Proof. By the Euler product formula,

hence

By the triangle inequality,

Hence,

Note that these products converge if and only if ${\sum }_p{p}^{-\sigma }$ converges, and this sum converges by the comparison test. □

Proof. Let $k\ge 0$ be an integer.

We claim that there exist polynomials $P_k$, $Q_k$ of degree $\le k+1$ and such that for $\mathrm{Re}(s)>1$,

First assume ($\ast$) holds.

Then, since $P_k(\{t\}){\ll }_{k}1$, for $\mathrm{Re}(s)>-k-\frac{1}{2}$, $|s-1|\ge \frac{1}{10}$, ($\ast$) gives $|\zeta (s)|{\ll }_k|s{|}^{k+2}+1$. So (ii) follows.

For (i), using analytic continuation and ($\ast$), it suffices to show that the RHS of ($\ast$) is meromorphic for $\mathrm{Re}(s)>-k$, with the only pole a simple one at $s=1$.

Suffices to show that

is analytic for $\mathrm{Re}(s)>-k$.

This follows from the following criterion: If $U\subseteq ℂ$ is open, $f:U\times ℝ\to ℂ$ is piecewise continuous, and if $s\mapsto f(s,t)$ is analytic in $U$ for any $t\in ℝ$, then ${\int }_{ℝ}f(s,t)dt$ is analytic in $U$, provided that ${\int }_{ℝ}|f(s,t)|dt$ is bounded on compact subsets of $U$.

Applying this with $f(s,t)=\frac{P_k(\{t\})}{t^{s+k+1}}{𝟙}_{[1,\infty )}$ concludes the proof of (i) assuming ($\ast$).

We are left with proving ($\ast$). We use induction on $k$.

Case $k=0$: By Partial Summation,

Take $P_0(u)=u$, $Q_0(u)=1$.

Case $k+1$ assuming $k$: Let

For $\mathrm{Re}(s)>1$, we have

Let

This is a polynomial of degree $\le k+2$. By integration by parts,

Substituting this into the previous equation, we get that case $k+1$ follows. □

Proof. Apply the same criterion for integral of analytic function being analytic as in the previous lemma, taking $f(s,t)=t^{s-1}{e}^{-t}{𝟙}_{[0,\infty )}(t)$.

Note that for $0<{\sigma }_1\le \mathrm{Re}(s)\le {\sigma }_2$,

Proof.

• (i) For $\mathrm{Re}(s)>0$, by integration by parts,

  $${\int }_0^{\infty }{t}^s{e}^{-t}dt=s{\int }_0^{\infty }{t}^{s-1}{e}^{-t}dt.$$

  This proves (i) for $\mathrm{Re}(s)>0$.

  Now for any $k\in \mathbb{N}$, for $\mathrm{Re}(s)>0$ we have

  $$\mathrm{\Gamma }(s)=\frac{\mathrm{\Gamma }(s+k)}{(s+k-1)\cdots (s+1)s}.$$

  The RHS is analytic for $\mathrm{Re}(s)>-k$, so can use analytic continuation to extend $\mathrm{\Gamma }(s)$ meromorphically to $\mathrm{Re}(s)>-k$, with the only poles being simple poles at $s=0,-1,\dots ,-k,-1$.

  Let $k\to \infty$.

• (ii) Since both sides are analytic in $ℂ\setminus \mathbb{Z}$, by analytic continuation, it suffices to prove the formula for $0<s<1$.

  Now, for any $t>0$,

  $$\begin{array}{llllll}\hfill t^{s-1}\mathrm{\Gamma }(s-1)& =t^{s-1}{\int }_0^{\infty }{u}^{-s}{e}^{-u}du& \hfill & & \hfill & \\ \hfill & ={\int }_0^{\infty }{v}^{-s}{e}^{-vt}dt& \hfill & (u=vt)& \hfill & & \hfill & \end{array}$$

  Multiply by $e^{-t}$ and integrate, and use Fubini to get

  $$\begin{array}{llllll}\hfill \mathrm{\Gamma }(s)\mathrm{\Gamma }(s-1)& ={\int }_0^{\infty }{\int }_0^{\infty }{v}^{-s}{e}^{-vt}dv{e}^{-t}dt& \hfill & & \hfill & \\ \hfill & ={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-(v+1)t}dt{v}^{-s}dv& \hfill & & \hfill & \\ \hfill & ={\int }_0^{\infty }\frac{v^{-s}}{1+v}dv& \hfill & & \hfill & \\ \hfill & ={\int }_{-\infty }^{\infty }\frac{e^{(1-s)x}}{1+e^x}dx& \hfill & (v=e^x)& \hfill & & \hfill & \end{array}$$

  Hence, the remaining task is to show

  $${\int }_{-\infty }^{\infty }\frac{e^{(1-s)x}}{1+e^x}dx=\frac{\pi }{\sin (\pi s)}.$$

  We will do this next time.

Proof. Let $\mathrm{Re}(s)>1$. Then,

Make the change of variables $f=\pi n^{2}u$ to get

Hence

Summing over $n\in \mathbb{N}$ and using Fubini,

By the functional equation

we have

Plugging this into ($\ast$), we get

Hence

Since $|𝜃(u)-1|\ll e^{-\pi u}$, applying the criterion for integrals of analytic functions being analytic, we see that $\xi (s)$ is entire. So by analytic continuation, ($\ast \ast$) holds for all $s\in ℂ$.

