The proof that $\Gamma(z)\pm \Gamma(1-z)$ only has zeros for $z \in \mathbb{R}$ or $z= \frac12 +i \mathbb{R}$ has been given here:

Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?

An obvious follow up question is whether $\zeta(s) \pm \zeta(1-s)$ also has zeros (other than its non-trivial ones that would induce 0+0 or 0-0).

This is indeed the case and $\zeta(s)^2 - \zeta(1-s)^2$ has the following zeros:

$\frac12 \pm 0.819545329 i$

$\frac12 \pm 3.436218226 i$

$\frac12 \pm 9.666908056 i$

$\frac12 \pm 14.13472514 i$ (the first non trivial)

$\frac12 \pm 14.51791963 i$

$\frac12 \pm 17.84559954 i$


These 'semi' trivial zeros appear to all lie on the critical line. I wonder if anything is known or proven about their location (I guess not, since a proof that they must have real part of $\frac12$ would automatically imply RH, right?).

EDIT: Two counterexamples found by Joro in the answers below. Both have real parts outside the critical strip, so I would like to rephrase my question as:

Are the 'semi' trivial zeros that are located within the critical strip all on the critical line?

  • 2
    $\begingroup$ By the functional equation of the zeta function, these are the values of $s$ at which either $\zeta(s)=0$ or $$ 2^s \pi^{s-1} \sin(\pi s/2) \phantom. \Gamma(1-s) = \pm 1. $$ The problem of locating the complex roots of the latter equation must be way easier than the Riemann Hypothesis (probably comparable with your earlier MO question). $\endgroup$ Feb 25, 2012 at 21:31
  • 1
    $\begingroup$ @Noam: We were thinking and typing the same thing! $\endgroup$
    – GH from MO
    Feb 25, 2012 at 21:38
  • $\begingroup$ There are other counterexamples, but so far none for your edit. $\endgroup$
    – joro
    Feb 26, 2012 at 13:36
  • $\begingroup$ @Joro: Could you share the other counterexamples you've found please? The French arxiv article referenced below does suggest that all zeros should be on $\Re s =\frac12$ except for a finite number that reside elsewhere. $\endgroup$
    – Agno
    Feb 26, 2012 at 16:21
  • $\begingroup$ I just proved that there are no nonreal solutions with $|\Re s|$ sufficiently large. See "EDIT 2" in my response. $\endgroup$
    – GH from MO
    Feb 26, 2012 at 18:00

5 Answers 5


If $\zeta(s)$ is nonzero, but $\zeta(s)\pm\zeta(1-s)=0$, then by the functional equation of the Riemann zeta function we have $$ \pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\pm \pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)=0.$$ That is, your question is just the Riemann Hypothesis plus a more elementary one similar to your earlier question about the zeros of $\Gamma(s)\pm\Gamma(1-s)$. I would expect that the exact same techniques work here, i.e. one can show by known estimates for the gamma function that all nonreal solutions of the displayed equation lie on $\Re s=1/2$.

EDIT 1. To keep up with new developments I now expect that within the critical strip all nonreal solutions of the displayed equation lie on $\Re s=1/2$. Moreover, it seems reasonable to believe that there are no nonreal solutions with $|\Re s|$ sufficiently large.

EDIT 2. It follows from a generalized Rouché's theorem and Stirling's approximation that there are no nonreal solutions with $|\Re s|$ sufficiently large. More precisely, consider the rectangular contour $C_n$ with vertices $2n\pm it$ and $2n+2\pm it$, where $n>0$ is a large integer and $t>0$ is sufficiently large in terms of $n$. It suffices to show that along $C_n$ we have $$ \left|\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\right|<\left|\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\right|,$$ because this implies that inside $C_n$ there is precisely one solution of the above displayed equation (which must be real by the reflection principle). One can show that the right hand side divided by the left hand side is $ \gg n^{2n-\frac{1}{2}}(\pi e)^{-2n}$ on the vertical sides of $C_n$, while it is $\gg_n t^{2n-\frac{1}{2}}$ on the horizontal sides of $C_n$. The claim follows.

  • $\begingroup$ Non-trivial solutions are not common for my search. And large Re(s) slows the computations. $\endgroup$
    – joro
    Feb 26, 2012 at 15:10
  • $\begingroup$ RE: large Re(s). Thank you for the advice, will try it. The current implementation uses zeta. $\endgroup$
    – joro
    Feb 26, 2012 at 19:15

Slightly off-topic : this article from Arxiv (in french) shows that zeros of functions f : s -> h(s) - h(1-s), where h is a meromorphic function satisfying appropriate growth conditions, are inclined to lie on the critical line, so this phenomenon is not specific to gamma functions.

