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Bear with me, I'm fresh out of high school so my level of mathematical knowledge is quite low (probably too low to be trying to understand the Riemann hypothesis, but at least I'm trying).

At this current time, I'm trying to make sense of John Derbyshire's (fantastic) book Prime Obsession. One thing that hasn't been explained in the book and to which I cannot find answers within the bounds of my understanding through research, is why exactly Riemann thought his hypothesis was true.

From what I understand, Bernhard Riemann was a very intuitive mathematician so it's quite possible that although he did not have a proof for his hypothesis, it made sense to him intuitively. I guess my question is, How does it intuitively make sense that all non-trivial zeros of the zeta function lie on the critical line and not somewhere else?

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    $\begingroup$ Riemann proved that $\zeta(s)$ has a non-trivial zero at every sign change of the real function $\xi(1/2+it)$ where $\xi(s) = s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s), \xi(s) = \xi(1-s),\xi(\overline{s})=\overline{\xi(s)}$, and that the first few zeros are on the critical line. Also he realized that the Riemann hypothesis was equivalent to the Möbius function behaving like a random i.i.d. function, which makes sense. $\endgroup$ – reuns Feb 19 '17 at 19:02
  • $\begingroup$ There's actually been a lot of debate about what Riemann really knew and conjectured. After all he mentioned the actual Hypothesis only in a half sentence. So maybe he didn't believe it very firmly (he only calls it "wahrscheinlich" - "probable"), or maybe he had more evidence than we even know today - some of his notes were destroyed. It's really hard to tell... $\endgroup$ – Markus Shepherd Feb 19 '17 at 19:13
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You'll probably get a better answer from someone else, but here's a fairly distant overview:

He takes a contour integral around the critical strip for the logarithm of (a relative of) the zeta function, and finds that it gives the same result, to the limit of his analysis, when you bring the contour down to the critical line. The residue theorem means that a contour integral is entirely defined by the poles inside the contour, and the poles of the logarithm would be the zeroes of the original function.

Take a look at his original paper, where the crux of the argument is found at the top of page 4 (not counting the cover).

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  • $\begingroup$ Not really. The only argument is One now finds indeed (numerically) approximately this number of real roots within these limits, and it is very probable that all roots are real. (p.5). In this only one paper he proves the functional equation and the Riemann explicit formula, not giving any clue for the Riemann hypothesis. $\endgroup$ – reuns Feb 19 '17 at 19:10
  • $\begingroup$ He doesn't say how; it seems to stand to reason that he found it by numerically evaluating the contour integral, since by numerical methods alone it would be difficult to positively identify a zero. $\endgroup$ – user361424 Feb 19 '17 at 23:50
  • $\begingroup$ There is a zero whenever $\text{arg } \zeta(s)$ isn't continuous or whenever $\int_C \frac{\zeta'(s)}{\zeta(s)}ds\ne 0$, both methods are equivalent and work by numerically evaluating $Im(\log \zeta(s)) =\text{arg } \zeta(s)$. Also there is a zero exactly on the critical line at every sign change of $\xi(1/2+it)$. This is what Riemann did. $\endgroup$ – reuns Feb 20 '17 at 0:00
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From Riemann's Zeta Function, By Harold M. Edwards:

enter image description here

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  • $\begingroup$ I don't understand really the argument, it doesn't seem very clear to me. See the Z function it isn't easy to understand and count its sign changes, the critical line theorem being proven much later by Littlewood and Selberg $\endgroup$ – reuns Feb 20 '17 at 0:07
  • $\begingroup$ @user1952009: What you are saying is fact. Nevertheless, Edwards stated clearly that (… we can never know …). Yet, he allowed himself to “guess” that Riemann suspected the hypothesis because of the nature of the tool that Riemann used to calculate the few first zeros (Riemann-Siegel formula). This (in my opinion) seems logic. $\endgroup$ – Hazem Orabi Feb 20 '17 at 0:31
  • $\begingroup$ I'm saying I don't understand what Edwards is referring to with "as long as the terms do not exhibit too much reinforcement" for approximating the number of sign changes of $Z(t)$ with $2 \cos \theta(t)$ $\endgroup$ – reuns Feb 20 '17 at 0:39
  • $\begingroup$ @user1952009: I think when he said "and as long as the terms do not exhibit too much reinforcement", he meant "the error terms": $$ Z(t)=2\,\sum_{n^2\lt\left(t/2\pi\right)}n^{-1/2}\,\cos\left[\vartheta(t)-t\log{n}\right]\color{red}{+R} $$ Where: $$ \color{red}{R}\sim(-1)^{N-1}\left(\frac{t}{2\pi}\right)^{-1/4}\times\left[C_{\small0}+C_{\small1}\left(\frac{t}{2\pi}\right)^{-1/2}+\cdots\right] $$ $\endgroup$ – Hazem Orabi Feb 20 '17 at 1:02
  • $\begingroup$ No, he said $Z(t) -2 \cos(\vartheta(t)) = 2\,\sum_{1 < n^2<(t/2\pi)}n^{-1/2}\,\cos\left[\vartheta(t)-t\log{n}\right]+ R(t)$ but there is no reason for it being negligible $\endgroup$ – reuns Feb 20 '17 at 1:45

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