Moreover, the expression for $\zeta (s)$ is symmetric with respect to $s\mapsto 1-s$, so $\xi (s)=\xi (1-s)$, $s\in ℂ$. □

Proof.

• (ii) – (iii) Follows from the lemma on polynomial growth of $\zeta$ on vertical lines.
• (ii) – (iii) We know $\zeta (s)\ne 0$ for $\mathrm{Re}(s)>1$. We want to show that if $\zeta (s)=0$ and $\mathrm{Re}(s)>0$ then $s\in \{-2,-4,-6,\dots \}$ and $s$ is a simple zero.

  Let $\mathrm{Re}(s)>0$. By the functional equation for $\zeta$,

  $$\zeta (s)=\underset{\ne 0}{\underbrace{{\pi }^{s-\frac{1}{2}}}}\frac{\mathrm{\Gamma }\left(\frac{1-s}{2}\right)}{\mathrm{\Gamma }\left(\frac{s}{2}\right)}\zeta (1-s).$$

  We claim that $\mathrm{\Gamma }(s)\ne 0$ for all $s\in ℂ$. By the Euler reflection formula,

  $$\mathrm{\Gamma }(s)\mathrm{\Gamma }(1-s)\sin (\pi s)=\pi ,$$

  for $s\in ℂ$.

  If $\mathrm{\Gamma }(s)=0$, then $\mathrm{\Gamma }$ has a pole at $1-s$. Hence, $1-s=-n$ for some $n\ge 0$ integer. But then $s=n+1$, and $\mathrm{\Gamma }(n+1)=n!\ne 0$.

  We conclude that for $\mathrm{Re}(s)<0$,

  $$\begin{array}{llll}\hfill \zeta (s)=0& ⟺\frac{s}{2}\text{ is a pole of }\mathrm{\Gamma }& \hfill & \\ \hfill & ⟺s=-2n,n\in \mathbb{N}& \hfill & \end{array}$$

  Since the poles of $\mathrm{\Gamma }$ are simple, $\zeta$ has a simple zero at $s=-2n$, $n\in \mathbb{N}$. □

Proof. This is Exercise 10 on Example Sheet 2. □

Proof. Let

Then $g$ is analytic and non-vanishing in $\overline{B(z_0,r∕2)}$. Note that

($\frac{{(f_1-f_n)}^{\prime }}{f_1-f_n}={\sum }_{i=1}^n\frac{f_1^{\prime }}{f_1}$).

Hence, it suffices to prove

for $z\in B(z_0,r∕4)$. Write

Then $h$ is analytic and non-vanishing in $\overline{B(0,r∕2)}$, and $h(0)=1$. We want to show

for $z\in B(0,r∕4)$.

For all $z_0\in \partial B(0,r)$ we have

since

for $z\in \partial B(0,r)$.

By the maximum modulus principle, $|h(z)|<e^M$ for $z\in \overline{B(0,r∕2)}$, so

By the Borel-Caratheodory Theorem with radii $\frac{3r}{8}$, $\frac{r}{4}$ we have for for $z\in B(0,3r∕8)$,

Now, for $z\in B(0,r∕4)$, Cauchy’s theorem gives us

Proof. We apply Landau with $z_0=2+it$, $r=50$, with $f(s)=(s-1)\zeta (s)$.

By the lemma on polynomial growth of $\zeta$ on vertical lines, for $\sigma +it\in B(z_0,50)$, we have

($|\zeta (z+it)|\asymp 1$).

Let $s\in B\left(z_0,\frac{25}{2}\right)$. Then Landau gives us

Since $B\left(z_0,\frac{25}{2}\right)$ contains all the points $s=\sigma +it$ with $|\sigma |\le 10$, it suffices to show

Substituting $s=z_0$ in ($\ast$), we get

since

Taking real parts,

Since $\mathrm{Re}(\rho )\le 1$,

This proves part (ii). It gives also ($\ast \ast$) since the sum there contains $O(\log (|t|+2))$ zeros. □

Proof. Recall that

where $\mathrm{\Lambda }$ is the von Mangoldt function function, and $\mathrm{Re}(s)>1$.

Taking linear combinations, the LHS of ($\ast \ast$) becomes

().

We are done by the inequality:

for $\alpha \in ℝ$. □

Proof. Let $\sigma \in [1,2]$ and $t\in ℝ$. Suppose $\zeta (\beta +it)$. uwe know that $\beta \le 1$. We know that $\zeta$ has no zeroes in some ball $B(1,r)$ for some $r>0$ (otherwise the entire function $(s-1)\zeta (s)$ would have an accumulation point for its zeros).

Choosing $c>0$ small enough, we can assume that $|t|\ge r$. By the key inequality for $\frac{{\zeta }^{\prime }}{\zeta }$,

Apply partial fraction decomposition of $\frac{{\zeta }^{\prime }}{\zeta }$. So

($t=\mathrm{Im}(s)$).

Since $\mathrm{Re}(\rho )\le 1$ for any zero $\rho$,

($\sigma >1$).

Discarding terms, we get

Take $\sigma =1+\frac{sc}{\log (|t|+2)}$, and assume $\beta \ge 1-\frac{c}{\log (|t|+2)}$ to get

Take $c=\frac{1}{16C}$ to get a contradiction. □

Proof. If $s=\sigma +it$ with $\sigma >10$, then

Assume then that $\mathrm{Re}(s)\in [-10,10]$. Apply ($\ast \ast$). Each term satisfies

We know that there are $O(\log (T+2))$ zeros with multiplicity having imaginary part $\in [T-2,T+2]$. The claim follows from triangle inequality. □