ}^2)/{\pi }^2$. Letting $ϵ\downarrow 0$ squeezes them together. ThereforeThis uses monotonicity of the summatory function; differentiating an asymptotic formula without such control would not justify the conclusion.