Much effort has been expended on a famous unsolved problem about the Riemann Zeta function $\zeta(s)$. Not surprisingly, it's called the Riemann hypothesis, which asserts:

$$ \zeta(s) = 0 \Rightarrow \operatorname{Re} s = \frac1 2 \text{ or } \operatorname{Im} s = 0 .$$

Now there are numerical methods for approximating $\zeta(s)$, but as I understand, no one knows any exact values except at even integers, which include the trivial zeroes for which $\operatorname{Im} s = 0 $. So I've always wondered: For the rest, how does one prove $\operatorname{Re} s = 1/2$ exactly?

(All I know that seems vaguely useful is the argument principle, which I'm not sure helps here, but I'd be happy to learn about other techniques that aren't too far advanced.)

Edit: Looking over this again, I found out I missed the closed-form values at odd negative integers. This doesn't affect the question but seemed worth correcting. Thanks to the people who contributed.

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    $\begingroup$ You mean, how can one be sure that a numerically found zero lies exactly on the critical line? $\endgroup$ Oct 25, 2010 at 9:12
  • $\begingroup$ That has to be it because otherwise the question essentially boils down to "how would one approach proving the Riemann Hypothesis". But if one is knowledgable enough to understand a decent answer to that question (and that's assuming that it's possible to give a decent answer since even partial results about zeroes of $\zeta$ in the critical strip are difficult to get at), then one wouldn't ask that question here to begin with. $\endgroup$
    – kahen
    Oct 25, 2010 at 9:22
  • $\begingroup$ Yes -- you have to know about the zero before you can nail it. By that, I mean that if you compute $ \zeta(1/2 + it_0 + \epsilon) = 0 $ for some $ \epsilon \in \mathbb{C} $ with $ | \epsilon | $ small, how do you show $ \text{Re } \epsilon = 0 $? $\endgroup$ Oct 25, 2010 at 9:22
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    $\begingroup$ I think Andrew Odlyzko is the main guru is this field. Maybe an answer can be dug out from his papers: dtc.umn.edu/~odlyzko/doc/zeta.html $\endgroup$ Oct 25, 2010 at 10:46
  • $\begingroup$ what does "Im s = 0" mean? Does it mean the imaginary part of s = 0? $\endgroup$ Sep 11, 2013 at 14:53

5 Answers 5


As long as there are no multiple zeros on the line up to height $T$ you can determine, rigorously and using a finite amount of computation:

  • the number of zeros (counted with multiplicity) off the critical line, i.e., the number of counterexamples to the Riemann hypothesis, of height less than $T$. This is done by contour integrals of $d \log \zeta$ -- the "argument principle" in complex analysis. For each zero one can locate it to any specified accuracy, and determine its multiplicity.

  • the number of zeros on the critical line, up to height $T$. For each zero, its ordinate on the line can be calculated to any desired accuracy. This is done by counting sign changes of real-analytic functions with a $\zeta (1/2 + it)$ factor, on the critical line. As Matt E mentioned, the ability to write down a real-valued function capturing the behavior of a complex function on a line is a special phenomenon particular to zeta and L-functions, coming from the functional equation relating $\zeta(s)$ to $\zeta(1-s)$.

If there is a multiple zero of order $n$ at some point on the line, all you can determine by computations of the above type is that a very small neighborhood of that point contains $n$ zeros (counted with multiplicity). This could mean zeros on the line, counterexamples to the Riemann conjecture, or both.

It is part of the package of conjectures surrounding the Riemann hypothesis, that there are no multiple zeros, and that the zeros in fact "repel" each other in a quantifiable sense. There is a lot of numerical and theoretical support for the zero repulsion, especially from random matrix theory. If the Riemann hypothesis is true but there are multiple zeros, that would be almost as surprising as finding a zero off the line.


First, one uses the argument principle from complex analysis to compute the number of zeroes with real parts between $0$ and $1$ and with imaginary parts between $0$ and some positive number $T$. Since the answer is a priori an integer, one can get a precise answer even though one can't compute the $\zeta$-function exactly; it is "just" a matter of bounding the error in all approximations carefully enough.

Then, one has to show that this number of zeroes actually lie on the line where $\Re s = 1/2$. For this, one factors $\zeta(1/2 + i t)$ as a product of a certain nowhere zero function, and a real valued function $Z(t)$ (the function appearing in J.M.'s answer). (That this can be done in a useful way is part of the theory of the $\zeta$-function.)

