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Riemann-Siegel's approximate functional equation
$\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}) $
is the starting point for accurate numerical estimates of $\zeta(s)$ as a function of the partial sums in s and 1-s (of its respective infinite series representations, which inside the critical strip are individually not convergent).
It represents also a best approach for estimating the asymptotic behavior as $t\rightarrow \infty$.
The asymptotic expression for the error term
$E(s) = O(x^{-\sigma}+ \ |t|^{\frac{1}{2}-\sigma}y^{\sigma - 1}) $
is obtained under the assumptions $0\leq \sigma \leq 1 , \; x,y,t>C>0$, and $2\pi xy=t$.

I am however interested in a somewhat different question:
fixed $t>0$ and $\sigma \neq \frac{1}{2}$, at what rate does the sums of said s and 1-s partial sums approach $\zeta(s)$ ?
In other words, I am investigating the rate of vanishing of
$\sum_{n\leq x}\frac{1}{n^s} \ + \ \chi(s) \ \sum_{n\leq y}\frac{1}{n^{1-s}} \ - \ \zeta(s) \; = \; O( ? ) \; \; \; \; \; \; x=y \rightarrow \infty $

I am having difficulties in searching through the available literature, as it appears that the assumption $2\pi xy=t$ is usually introduced quite early in the proof of the above asymptotic expression (see for example A. Ivic, The Riemann Zeta-Function, Theorem 4.1), so that I am not quite sure on how to proceed if said assumption is instead left out ...
Could anybody help with suggestions on how to proceed, or about where to look for relevant literature? Thanks.

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Did you look in Titchmarsh? –  Micah Milinovich Jul 7 '11 at 14:02
    
No, I have not looked in Titchmarsh. If you know that it contains the answer, I will purchase it. –  Luca Jul 7 '11 at 14:29
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2 Answers

up vote 2 down vote accepted

The assumption $2\pi xy=t$ (or something similar) is certainly necessary to guarantee that you get an approximation to the zeta-function and not some other function. For instance, if we choose $x=y=t/2\pi$ then the approximate functional equation "approximates" twice the zeta-function. One can see this as follows:

Inside the critical strip ($\frac{1}{4}\le\sigma\le \frac{3}{4}$, say), it is known that $$ \zeta(s) = \sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^s} +O(t^{-\sigma})$$ for sufficiently large $t.$ Hence, by the functional equation and Stirling's formula, we have $$ \zeta(s) =\chi(s)\zeta(1-s) = \chi(s)\sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^{1-s}} +O(t^{-1/2}),$$ as well. Therefore, $$ 2\zeta(s) = \sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^s} + \chi(s)\sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^{1-s}} +O(t^{-\sigma}) +O(t^{-1/2}),$$ as claimed.

Edit: You can also see this choosing $x=\frac{t}{2\pi}$ and $y=1$ (and vice versa) in formula that you stated in your question.

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Thank you, Micah also your suggestion, reinforced the realisation that this is the reason why we need "to keep x in check" with $2\pi xy=t$, as $t\rightarrow \infty$. Rewriting the first formula in my question in terms of converging partial sums of the alternating zeta function, the end result seems to imply that the residual term (see Edwards, par. 7.2, eq. 4) $\frac{\Gamma(-s)}{2\pi i} \int_{C_M} \frac{(-x)^s e^{Nx}}{e^x - 1} \frac{dx}{x}$ shall asymptotically compensate the non converging terms. I'll detail this in a new question. I will work at it this weekend. –  Luca Jul 14 '11 at 21:05
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Assuming $\sigma$ is between 0 and 1, neither of the two sums you're adding actually converge, so I doubt your left-hand side converges at all, much less converges to 0 at a nice rate. In fact, the second sum will have order of magnitude $y^\sigma = x^\sigma$, while the first sum will have order of magnitude $x^{1-\sigma}$; this basically proves that the left-hand side is steadily increasing in size as $x$ goes to infinity.

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Thank you, Greg you made me realize that this is probably the reason why we need "to keep x in check" with $2\pi xy=t$, as $t\rightarrow \infty$. I have hence rewritten the formula in terms of converging partial sums of the alternating zeta function. The end result seems to imply that the residual term (see Edwards, par. 7.2, eq. 4) \frac{\Gamma(-s)}{2\pi i} \int_{C_M} \frac{(-x)^s e^{Nx}}{e^x - 1} \frac{dx}{x}$ shall asymptotically compensate the non converging terms. But I believe it is more appropriate if I detail about this in a new question. I will work at it this weekend. –  Luca Jul 14 '11 at 20:58
    
sorry, that was (see Edwards, par. 7.2, eq. 4) $\frac{\Gamma(-s)}{2\pi i} \int_{C_M} \frac{(-x)^s e^{Nx}}{e^x - 1} \frac{dx}{x}$ –  Luca Jul 14 '11 at 21:01
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