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If non-trivial zeroes of the Riemann Zeta function are irrational, and we have to calculate $\zeta[t]$ numerically, how is it ever possible to prove that a particular zero of $\zeta[t]$ falls exactly on the $Re[t]=1/2$ line? Are we really only able to prove that they're within our computational precision of that line?

Jerry Guern
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    That's why you ought to prove the claim analytically – b00n heT Mar 27 '17 at 21:31
  • @b00nheT But that would take me all day! – Jerry Guern Mar 27 '17 at 21:54
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    Stop wasting people's time – b00n heT Mar 27 '17 at 21:57
  • See this answer. You know that the zeros of $\cos(z)$ fall exactly on $Im(z) = 0$ because $\cos(t)$ is a function $\mathbb{R} \to \mathbb{R}$ having a zero at each sign change. It works the same way for Dirichlet series with functional equation : obtaining a function $\Xi : \mathbb{R} \to \mathbb{R}$ with a real zero at every sign change. Here $\Xi(t) =(t+i/2)(t-i/2) \xi(1/2+it), \xi(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$ – reuns Mar 29 '17 at 10:15

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You can numerically evaluate the contour integral:

$$\frac{1}{2\pi i}\oint_C\frac{\zeta'(s)}{\zeta(s}ds$$

where $C$ is a small counterclockwise circle around the zero with its center placed on the critical line. This yields the total number of zeros (counting multiplicities) that are within the circle. Then suppose that the zero is not on the critical line. We then know from the reflection formula that there there should be two zeros on either side of the line at equal distances from the line. So, if the above integral numerically evaluates to 1 such that the numerical error is much smaller than 1 so that you know for sure it cannot be 2, you've proven that the zero lies exactly on the critical line.

Count Iblis
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