I didn't check that the main result in the article apply to the function in GH's answer above.

  • 2
    $\begingroup$ +1: nice catch. wow! $\endgroup$
    – Suvrit
    Feb 26, 2012 at 15:36
  • $\begingroup$ @js: Many thanks. Don't think this is slightly off topic at all, but spot on. Despite my French being a bit rusty, I believe the author generalizes the claim to: $f(s) = h(s)±h(2a−s)$ and the associated locations of zeros. He shows that all zeros for these functions should be on $\Re s = a$, save for a finite number of exceptions. Apparently the number of exceptions can also be zero, e.g. when $h(s)=\Gamma(s)$ that was proven by GH et al. For $h(s)=\zeta(s)$, there are exceptions (found by Joro), but if the author is right these should be finite. $\endgroup$
    – Agno
    Feb 26, 2012 at 16:36
  • 2
    $\begingroup$ @Agno: I do not think that $a=1/2$ is any more general than $a$ arbitrary. $\endgroup$
    – Marc Palm
    Feb 27, 2012 at 15:49
  • $\begingroup$ @pm experimentally zeta(s)-zeta(6-s) has much more zeros with Re(s) \ne 3, including 0<Re(s)<6. Don't know if the paper applies to this case. $\endgroup$
    – joro
    Feb 27, 2012 at 17:10

This might be a counter example to $\zeta(s)-\zeta(1-s)$.

Newton's method starting from $13 + 3i$ converges to $s \approx 13.16278786499+2.580464971850i$ which appears a zero.

[Added] A counterexample for $\zeta(s)+\zeta(1-s)$ appears $$ s \approx 14.870309115978233377 + 1.6450192905454179639i $$

[Added later] Per Agno's request the additional zeros found so far for $\zeta(s)-\zeta(1-s)$ are $16.478090665944547285 \pm 0.67940600947784773819i$ and $8.9909145336149198065 \pm 4.5105941406991465448i$

[Added later 2] The additional zeros of $\zeta(s)+\zeta(1-s)$ are $6.0002215061605926659 \pm 5.5128690434285266557$ and $11.248198934515946877 \pm 3.5349340823965337997$.

In all cases if $s$ is a zero so is $1-s$.

Here is a pari session with big precision:

\p 120
a=13.1627878649910358141631713461028903502674045760701701738298467140072119876746309296270651170803651644473445990732853741 + 2.58046497185066957108186074406714794119615081581034778843315640867385970134939303047779039837771630019890293928089781416*I
%2 = 3.667106168519458724 E-119 - 5.4882932400387681018240947190706996903 E-119*I

The starting value was opportunistically chosen from this plot. The colors are explained here

  • $\begingroup$ @Joro: You are right. These are both counterexamples. Fortunately the real part of both lies outside the critical strip, so not all is lost yet. I will tighten my opening question a bit further (see Edit). $\endgroup$
    – Agno
    Feb 26, 2012 at 11:32
  • $\begingroup$ What exactly is being plotted? $\endgroup$ Feb 26, 2012 at 11:42
  • $\begingroup$ @Gerry just added a legend to the plot (may need to reload). This is sage complex_plot of $\zeta(s)-\zeta(1-s)$. $\endgroup$
    – joro
    Feb 26, 2012 at 12:04
  • $\begingroup$ Why does large Re(s) slow the computation? Note that you don't need to calculate the zeta function, the question is really about the gamma function (cf. my response). $\endgroup$
    – GH from MO
    Feb 26, 2012 at 16:41
  • $\begingroup$ I just proved that there are no nonreal solutions with $|\Re s|$ sufficiently large. See "EDIT 2" in my response. $\endgroup$
    – GH from MO
    Feb 26, 2012 at 18:00

Here are two images to complement joro's image.

Zeroes of $ \zeta(s)^2 - \zeta(1-s)^2 $:

Zeroes of $\left ( 2^s \pi^{s−1} \sin \left ( \frac{\pi s}{2} \right ) \Gamma (1-s) \right ) ^2 - 1$:

In particular it seems to be the case that all zeroes inside the critical strip are on the critical line, and that all other zeroes don't stray too far from the origin, as hinted at by GH's answer.

