Return to Answer

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 $<<_{\sigma-1/2} \to \infty, \sigma \to 1/2$

(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 $g(X) \ll _{\varepsilon}X^{2\Theta-1+\varepsilon}\ll X^{\sigma+\Theta-1}$ 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 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 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 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) 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 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 $<<_{\sigma-1/2} \to \infty, \sigma \to 1/2$