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