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

1$\begingroup$ 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. $\endgroup$ – Will Sawin Jul 3 at 1:32

$\begingroup$ @WillSawin, i heard that Littlewood was skeptical of it...Also, what do you think of my own heuristic above ? $\endgroup$ – OneTwoOne Jul 3 at 1:35

3$\begingroup$ 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. $\endgroup$ – Will Sawin Jul 3 at 1:36

7$\begingroup$ There is an article of Ivic discussing this: arxiv.org/pdf/math/0311162.pdf $\endgroup$ – Mayank Pandey Jul 3 at 2:48

1$\begingroup$ 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. $\endgroup$ – Sylvain JULIEN Jul 3 at 6:00
(note that the original question before being edited out had an argument about why RH is false and the post below was a refutation of that); the Ivic paper linked by @Mayank contains some good arguments and is worth reading for sure
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 $g(X) \ll _{\varepsilon}X^{2\Theta1+\varepsilon}\ll X^{\sigma+\Theta1}$ the $<<_{\epsilon}$ carries through so the inequality $g(x) << X^{\sigma+\Theta1}$ is not absolute but it in fact is $g(x) <<_{\sigma\theta} X^{\sigma+\Theta1}$ 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 $f(\sigma)=1/(2\sigma1)$ (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 $\sigma1/2$ so there is no contradiction in $(2\sigma1)f(\sigma) \to 1, \sigma \to 1/2$ and $f(\sigma)=(2\sigma1)\int_1^{\infty} (\log x) x^{2\sigma+1}dx <<_{\sigma1/2}\int_1^{\infty} x^{\sigma+1/2}dx <<_{\sigma1/2}1/2$ as the $<<_{\sigma1/2} \to \infty, \sigma \to 1/2$