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 $\sum (-1)^n f(n) = \sum (-1)^n \hat f(n/2)$, 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.

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.

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 $\sum (-1)^n f(n) = \sum (-1)^n \hat f(n/2)$, 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 $\zeta(s)$.

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 $\sum (-1)^n f(n) = \sum (-1)^n \hat f(n/2)$, 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.