  • $\begingroup$ You plotted only 6 basic non-trivial non-real zeros? (not counting +/- and 1-s) $\endgroup$
    – joro
    Feb 27, 2012 at 11:48
  • 2
    $\begingroup$ I like these plots a lot. How did you make them? $\endgroup$ Mar 8, 2015 at 22:50

We now have strong results from GH for $\Gamma(s) \pm \Gamma(1-s)$ and $\zeta(s) \pm \zeta(1-s)$ (when both are nonzero), as well as a more generalized claim outlined in this article http://arxiv.org/abs/0712.1266, that zeros for $f(s) = h(s) - h(2a-s)$, with $h$ being a meromorphic function satisfying appropriate growth conditions, are all inclined to lie on a critical line $a$, with only a finite number of exceptions.

However, despite the encouraging outcomes that all zeros are on the critical line and only a finite number of exceptions residing outside the critical strip, there doesn't seem to be any way to extrapolate these results to the non trivial zeros ($\rho$). No new constraints seem to be imposed on their location and they really seem to originate "from deep down the unaccessible cave" of $\zeta(s)$ itself. They therefore could still reside anywhere on the critical strip.

Just to share some 'futile attempts' I made to make a link. Firstly I had hoped that somehow the results would force the following outcome for all $\Re s$ in the strip:


and thereby seriously limit the possible values $\Re(\rho)$ could assume. But they don't.

In an attempt to find another relation between $\zeta(s)2$ and $\zeta(1-s)$, I converted both of them to the alternating $\eta(s)$ function and then paired up the individual terms for each $n$ (this is allowed since $\eta(s)$ is valid for $\Re s>0$). This gives:

$\displaystyle \sum _{n=1}^{\infty } \left({\frac { \left( -1 \right) ^{n-1}}{(1-{2}^{1-s}) {n}^{s}}} \pm{\frac { \left( -1 \right) ^{n-1}}{ (1-{2}^{s}) {n}^{1-s}}}\right)$

For each individual $n$, this yields a wave with a fixed frequency and amplitude, that only has zeros when $\Re s=1/2$. These waves nicely sum up to a curve that produces all the 'non & semi' trivial zeros from the OP. However, the $\rho$s obviously do arise from summing up the individual terms as well and I could not find a meaningful way to smartly swap out left and right terms, so that maybe new information about the non trivial zeros would be revealed.

As a last attempt, I also experimented with the PrimeZeta function $P(s)$.

For $P(s)-P(1-s)$ I found:

$\displaystyle \sum _{k=1}^{\infty } \frac{\mu \left( k \right)}{k} \ln \left( {\frac {\zeta \left( ks \right) }{\zeta \left( k \left( 1-s \right) \right) }} \right)$

and for $P(s)+P(1-s)$:

$\displaystyle \sum _{k=1}^{\infty } \frac{\mu \left( k \right)}{k} \ln \left( {\zeta \left( ks \right) \zeta \left( k \left( 1-s \right) \right) } \right)$

For both functions, all zeros appear to lie on the critical line $\Re s=\frac12$, but for the first function the non-trivial zeros have now turned into poles. Interesting to note that the Wolfram site on the PrimeZeta states: "According to Fröberg (1968), very little is known about the roots", so maybe there is something new here to proof for $P(s) \pm P(1-s)$... :-)

  • $\begingroup$ Does the finiteness of the exceptions assumes RH or is unconditional? $\endgroup$
    – joro
    Feb 29, 2012 at 6:41
  • $\begingroup$ @joro: I don't think the finiteness of the exceptions assumes RH since it only applies to $h(s)-h(2a-s)$ and not to the terms individually. I believe the most promising route is to use $h(s)=\eta(s)$. E.g. swapping items between both terms could force the non trivial zeros for each term to disappear, whilst potentially still retaining a form of $h(s)-h(2a-s)$ with all zeros (now including the non trivial ones) on a critical line $a$ (but no longer at $a = \frac12$). A mixing and matching approach could f.i. be to swap all even numbers or better: swap out all the prime numbers. $\endgroup$
    – Agno
    Feb 29, 2012 at 9:36
  • $\begingroup$ I proved unconditionally that if $s$ is not real and not a zero of $\zeta(s)$, but a zero of $\zeta(s)\pm\zeta(1-s)$, then the real part of $s$ is bounded. One still needs to prove that the imaginary part of $s$ is bounded, in order to conclude that the number of such values $s$ is finite. I expect that this can be established rather easily via some horizontal monotonicity property of the gamma function, similar to the mentioned treatment of the zeros of $\Gamma(s)\pm\Gamma(1-s)$, but I had not time and mood to work out the details. $\endgroup$
    – GH from MO
    Feb 29, 2012 at 13:36

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