Hi,
referring to the Riemann-Siegel approximate functional equation for Riemann's Zeta  

$ \zeta(s) = \sum\_{n\leq x}\frac{1}{n^s} \ + \ \chi(s) \ \sum\_{n\leq y}\frac{1}{n^{1-s}} \ + \ O(x^{-\sigma}+ \ |t|^{\frac{1}{2}-\sigma}y^{\sigma - 1}) $  

would anybody know of a similar result applying to the Dirichlet Eta function ?  

$ \eta(s) = \sum\_{n=1}^\infty\frac{(-1)^{n-1}}{n^s} = 1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+-\ldots $  

I am interested in expressing the Dirichlet Eta function in terms of its partial sums, as well as in terms of the partials sums of its critical line symmetrical "twin". So, I am looking for an expression of this kind  

$ \eta(s) = !(s) \ \sum\_{n\leq x}\frac{(-1)^{n-1}}{n^s} \ + \ ?(s) \ \sum\_{n\leq y}\frac{(-1)^{n-1}}{n^{1-s}} \ + \ O( ........) $  

where !(s) and ?(s) are functions yet unknown to me, and I am not even sure whether such an approximate functional equation might actually exist. I will greatly appreciate suggestions from anybody familiar with the subject. Many Thanks.



Approximate functional equation for Dirichlet eta, does any exist?
First of all, if you want just a single sum up to T, then just like with the zeta function you have an approximation:
$$\eta(s)\sim\sum\_{n=1}^{T/2\pi}\frac{(-1)^{n-1}}{n^s}$$


It turns out additionally that the full approximate functional equation (aka Riemann-Siegel formula) holds for really anything where you have a functional equation. You can see, for example, "The approximate functional equation for a class of zeta-functions" by Chandrasekharan and Narasimhan.


In this case, you can write the functional equation as relating $1-2^{-s}+3^{-s}-\cdots$ and $0.5^{-s}-1.5^{-s}+2.5^{-s}-\cdots$ (this form does not have an annoying $1-2^s$ term in it) so this latter term is what you find in the final formula:
$$\eta(s) \sim \sum\_{n\le x}\frac{(-1)^{n-1}}{n^s} - \chi(s)\sum\_{n\le y}\frac1{(n-1/2)^s}$$


Also, any of the methods for proving the formula for $\zeta$ carry over naturally to $\eta$ (sometimes more naturally), so let me know if you had a particular one in mind.


