Is there a collection of evidence and heuristic arguments against the Riemann hypothesis? [closed]

There is undoubtedly an overwhelming collection of evidence for the Riemann hypothesis. However, is there any evidence against it ?

• I don't know of any such list. I would imagine most entries on it would be controversial, with various counterarguments given to each piece of evidence. – Will Sawin Jul 3 '20 at 1:32
• @WillSawin, i heard that Littlewood was skeptical of it...Also, what do you think of my own heuristic above ? – user160539 Jul 3 '20 at 1:35
• I do not think people here will want to discuss your heuristic argument, because of MO's policy that it is not a place to discuss original research. – Will Sawin Jul 3 '20 at 1:36
• There is an article of Ivic discussing this: arxiv.org/pdf/math/0311162.pdf – Mayank Pandey Jul 3 '20 at 2:48
• Newman's conjecture (now a theorem proven in 2018 by Terry Tao and Brad Rodgers) may be seen as a clue that RH might be false. – Sylvain JULIEN Jul 3 '20 at 6:00

The mistake in the argument above is quite subtle and is easily seen by doing the above with $$M^2(x)=x$$, namely that while indeed $$g(X) \ll _{\varepsilon}X^{2\Theta-1+\varepsilon}\ll X^{\sigma+\Theta-1}$$ the $$<<_{\epsilon}$$ carries through so the inequality $$g(x) << X^{\sigma+\Theta-1}$$ is not absolute but it in fact is $$g(x) <<_{\sigma-\theta} X^{\sigma+\Theta-1}$$ so the final result is actually $$f(\sigma) <<_{\sigma-\Theta} \frac{2\sigma - 1}{\sigma-\Theta}$$ and obviosuly there is no contradiction since the $$<<_{\sigma-\Theta}$$ can (and actually does at least when $$\Theta=1/2$$) very well go to infinity too when $$\sigma \to \Theta$$.
As noted with $$M^2(x)=x$$ (which satisfies all the necessary stuff) we get $$f(\sigma)=1/(2\sigma-1)$$ (and indeed we get the required singularity at $$\Theta=1/2$$).
Then $$g(x) =\log x$$ and while indeed $$\log x << x^{\sigma -1/2}$$ the $$<<$$ depends on $$\sigma-1/2$$ so there is no contradiction in $$(2\sigma-1)f(\sigma) \to 1, \sigma \to 1/2$$ and $$f(\sigma)=(2\sigma-1)\int_1^{\infty} (\log x) x^{-2\sigma+1}dx <<_{\sigma-1/2}\int_1^{\infty} x^{-\sigma+1/2}dx <<_{\sigma-1/2}1/2$$ as the $$<<_{\sigma-1/2} \to \infty, \sigma \to 1/2$$
• the first inequality depends on $\epsilon$ so $g(X) \le C_{\epsilon}X^{2\Theta-1 + \varepsilon}$ – Conrad Jul 3 '20 at 3:36
• $D=D_{\sigma-\Theta}$ is not absolute unless $\sigma-\Theta > \delta >0$ fixed – Conrad Jul 3 '20 at 3:40