The right-hand side of the well known equation:

$$\displaystyle \pi^\frac{-s}{2}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s) =\int_1^{\infty} \left({x}^{\frac{s}{2}} + {x}^{\frac{1-s}{2}}\right)\,\frac{\psi(x)}{x} \text{d}x - \frac{1}{s\,(1-s)}$$

with $\displaystyle \psi(x)=\sum_{n=1}^{\infty}e^{-\pi\,n^2\,x}$, can be decomposed as $A(s)+A(1-s)$ when:

$$A(s):=\int_1^{\infty} \frac{{x}^{\frac{s}{2}} \,\psi(x)}{x} \text{d}x - \frac{1}{s}$$

Note that $A(s)=\overline{A(\overline{s})}$ and the reflection formula is $A(s)=\pi^\frac{-s}{2}\,\Gamma\left(\frac{s}{2}\right)\, \zeta(s)-A(1-s)$. This implies that when $s=\rho$, it follows that $A(\rho)=-A(1-\rho)$.

$A(s)$ also has complex roots, that seem to increase quite regularly (truncated at 3 decimals):

$15.809\pm11.848\,i, 19.574\pm19.240\,i, 22.660\pm25.536\,i, 25.364\pm31.263\,i, 27.815\pm36.621\,i, 30.083\pm41.713\,i, 32.210\pm46.600\,i, 34.224\pm51.322\,i, 36.145\pm55.907\,i, 37.988\pm60.377\,i, 39.763\pm64.742\,i, 41.480\pm69.027\,i, 43.145\pm73.230\,i, 44.766+77.552\,i, 46.375\pm81.286\,i, 47.511\pm84.415\,i,\dots $

Note that since these zeros seem to grow increasingly further away from the critical strip, it follows from the reflection formula above ($\rho$'s exist only in the strip) that $A(s) \ne A(1-s)$ at these zeros.


Is anything known about these zeros? Have these been studied before?

  • $\begingroup$ This is essentially the Riemann-Siegel formula, well, you're integrals are the starting point. Check the references, especially Edwards- he guides you through the derivation step by step- you might find something interesting. Also, you can view this as a sort of statement about eigenfunctions for the Fourier transform on $\mathbb{R}^+$. Might help too. $\endgroup$ Oct 8, 2014 at 6:23
  • $\begingroup$ I mean $\mathbb{R}^X$ $\endgroup$ Oct 8, 2014 at 6:44

1 Answer 1


Just to visualize your zeros. This shows that their distribution seems indeed "quite regular", but nevertheless with some lettle "dents", e.g. they are not on a concave/convex curve, and the distances of consecutive ones do not decrease monotonously.

enter image description here

  • $\begingroup$ You are right, Wolfgang. I did not spot these "dents" before. Could they be fluctuating around a line/curve? $\endgroup$
    – Agno
    Oct 4, 2014 at 14:58
  • $\begingroup$ @Agno Well, there is your related one mathoverflow.net/questions/165779 for the zeros of $A(s)+A(1−s)$. I'd guess they have similar dents, too, and even if they don't, what Alex has said there in his answer should apply here, too. Of course it's intriguing that for both, the zeros are so close to curves with supposedly simple formulas. But I don't think those formulas are easier to find (with proof) than the one for the zeros of $\zeta(s)$, which is as simple as $\Re(\rho)=\dfrac12$... and moreover with the zeros not only fluctuating around it, but right ON it! $\endgroup$
    – Wolfgang
    Oct 4, 2014 at 16:07
  • $\begingroup$ Thanks Wolfgang. Agree with your comment and indeed nothing compares to this "deceitfully simple beauty" of all non-trivial zeros lying on the line $\Re(s)=\frac12$... :-) $\endgroup$
    – Agno
    Oct 4, 2014 at 16:25

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