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 $B_4=-1\mathrm{/}30$.

For $\mathrm{\mid }z\mathrm{\mid }<1$, expand each term of [L2] into an absolutely locally convergent geometric series and use [L3]. The coefficient of $z^4$ in $\pi z\cot (\pi z)$ is then $-2{\sum }_{n\ge 1}{n}^{-4}=-2\zeta (4)$. On the other hand, setting $t=2\pi iz$ in [L1] gives $\pi z\cot (\pi z)=t\mathrm{/}(e^t-1)+t\mathrm{/}2$; the coefficient of $z^4$ is $B_4(2\pi i{)}^4\mathrm{/}4!=-{\pi }^4\mathrm{/}45$ by step 1.1. Equating these coefficients yields $\zeta (4)={\pi }^4\mathrm{/}90$.

Put $q(x)=\{x\}-1\mathrm{/}2$ and $Q(x)={\int }_1^{x}q(t)dt$. The function $q$ has integral zero over each interval $[n,n+1]$, so $Q$ is bounded. Integration by parts shows that, for $\mathrm{Re}s>-1$, ${\int }_1^{\mathrm{\infty }}q(x)x^{-s-1}dx=(s+1){\int }_1^{\mathrm{\infty }}Q(x)x^{-s-2}dx\mathrm{.}$ The integral on the right converges locally uniformly in this half-plane and defines a holomorphic function there.

For $\mathrm{Re}s>0$, ${\int }_1^{\mathrm{\infty }}\{x\}{x}^{-s-1}dx=\frac{1}{2s}+{\int }_1^{\mathrm{\infty }}q(x)x^{-s-1}dx\mathrm{.}$ Substituting this into [L1] gives $\zeta (s)=\frac{s}{s-1}-\frac{1}{2}-s{\int }_1^{\mathrm{\infty }}q(x)x^{-s-1}dx\mathrm{.}$ By step 1.1, the right-hand side extends holomorphically across $s=0$ and has value $-1\mathrm{/}2$ there. Uniqueness of analytic continuation identifies this value with $\zeta (0)$.

Start from [L1] and split the integral at $1$. The piece on $(0,1)$ is exactly where the theta transformation is used under [A1] 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 $P_{\rho }(1-s)=P_{\rho }(s)$. Under countable choice, [L1] supplies the full xi product over nontrivial zeros; its conjugation and reflection symmetries repeat this finite root pattern.

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.

The $N$th partial sum of $1+2+3+\cdots$ is $N(N+1)\mathrm{/}2$, which tends to $+\mathrm{\infty }$ as $N\to \mathrm{\infty }$. Thus the series has no ordinary sum.

Since the right-hand side has no ordinary sum, it cannot equal any value assigned to $\zeta (-1)$ by analytic continuation. This refutes the claim independently of that value.

By [L1], any global extension of zeta still has a pole at $1$, because it agrees with this local meromorphic function near $1$. Therefore it 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.

Take the entire function $F(s)=0$. Both sides of the given equation are identically zero wherever its displayed coefficient is finite. At its poles the product with the zero function has the zero meromorphic extension. Thus $F$ satisfies the equation as a meromorphic identity.

By [L1], $\zeta (2)={\sum }_{n\ge 1}{n}^{-2}\ge 1$, whereas $F(2)=0$. Hence a meromorphic function different from zeta satisfies the equation, and the equation alone cannot determine zeta. No assertion that zeta itself satisfies the equation is needed for this counterexample.