To proove that the so called Dirichlet eta function \begin{equation} \eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} \end{equation} is part of the extended Selberg class $\mathcal{S}^\#$, I'm looking for a way to write \begin{equation} \eta(s)=\omega Q^{1-2s}\overline{\eta}(1-s)\prod_{j=1}^r \frac{\Gamma(\lambda_j (1-s)+\overline{\mu_j})}{\Gamma(\lambda_j s+\mu_j)} \end{equation} with $\lambda_j,Q>0$, $\mu_j,\omega\in\mathbb{C}$, $\Re{\mu_j}\geq0,|\omega|=1$.

Starting with $\eta(s)=(1-2^{1-s})\zeta(s)$ and using \begin{equation} \zeta(s)=\left(\frac{1}{\sqrt{\pi}}\right)^{1-2s}\overline{\zeta}(1-s)\frac{\Gamma(\frac12(1-s))}{\Gamma(\frac12 s)} \end{equation} we get \begin{equation} \eta(s)=\left(\frac{1}{\sqrt{\pi}}\right)^{1-2s}\overline{\eta}(1-s)\frac{(1-2^{1-s})}{(1-2^s)}\frac{\Gamma(\frac12(1-s))}{\Gamma(\frac12 s)} \end{equation} which doesn't look too bad, but I don't know how to go on.


1 Answer 1


With the Poisson summation formula we show that $$\theta(x) = \sum_{n=-\infty}^\infty e^{-\pi n^2 x} = x^{-1/2} \theta(1/x)$$ So that $$\Lambda(s) = \pi^{-s/2}\Gamma(s/2) \zeta(s) = \int_0^\infty x^{s/2-1} \frac{\theta(x)-1}{2}dx = \Lambda(1-s)$$ Thus $$\lambda(s) = 2^{s}\pi^{-s/2}\Gamma(s/2) (1-2^{-s})\eta(s) =(2^s-1)(1-2^{1-s})\Lambda(s)\\ = 2^{s}(1-2^{-s})(1-2^{1-s})\Lambda(1-s)= (2^s-1)(1-2^{1-s})\Lambda(1-s)=\lambda(1-s)$$

Which shows that $(1-2^{-s})\eta(s)$ is in the $S^\#$ class with the Gamma factor $\Gamma(s/2)$ and $Q = 2\pi^{-1/2}$. It is almost in the $S$ class because it has an Euler product, it only one problem being its $\log$-Euler product : $$\log((1-2^{-s})\eta(s)) = \log(1-2^{1-s})-\sum_{p \ge 3} \log(1-p^{-s})$$ and $\log(1-2^{1-s})$ isn't analytic for $\Re(s) \ge 1/2$ as required.

  • $\begingroup$ If I'm not much mistaken your equation $\lambda(s)=\lambda(1-s)$ is nothing else than my last equation. This is in fact a functional equation for $\eta(s)$, but not of the for $\mathcal{S}^\#$ required type. $\endgroup$
    – DonFuchs
    Jun 8, 2017 at 20:29
  • $\begingroup$ And I don't get the point of your last statement: At first I'm sure you mean $\log\eta(s)$ instead of $\eta(s)$. And then this is not the condition I know for being in $\mathcal{S}$. $\endgroup$
    – DonFuchs
    Jun 8, 2017 at 20:37
  • $\begingroup$ The commonly used axiom is, that $\log\eta(s)=\sum_{n=2}^\infty\frac{b(n)\Lambda(n)}{\log n}\frac{1}{n^s}$ for $\sigma$ sufficiently large and $b(n)\ll n^\vartheta,\vartheta<\frac12$. $\endgroup$
    – DonFuchs
    Jun 8, 2017 at 20:42
  • $\begingroup$ @M.Charbonnier Of course I meant $\log\eta(s)$. Yes the condition is that $\log F(s) = -\sum_p \sum_{k \ge 1} \frac{b_{p^k}}{k} p^{-sk}$ where $b_{p^k} = O(p^{k(1/2-\epsilon)})$ which means that $\sum_{k \ge 1} \frac{b_{p^k}}{k} p^{-sk}$ is analytic on $\Re(s) \ge 1/2$ $\endgroup$
    – reuns
    Jun 9, 2017 at 5:04
  • $\begingroup$ @M.Charbonnier The problem in what you wrote is the quotient of Gamma factors, you need to use the duplication formula to make it a single Gamma factor. And the axioms of the Selbergs class say that $F(s) = (1-2^{-s})\eta(s)$ is in it $\endgroup$
    – reuns
    Jun 9, 2017 at 5:09

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