To count zeroes of $\zeta$ with $\Re s = 1/2$ is now the same as counting zeroes of $Z(t)$, which can be determined by sign changes in $Z(t)$ (provided one estimates $Z(t)$ accurately enough).

Assuming that all works out (which it will if RH is true!) one gets a rigorous proof that all zeroes with imaginary part $\leq T$ lie on the line $\Re s = 1/2$. (In fact, for this to work out, one also needs the zeroes of $\zeta$ to be simple, which they are conjectured to be; see T..'s answer for an elaboration on this.)

  • $\begingroup$ I should probably have mentioned the splitting equation for $\zeta\left(\frac12+it\right)$ into $Z$ and $\vartheta$ somewhere in my answer, but thanks for this! $\endgroup$ Oct 25, 2010 at 22:30
  • $\begingroup$ Should you really have "assuming RH is true" in the last paragraph? If RH is true, we don't need to compute anything numerically to know that the zeros are on the critical line... $\endgroup$ Oct 29, 2010 at 9:20
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    $\begingroup$ @Hans: Dear Hans, Thanks for your comment. What I meant is that if RH is false then at some point this won't all work out. The phrasing is not very good; I'll fix it. $\endgroup$
    – Matt E
    Oct 29, 2010 at 15:52

From what I've found, it's mostly presented as a fait accompli, for instance, the Riemann-Siegel function $Z(t)$, a function frequently used in the zero-finding, is defined with a $\zeta\left(\frac12+it\right)$ factor. What now one looks for when using constructs like these is if the function ever "turns away" from the horizontal axis without crossing it (if this does happen, the hypothesis is false).

This is why things like Lehmer's phenomenon are interesting behavior for the ζ function. Put simply, these are sections of the critical strip where there is (very!) nearly no crossing.

Here are two traditional ways of visualizing the Lehmer phenomenon: one can look at the graph of the Riemann-Siegel function:

Riemann-Siegel Z(t)

(Mathematica code: Plot[RiemannSiegelZ[z], {z, 0, 100}, AspectRatio -> 1/5, Frame -> True])

or the so-called "zeta spiral" $z(t)=\zeta\left(\frac12+it\right)$ in the complex plane:

zeta spiral

(Mathematica code: ParametricPlot[Through[{Re, Im}[Zeta[1/2 + I t]]], {t, 0, 40}, AspectRatio -> Automatic, Frame -> True]).

Now, here is the first instance of the Lehmer phenomenon, seen using both viewpoints:

Lehmer for Riemann-Siegel

(Mathematica code: Plot[RiemannSiegelZ[z], {z, 7004, 7006}, Frame -> True, PlotRange -> {-2, 2}])

Lehmer for zeta spiral

(Mathematica code: ParametricPlot[Through[{Re, Im}[Zeta[1/2 + I t]]], {t, 7004 + 1/2, 7005 + 1/2}, AspectRatio -> Automatic, Frame -> True])

The hypothesis would have been false if for the Riemann-Siegel function, the function displayed a local extremum without crossing the x-axis, or for the zeta spiral, the apparent "cusp" was actually a cusp or did not even pass through the origin. (I won't spoil the fun by posting zoomed-in versions of those last two images, you can do it yourself with Mathematica or some other computing environment that can evaluate ζ(s) for complex values).

The two zeroes are in fact quite close: FindRoot[RiemannSiegelZ[x], {x, ##}, WorkingPrecision -> 20] & @@@ {{7005 + 1/50, 7005 + 7/100}, {7005 + 9/10, 7005 + 11/100}} returns the two approximate zeroes 7005.0628661749205932 and 7005.1005646726748389.

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    $\begingroup$ J.M - has it been proven that the Riemann hypothesis is equivalent to Z(t) having no negative local maxima and no positive local minima? $\endgroup$ Oct 25, 2010 at 23:16
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    $\begingroup$ @George: I've forgotten the provenance of what I'm about to say, but I do recall a statement that the negative of the logarithmic derivative of the Riemann-Siegel function ought to be strictly increasing in between consecutive zeroes for large enough argument, assuming that the hypothesis is true. $\endgroup$ Oct 25, 2010 at 23:20

You should prove that for example $ \frac{\xi(s)}{\xi(0)} = \frac{\det(H+1/4+s(s-1)}{\det(H+1/4)}$ with $H$ being a Hamiltonian operator with potential $ V^{-1}(x) $ proportional to the half-derivative of the Eigenvalue Staircase $ N(x)\pi = Arg \xi(1/2+ i \sqrt x) $ (see here).

