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I was wondering whether we can prove that $\zeta$ has at least one zero on the critical line $\mathrm{Re}(s)=1/2$.

For instance, Hardy proved in 1914 that $\zeta$ has infinitely many zeros on the critical line. See also here for the critical strip. I'm asking a much easier question, which to prove that $\zeta$ has at least one zero on the critical line.

The point of my question is that I don't need a constructive answer. I know that we have many examples of zeros with $\mathrm{Re}(s)=1/2$, but we can only compute approximations of these zeros (see here), because the imaginary parts of these zeros are conjecturally transcendental numbers, and probably algebraically independent with usual constants as $\pi,e$, etc (so that there is probably no closed formula...). So my question is different from that one, because I'm talking about any "known" or any "specific" zero, or any kind of numerical computation/approximation.

My question is similar but more specific than this one, which is just asking about the existence of at least one non-trivial zero. I would like an argument from analytic number theory (without approximations nor numerical computations) which shows the existence of a zero of $\zeta$ with $\mathrm{Re}(s)=1/2$.

I hope that my question is clear enough. Thank you.

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    $\begingroup$ Does computing $1+1=2$ count as a numerical computation? Numerical computations really are computer-assisted proofs of inequalities. You can do them by hand; it just takes more time. $\endgroup$
    – Bart Michels
    Commented Jan 2, 2018 at 12:58
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    $\begingroup$ As explained in this answer to a question you linked to, $\xi(0.5+it)$ is real-valued, so we just have to estimate it(s real part), by hand or by a faster method. $\endgroup$
    – Bart Michels
    Commented Jan 2, 2018 at 13:02
  • $\begingroup$ @barto : the most important is that I don't want any approximations. Computations are fine (as $1+1=2$), provided that they are dealing with exact values (but the problem is that there is no closed formula for the imaginary parts of any zero of $\zeta$). $\endgroup$
    – Alphonse
    Commented Jan 2, 2018 at 13:03
  • $\begingroup$ More than one third of zeros of Riemann's zeta-function are on the critical line. $\endgroup$
    – user436658
    Commented Jan 2, 2018 at 13:03
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    $\begingroup$ @Alphonse Altough I think I see what you're looking for, I'd like to stress that a numerical approximation is a proof, a computer-assisted one, which uses inequalities like the mean value theorem or the triangle inequality in a clever way to prove that $\pi$ lies between $3.1$ and $3.2$. There's no magic, it's just too long to do by hand. $\endgroup$
    – Bart Michels
    Commented Jan 2, 2018 at 13:11

1 Answer 1

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It is because $\zeta(s)$ has a functional equation that we can prove some of its zeros have real part exactly $1/2$.

The functional equation implies $$\Lambda(t) = \pi^{-(1/2+it)/2} \Gamma((1/2+it)/2) \zeta(1/2+it)$$ is $\mathbb{R} \to \mathbb{R}$, thus it has a zero at every sign change. Proving $\Lambda(t)$ changes of sign around $t \approx 14.15$ is not hard. See this plot, together with some bounds for the approximation it is a proof that $\Lambda(t)$ changes of sign.

Usually we look instead at the Hardy Z-function $Z(t) = \frac{\Lambda(t)}{|\pi^{-(1/2+it)/2} \Gamma((1/2+it)/2)|} = \zeta(1/2+it) e^{i \vartheta(t)}$ because it doesn't decay as fast as $\Lambda(t)$.

In practice $\Lambda(t) = \lambda(1/2+it)$ where $$\lambda(s) = \frac{1}{s-1} - \frac{1}{s}+ \int_1^\infty (x^{s/2-1}+x^{(1-s)/2-1})\theta(x)dx$$ where $\theta(x) = \sum_{n=1}^\infty e^{-\pi n^2 x}$. Approximating $\theta(x)$ by its first few terms gives an approximation with an arbitrary precision to $\Lambda(t)$.

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  • $\begingroup$ Looking at a plot doesn't give a proof at all, for me. Please provide a precise proof that shows that $\Lambda(t)$ has at least one change of sign. (I don't know what $t \approx 14.15$ means exactly — I mentioned in my question that I don't want any approximation anyway ; for instance showing that $\Lambda$ has (at least) a change of sign on the interval $[14,15]$ would be fine). $\endgroup$
    – Alphonse
    Commented Jan 2, 2018 at 18:57
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    $\begingroup$ @Alphonse It suffices to show $\Lambda(t_0) > a$ and $ \Lambda(t_1) < -a$ using any series or integral representation of $\Lambda(t)$, for example the one I added $\endgroup$
    – user798409
    Commented Jan 2, 2018 at 19:03
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    $\begingroup$ @Alphonse But writing all the details will be fastidious, we need an algorithm to check to what precision we are capable to compute $\frac{1}{s-1}$, $x^{s-1}, e^{-\pi n^2 x}$, the integral.. All those things are of course implemented in the softwares checking the zeros of $\zeta(s)$ up to $t < 10^{20}$. What Riemann probably did is using the argument principle to show $\zeta(s)$ has only one zero on $\Im(t) \in [14,14.5], \Re(s) \in [1/4,3/4]$, if the zero was off the line, there would be at least two zeros. $\endgroup$
    – user798409
    Commented Jan 2, 2018 at 19:13
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    $\begingroup$ @Alphonse The series for $\theta(x)$ decays very fast in $n$ and $\theta(x)$ decays very fast in $x$, so you can approximate the integral by $\int_1^M \sum_{n=1}^N$ (ie. find an upper bound for the error term). Then it reduces to evaluate $\int_a^b f(x)dx$ with $f$ continuous, using Riemann sums, keeping track of all the error terms. We can do all this, by it takes quite some time, and the obtained formulas won't be very useful. It is much easier to trust mathematica when it says its algorithm for $\Lambda(t)$ has a precision $< 10^{-7}$. $\endgroup$
    – user798409
    Commented Jan 2, 2018 at 19:28
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    $\begingroup$ @Alphonse Sure there is a proof that $\Lambda(t)$ changes of sign infinitely many times, by looking at the sign of $\int_0^\infty \Lambda(t) t^k dt$, this is done in Titchmarsh's book chapter X, the proof is 2 pages long. See also the last chapter about the computations for the zeros. $\endgroup$
    – user798409
    Commented Jan 2, 2018 at 20:20

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