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Starting from the alternating Abel Plana summation formula (presented here):

$$\sum_{n=0}^\infty (-1)^nf(n) = \frac{f(0)}{2} + i\int_0^\infty\frac{f(iy)−f(−iy)}{2\sinh(πy)} dy \tag{1}$$

and putting $f(t)=\xi\left(\frac12+it\right)$, where $\xi(s)$ is the Riemann $\xi-$function. Since $\xi(s) = \xi(1-s)$ the integral part reduces to zero and we obtain:

$$2\,\sum_{n=0}^\infty (-1)^n\,\xi\left(\frac12+in\right) = \xi\left(\frac12\right) \tag{2}$$

Questions:

  1. Numerically this appears to work fine, but is it formally allowed in terms of the required growth conditions for the Abel Plana formula?

  2. This is rather philosophical, but the series on the LHS "samples" its values at the integer imaginary parts on the critical line. We know that $\xi\left(\frac12+it\right)$ oscillates around $0$ and has its sign changes at the non-trivial zeros. Given the unpredictable nature of these zeros, the signs of the integer samples are expected to fluctuate rather randomly as well. Does this imply that the complexity of the distribution of the non-trivial zeros is fully "encoded" in $\xi\left(\frac12\right)$

ADDED 1:

Numerical evidence strongly suggests that the above could be generalised as follows:

$$\xi(s) = \sum_{n=1}^\infty (-1)^{n+1}\,\big(\xi\left(s+in\right)+\xi\left(1-s+in\right)\big) \qquad s \in \mathbb{C} \tag{3}$$

or after expanding to the $d$-th derivative:

$$\xi^{(d)}(s) = \sum_{n=1}^\infty (-1)^{n+1}\,\big(\xi^{(d)}\left(s+in\right)+(-1)^d\,\xi^{(d)}\left(1-s+in\right)\big) \qquad s \in \mathbb{C} \tag{4}$$

ADDED 2:

Numerical evidence also strongly supports the alternating series with multiplication of $\xi(s)$:

$$\frac12\,\xi(s)^2 = \sum_{n=1}^\infty (-1)^{n+1}\,\xi\left(s+in\right)\,\xi\left(1-s+in\right) \qquad s \in \mathbb{C} \tag{5}$$

Gary
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Agno
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  • note that $f(-iy)=\xi(y-i/2)$ so $\log f(-iy) \approx y \log y$ for large $y>0$ so the Abel-Plana assumptions are not satisfied ($f(-iy)/\sinh(πy)$ is highly unbounded while for Abel-Plana you need that to go to zero as $|y| \to \infty$); this being said, $\xi$ decrease very fast on vertical lines so $\sum_{n=0}^\infty (-1)^n,\xi\left(\frac12+in\right)$ is absolutely convergent – Conrad Dec 24 '23 at 04:05
  • @Conrad Agreed. However, the numerator in the integrand $f(iy)-f(-iy) = 0$. Since the numerical evidence is so strong (I've added a generalisation), I start to wonder whether $f(iy)=f(-iy)$ is an overriding condition for alternating Abel Plana to hold, i.e. it is independent of the boundedness of $f(x)$. – Agno Dec 24 '23 at 12:22
  • Try and see what you get when you do the residue integral on large rectangles and take the limit; it may or may not work but as Hardy noted in his work on Ramanujan formula that usually the condition $f$ entire exponential of type at most $2\pi$ is necessary as for higher type like here (order $1$ but maximal or infinity type due to $\Gamma$) formula fails badly – Conrad Dec 24 '23 at 17:02
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    I actually did the residues and while the given symmetry cancels the vertical integrals, one remains with horizontal integrals $\int_0^{\infty}(\xi(i\sigma +1/2\pm K)/\sin \pi (\sigma \pm iK)d\sigma$ which are independent of $K$ large and absolutely convergent but unclear why they would add to zero - here the growth condition on $f$ that makes Abel plana work ensures these guys go to zero as $K\to \infty$ but in our case that fails since $\xi$ grows fast on horizontal lines so the above grows fast with $K$ while the sine term doesn't decay fast enough; so the formula in the OP is not correct – Conrad Dec 26 '23 at 01:56
  • @Conrad the strange thing is that it works so well numerically. I've also tried it for some completed Dirichlet L-series equivalents of $\xi(s)$ and these work fine as well. However, when trying other entire functions that are symmetric around the line $\Re(s)=\frac12$ like ${}_2F_0 \left([s, 1-s],[], -\frac12\right)$ or $\frac{1}{\Gamma(s)}+\frac{1}{\Gamma(1-s)}$ it fails. Have also tried to use the integral expression for $\xi(s)$ and swapping the integral and summation to see whether a path to a proof would emerge, however this violates Fubini's rule and leads to divergent series. – Agno Dec 26 '23 at 13:45
  • It may be that the integral error is very small as $\xi$ decays very fast on vertical lines, while integral is oscillating too; it is very instructive to look at Hardy's lecture 11 in his monograph on Ramanujan work to see how and why integral formulas like here can and do fail when growth conditions are not satisfied (in a sense it is a reflection of the fact that entire functions that vanish at integers are identically zero precisely when they grow slower than than $\sin \pi z$ which $\xi$ doesn't – Conrad Dec 26 '23 at 15:24
  • @Conrad Found a link to how the alternating Abel Plana formula should be derived. I looks different from the one I used on the OP. https://math.stackexchange.com/questions/4702171/abel-plana-formula-for-int-0-infty-fracfxe2-pi-x-1dx-and-int-0 Maybe it sheds some light on the problem. – Agno Dec 27 '23 at 12:41
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    Regardless how you derive it the formula I mentioned follows with residues (take a large rectangle with vertices at $\pm iK, N+1/2 \pm iK$ indent it a little near $0$ on either side and apply Cauchy to $f/\sin \pi z$ so on one side you get the alternating sum and on the other the integrals (the indent at zero gives you half the value with the appropriate orientation if you do it left or right and get precisely the term in the sum in the post) - the vertical guys give the usual integral terms when they converge, while here they cancel by functional equation etc – Conrad Dec 27 '23 at 15:18
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    So the formula has those horizontal integrals at $\pm iK$ from $0$ to infinity when you let $N$ go to infinity - to me it's unclear if those cancel; when the growth condition on $f$ is satisfied those go to zero as $K$ goes to infinity by the dominated convergent theorem but here they do not; similarly here the vertical integrals generally diverge at infinity but here they cancel from the functional equation – Conrad Dec 27 '23 at 15:21
  • I have asked an extended version of this question on MO: https://mathoverflow.net/questions/461316/a-conjectured-series-expression-for-the-riemann-xi-function-and-or-completed and the conjecture seems to be proven. – Agno Dec 30 '23 at 22:51
  • Excellent - it seems then that the integral term cancels despite its growth – Conrad Dec 31 '23 at 00:31

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