Since the Hamiltonian operator is Hermitian then all the roots of the Riemann zeta have real part $ 1/2$. This proof is similar to the proof for the sine function

$ \frac{sin(x)}{x} $ with the potential $ V(x)=0 $ and boundary conditions $ y(0)=y(1)=0 $


If You like a computational proof of that make use of

Zeta (in Mathematica):

Zeta[1 - s] == \[Pi]^(1/2 - s) Gamma[s/2]/
   Gamma[(1 - s)/2] Zeta[s] // FullSimplify

This is published by G. F. Bernhard Riemann himself in his 1859 original paper. Look for example at Riemann Hypothesis and references there in.

$\zeta (1-s)=\frac{\pi ^{\frac{1}{2}-s} \zeta (s) \Gamma \left(\frac{s}{2}\right)}{\Gamma \left(\frac{1-s}{2}\right)}$

Do two limits. One from $0$ to $\frac{1}{2}$, and the other from $1$ to $\frac{1}{2}$. Or do variable change and look at the identity given: $t=1-s$. This proves immediately the question.

Often this is written as

$\Gamma \left(\frac{1-s}{2}\right) \zeta (1-s)=\pi ^{\frac{1}{2}-s} \zeta (s) \Gamma \left(\frac{s}{2}\right)$

to show the symmetry to the critical line with $Re(s)=\frac{1}{2}$.

Now know that for $Re(s)=1$ there is no zero and for $Re(s)=0$ there is no zero. For $s=1$ on the real axis there is a singularity of order $1$.

So in the critical stripe there can only exist zero for $Re(s)=\frac{1}{2}$ following from Riemann identity for the Riemann $\zeta$ function.

Graphical the statement at the borders of the critical stripe can be shown with (in Mathematica)

Plot[{Abs@Zeta[I z], Abs@Zeta[1 + I z], 
  Abs@Zeta[1 + I Im@ZetaZero[1]]}, {z, 0, 20}]

Riemann Zeta on the borders of the critical stripe

The green line is the functions value of the $\zeta$ function on the border $1$ for the first Riemann $\zeta$ on the critical stripe $Re(s)=\frac{1}{2}$. On the critical stripe the Riemann $\zeta$ is symmetrical to the real axis. So this is the first positive and negative zero. On the lower border $Re(s)=0$ the Riemann $\zeta$ functions is divergent. That is not so on the upper border. There the function remains bounded between the given lower limit and a small divergence of the maxima if You accept Riemann's hypothesis.

An exemplary and fun look at Riemann $\zeta$ function is

Plot[{Re@Zeta[I Im@ZetaZero[1] + z], Im@Zeta[z + I Im@ZetaZero[1]], 
  Abs@Zeta[z + I Im@ZetaZero[1]]}, {z, 0, 1}]

across the critical stripe at the first Riemann zeta zero

This is dominated by the real part and both get negative on the lower boundary. Mind the given identity above holds for both together.

A contourplot shows this

contourplot at the first Riemann zeta zero

This gives a nice idea of how the Riemann $\zeta$ behaves at the first zero over the critical line. This is not so symmetric as expected.

A the twelves zero this looks

Riemann zeta over the critical stripe at his zeros

And at the 120's

Riemann zeta over the critical stripe at his zeros

This shows that the divergence on the lower border of the critical stripe dominates the behavior across the stripe very much.

This is very exemplary and just for fun. But is shows up how complicated the situations is. And it allows hypothesizing that every zero of the Riemann $\zeta$ is individual and the behavior of the critical stripe varies a lot.

The Riemann $\zeta$ is steady and infinitely often differentiable up to the singularity at $s=1$.

This is not considered chaos. The Riemann $\zeta$ is an L-function and behaves well.

There is only the Riemann hypothesis open, whether there are infinitely many zero on the critical line $Re(s)=\frac{1}{2}$.

It is possible to transform the average line approximating, the Mangoldt lambda function, into the stair plot of the number of positive primes, in Mathematica this is the PrimePi function, with the Riemann $\zeta$ zeros so that the stairs are reproduced exactly. But this is an infinity process. For finite numbers of Riemann $\zeta$ zeros used, the approximation remains undulated with high deviation at the steps.

