0
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.