# Functional equation for $\eta(s)$ following Riemann's $2^{nd}$ method

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.

• A.Neves, Can I get your email id to contact you by email? – Sourangshu Ghosh Apr 16 at 17:24

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.
• 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. – Mark Jul 29 '20 at 18:26