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I'm crossposting.

Being \begin{equation*} \eta(s)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}=\frac{1}{1^{s}}-\frac{1}{2^{s}}+\frac{1}{3^{s}}-\frac{1}{4^{s}}+\cdots \end{equation*} and following Riemann's second method, (Edwards p.15), to obtain the functional equation for $\zeta(s)$ one can think the same way and try the same aproach for $\eta(s)$. Thus from \begin{equation*} \int_{0}^{\infty} \operatorname{exp}\left(-n^{2} \pi x\right) x^{s / 2-1} d x=\pi^{-s / 2} \Gamma\left(\frac{s}{2}\right)\frac{1}{n^{s}} \text { for } s>0 \end{equation*} one can express $\eta(s)$ as \begin{equation*} \pi^{-s / 2} \Gamma \left(\frac{s}{2}\right)\underbrace{\left(1-\frac{1}{2^{s}}+\frac{1}{3^{s}}+\cdots\right)}_{\eta(s)} =\int_{0}^{\infty}\left(e^{-\pi 1^2 x}-e^{-\pi 2^2 x}+e^{-\pi 3^2 x}+\cdots\right)x^{s/2}\text{ }\frac{dx}{x} \end{equation*} How would one proceed from here to craft a functional equation for $\eta(s)$?

I'm interested in refferences and/or answers. Any of them will be very much apreciated.

Thanks.

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  • $\begingroup$ A.Neves, Can I get your email id to contact you by email? $\endgroup$ Apr 16 at 17:24
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Ignoring technicalities of convergence, in Riemann's second proof, you start with the Poisson summation formula $\sum_{n\in\mathbb Z} f(n / x) = x \sum_{n\in\mathbb Z} \hat f (n x)$, take the Mellin transform of both sides, and use the self-dual function $f(x)=e^{-x^2}$.

To get the alternating sum you want, you could either change the function or change the summation formula. For the function, you could use something like $\sum_{n\in\mathbb Z} f(n / x) \exp(\pi i n)$, and do some computations. You could also take a twisted Poisson summation formula $\sum (-1)^n f(n) = \sum_{n \textrm{ odd}} \hat f(n/2)$, but the steps for proving that are identical to the manipulations done to derive the functional equation for $\eta(s)$ from the functional equation for $\zeta(s)$.

Furthermore, an inverse Mellin transform allows you to go in the converse direction: a functional equation of Dirichlet series gives a summation formula. If the gamma factor is different then it will not be a Fourier transform but a generalization. If the degree of the functional equation is $d$ then the sum will be over $d$-th roots of natural numbers instead of over natural numbers.

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    $\begingroup$ Isn't it $f(x) = \exp(-\pi x^2)$ which is self-dual? I think if one uses $\exp(-x^2)$ a slightly different form of Poisson summation is required, see here. $\endgroup$
    – Mark
    Jul 29 '20 at 18:26

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