Thanks!. I changed the title as you suggested!

In my opinion, "methods outside of number theory" means that they do not use any arithmetic functions like Mobius function $\mu(n)$. I would like to mention the Polya-Hurwitz approach. Please see the 2017 preprint by Shi (arxiv.org/abs/1706.08868). By truncating the Fourier Kernel $\Phi(t)$ and Fourier transformation integration range, Shi constructed a family of functions $\{F(n,z)\}_{n=9}^{\infty}$ that uniformly converge to the Riemann $\Xi(z)$ function in the critical strip $|Im(z)|<1/2$; Shi then proved that all the zeros of $W(n,z)$, a variant of $F(n,z)$, are real.

@reuns I read it several years ago and got the impression that is exactly like you said "the correct citation of Weil was "the Riemann Hypothesis would be settled by analysis rather than prime number theory""". But when I read it again today, I was not able to find the relevant part anymore. I was hoping that someone else can find the relevant part from reading it and thus consider it as a reference.

@AntonioVargas Thanks a lot for the info and continuing help. I will definitely read it. After I posted this question, I completed a MS in 2017 and uploaded it to arxiv.org/abs/1706.08868 . It is about using complex analysis method to study the zeros of Riemann $\Xi(z)$ function. I would like to ask you to briefly go through it see if there is any major gap. Best regards-

@user1952009. Thanks for the reference. I read his thesis before. It is nice one.

Thanks for the solution. What I really need is only the second part: If $f(z)$ and $g(z)$ are expressed by (1) and (2), then $f'(x)g(x)-g'(x)f(x)>0,x\in\mathbb{R}$ is sufficient for $f(z)-g(z)$ to have real zeros only. Is this statement correct?

@ChristianRemling. You are right. I only need to prove that $h(z)=f(z)-g(z)$ has only real zeros. Let $f_m(z),g_m(z)$ be the cut off version of the products up to $m$ terms. I think that I might be able to prove that $f'(x)g(x)-g'(x)f(x)>0,x\in\mathbb{R}$. But it might be hard for me to prove that $f'_m(x)g_m(x)-g'_m(x)f_m(x)>0$. Thanks again for the advice.