The Riemann Zeta Function — Examples

Evaluating the Euler factors from [L1] at $s=2$ gives ${(1-\frac{1}{4})}^{-1}{(1-\frac{1}{9})}^{-1}{(1-\frac{1}{25})}^{-1}{(1-\frac{1}{49})}^{-1}=\frac{4}{3}\cdot \frac{9}{8}\cdot \frac{25}{24}\cdot \frac{49}{48}=\frac{1225}{768}\mathrm{.}$

By [L2], the target value is ${\pi }^2\mathrm{/}6\approx 1.64493$. Comparing with step 1.1 shows that even this short prime truncation already lands within about five hundredths of $\zeta (2)$.

Expanding $e^t-1=t+t^2\mathrm{/}2+t^3\mathrm{/}6+t^4\mathrm{/}24+\cdots$ and solving for the quotient in [L1] gives $\frac{t}{e^t-1}=1-\frac{t}{2}+\frac{t^2}{12}-\frac{t^4}{720}+\cdots ,$ so $B_4=-1\mathrm{/}30$.

Substitute $m=2$ and $B_4=-1\mathrm{/}30$ into [L2]: $\zeta (4)=-\frac{(-1\mathrm{/}30)(2\pi {)}^4}{2\cdot 4!}=\frac{{\pi }^4}{90}\mathrm{.}$

Let $s\to 0$ in [L1]. Then $2^s\to 1$, ${\pi }^{s-1}\to {\pi }^{-1}$, $\sin (\pi s\mathrm{/}2)\sim \pi s\mathrm{/}2$, and $\mathrm{\Gamma }(1-s)\to 1$. Also [L2] gives $\zeta (1-s)\sim -1\mathrm{/}s$.

Multiplying the limits from step 1.1 yields $\zeta (0)={\lim }_{s\to 0}{2}^s{\pi }^{s-1}\sin (\pi s\mathrm{/}2)\mathrm{\Gamma }(1-s)\zeta (1-s)=-\frac{1}{2}\mathrm{.}$

Start from [L1] and split the integral at $1$. The piece on $(0,1)$ is exactly where the theta transformation is used in the proof of the completed functional equation.

The resulting rewritten form is the symmetric identity recorded in [L2], $\mathrm{\Lambda }(s)=\frac{1}{s(s-1)}+\frac{1}{2}{\int }_1^{\mathrm{\infty }}(\theta (t)-1)(t^{s\mathrm{/}2-1}+t^{-s\mathrm{/}2-1\mathrm{/}2})dt,$ so the two polar terms are isolated explicitly in the factor $1\mathrm{/}(s(s-1))$.

Apply [L1] with $m=1$. Then $\zeta (-2)=0$.

This is exactly the first trivial zero.

The four listed roots of $P_{\rho }$, counted with multiplicity, are closed under complex conjugation and under $\alpha \mapsto 1-\alpha$. Therefore the coefficients of $P_{\rho }$ are real, and replacing $s$ by $1-s$ permutes this root multiset.

A polynomial is determined by its roots together with the leading coefficient, and both remain unchanged in step 1.1. Hence $P_{\rho }(1-s)=P_{\rho }(s)$. This finite symmetric zero set is exactly the pattern that the full xi product repeats over all nontrivial zeros.

Apply [L1] at $s=1\mathrm{/}2$. Then ${\sum }_{n\ge 1}\frac{(-1{)}^{n-1}}{\sqrt{n}}=(1-\sqrt{2})\zeta (1\mathrm{/}2),$ so the eta series represents the continued zeta value at $1\mathrm{/}2$.

By [L2], the defining Dirichlet series ${\sum }_{n\ge 1}{n}^{-1\mathrm{/}2}$ diverges. Therefore step 1.1 gives a point where the continuation is represented by the eta series even though the defining Dirichlet series diverges.

At $s=1$, the defining series in [L1] becomes ${\sum }_{n\ge 1}1\mathrm{/}n$.

By [L2], this is the harmonic series and it diverges. Therefore the defining Dirichlet series does not converge at $s=1$.

By [L1], the claim already fails at $s=1\mathrm{/}2$: the continuation exists there, but the Dirichlet series does not converge.

By [L2], the excluded point $s=1$ is also a divergence point for the defining series. Therefore the slogan "the same Dirichlet series works for every $s\mathrm{\ne }1$" is false.

By [L1], the continued zeta value at $-1$ is $-1\mathrm{/}12$.

The ordinary partial sums of $1+2+3+\cdots$ are $1,3,6,\dots$, so they do not equal the fixed number $-1\mathrm{/}12$. Step 1.1 and [L2] therefore refute the claim.

By [L1], zeta has a pole at $1$ and therefore is not holomorphic there.

An entire function is holomorphic on all of $\mathbb{C}$, so step 1.1 rules that out. Hence zeta is not entire.

The factor $e^{(s-1\mathrm{/}2{)}^2}$ is entire, nonconstant, and invariant under $s\mapsto 1-s$ because $(1-s-1\mathrm{/}2{)}^2=(s-1\mathrm{/}2{)}^2$. Therefore multiplying the functional equation in [L1] by this factor shows that $F$ satisfies the same functional equation as zeta.

The function $F$ is not equal to zeta, since for example $F(2)=e^{9\mathrm{/}4}\zeta (2)\mathrm{\ne }\zeta (2)$. Thus step 1.1 gives a different meromorphic function obeying the same functional equation, so that equation alone cannot characterize zeta.