Mind this is a requirement. Not a proof. This on the other defines the need to prove that there are infinitely many Riemann $\zeta$ zeros on the critical line, and that the proof give here holds on the critical stripe as well so that there are no other zeros present. As mentioned in a remark to another answer to the question, there are many zeros numerically calculated, but calculation is not a proof. This makes on the other hand sure that up very high values the Riemann $\zeta$ zeros are enumerated and calculated.

enter image description here

These zeros can get very close together as this example shows:

zero of the Riemann <span class=$\zeta$" />

This is the 23999 and 24000's zero. They are already close together. They differ by $0.498732$. The Plot3D surface of the absolute value shows small narrow dips toughing the plane $\zeta[s]==0$ is single points only. In this region the values on the lower border of the critical strip are very large and drop steeply before the critical stripe. On the upper border of the critical stripe the function is nearly flat and not wavy.

The shown pairing might get very important. If there is a tendency in this pairing that the values get closer and closer for larger values. The http://www.dtc.umn.edu/~odlyzko/doc/zeta.html study did an extra check for the separateness of the zeros. They did not find something spectacular by this.

The distribution that this study confirmed was a normal Gaussian. This does not inhibit such touching, and showed that the class of very small distances get more and more filled the larger the values are along the imaginary line.

 Abs[Zeta[x + I y]], {x, 0, 1}, {y, Im[ZetaZero[24000]] - 0.75, 
  Im[ZetaZero[24000]] + 0.25}, PlotPoints -> 300, 
 ColorFunction -> "Aquamarine", 
 Contours -> {0.02, 0.03, 0.1, 0.2, 0.3, 0.4, 0.5}]

countour plot of a paired zero of the Riemann zeta

The structure is superposed the repellance from the lower border of the critical line and a attraction of the contour lines for the absolute value. For this behavior both imaginary part and real part need to be zero. The attraction shows that the imaginary part dominates.

These dips for the zero point are very narrow and really points on the critical line. The closer the point is, the more are the contour lines circle. The disturbances are not as strong as the zero is.

But these are presumable transcendent zeros. The first derivative for example does not have much zero at the points where the Riemann $\zeta[s]$ has. Since the values of the derivatives tell about the multiplicity of the zero, these zeros are mostly "transcendent", L-function-type zeros of the infinitely many factors type, in the literature there is sometime used the attribute simple. This is in contrast to my use.

If the absolute function is still considered, but the negative values are used in between the consecutive zeros than this looks steady still:

enter image description here

real part and imaginary part

The plot of the real part and imaginary part show that along the critical line between these two zeros both contribute to the absolute value.

From the distribution point of view the extreme value have to be taken and confirm the validity of the contribution. It is possible to show that for larger values the narrower pairs occur more often.

ListPlot[ListConvolve[{1, -1}, 
    ZetaZero[#] & /@ (Range[2500] + 1000)] // Im // N]

list plot of the distances between Riemann zeta zeros

This show already the very narrow zeros occur already for lower values of the Riemann $\zeta[s]$.

ListPlot[ListConvolve[{1, -1}, 
    ZetaZero[#] & /@ (Range[2500] + 100000)] // Im // N]

list plot of the distances between Riemann zeta zeros

The same sample but at very much higher indices for the Riemann $\zeta[s]$. The trend is clear. The are less bigger distances between following each other zeros and much more smaller ones.

But the smallest one has not gotten much smaller. It only a little bit smaller.

This trend does not hold steadily. There are exceptions that not outliers. The distribution for the smaller values of the indices is somewhat wavy in the median. That is a fact all over the critical line. So the second distribution picture is a lucky choice.

As already mentioned, a distribution is for probabilities and the possible is possible. The bigger the number of considered zeros the more the distribution becomes valid. That is a proof for small ensembles known to statistician very well.

For the packed array of $\zeta$ zeros in Mathematica the end shows even smaller distances between pairs of zeros:

zeros even larger indices.

I hope that this gave some insights into the delicate situation on the critical stripe and the critical for the Riemann $\zeta$. This shows up what enormous deed, the Riemann identity. Riemann came from the functional equation approach not from the heuristic, empirical perspective as these examples are, and the work of Odlyzhko is. So statistics is mathematics, and this is the exemplary part of this proof approach.

Perhaps many of these coinside for very large indices. Maybe there are real intervals between the zeros in which the absolute function remains zeros all the way. As shown in the contour plot, and with the plots on the critical line, and the difference convolution of the consequtive zeros, the zeros get some attraction for each other from an unknown source.


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