I have been reading about Riemann Zeta function and have been thinking about it for some time.
Has anything been published regarding upper bound for the real part of zeta function zeros as the imaginary part of the zeros tend to infinity?
Thanks
[Math] On Zeta function zeros in the critical strip
analytic-number-theoryriemann-zeta
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Hardy proved in 1914 that an infinity of zeros were on the critical line ("Sur les zéros de la fonction $\zeta(s)$ de Riemann" Comptes rendus hebdomadaires des séances de l'Académie des sciences. 1914).
Of course other zeros could exist elsewhere in the critical strip.
Let's exhibit the main idea starting with the Xi function defined by : $$\Xi(t):=\xi\left(\frac 12+it\right)=-\frac 12\left(t^2+\frac 14\right)\,\pi^{-\frac 14-\frac{it}2}\,\Gamma\left(\frac 14+\frac{it}2\right)\,\zeta\left(\frac 12+it\right)$$ $\Xi(t)$ is an even integral function of $t$, real for real $t$ because of the functional equation (applied to $s=\frac 12+it$) : $$\xi(s)=\frac 12s(s-1)\pi^{-\frac s2}\,\Gamma\left(\frac s2\right)\,\zeta(s)=\frac 12s(s-1)\pi^{\frac {s-1}2}\,\Gamma\left(\frac {1-s}2\right)\,\zeta(1-s)=\xi(1-s)$$ We observe that a zero of $\zeta$ on the critical line will give a real zero of $\,\Xi(t)$.
Now it can be proved (using Ramanujan's $(2.16.2)$ reproduced at the end) that : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}\cos(x t)\,dt=\frac{\pi}2\left(e^{\frac x2}-2e^{-\frac x2}\psi\left(e^{-2x}\right)\right)$$ where $\,\displaystyle \psi(s):=\sum_{n=1}^\infty e^{-n^2\pi s}\ $ is the theta function used by Riemann
Setting $\ x:=-i\alpha\ $ and after $2n$ derivations relatively to $\alpha$ we get (see Titchmarsh's first proof $10.2$, alternative proofs follow in the book...) : $$\lim_{\alpha\to\frac{\pi}4}\,\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh(\alpha t)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$ Let's suppose that $\Xi(t)$ doesn't change sign for $\,t\ge T\,$ then the integral will be uniformly convergent with respect to $\alpha$ for $0\le\alpha\le\frac{\pi}4$ so that, for every $n$, we will have (at the limit) : $$\int_0^\infty\frac{\Xi(t)}{t^2+\frac 14}t^{2n}\cosh\left(\frac {\pi t}4\right)\,dt=\frac{(-1)^n\,\pi\,\cos\bigl(\frac{\pi}8\bigr)}{4^n}$$
But this is not possible since, from our hypothesis, the left-hand side has the same sign for sufficiently large values of $n$ (c.f. Titchmarsh) while the right part has alternating signs.
This proves that $\Xi(t)$ must change sign infinitely often and that $\zeta\left(\frac 12+it\right)$ has an infinity of real solutions $t$.
Probably not as simple as you hoped but a stronger result! $$-$$
From Titchmarsh's book "The Theory of the Riemann Zeta-function" p. $35-36\;$ and $\;255-258$ :
For $\mathrm{Re}(s)\gt1$, $$ \zeta(s)=\sum_{n=1}^\infty\frac1{n^s}\tag{1} $$ For $\mathrm{Re}(s)\le1$, we need to use Analytic Continuation and another formula for $\zeta(s)$ since $(1)$ does not converge for $\mathrm{Re}(s)\le1$.
One formula that can be used is the Functional Equation for $\zeta$ which says $$ \zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}=\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}\tag{2} $$ The symmetric form in $(2)$ is given as $(14)$ on MathWorld.
Since the recurrence for $\Gamma$ says that for $n\in\mathbb{Z}$ and $n\le0$, $\frac1{\Gamma(n)}=0$, we get that $\zeta(2n)=0$ for $n\in\mathbb{Z}$ and $n\lt0$.
Analytic Continuation of $\boldsymbol{\zeta}$ Using $\boldsymbol{\eta}$ and Integration by Parts
Define $\eta$, the alternating $\zeta$ function, as $$ \begin{align} \eta(s) &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\\ &=\sum_{n=1}^\infty\frac1{n^s}-2\sum_{n=1}^\infty\frac1{(2n)^s}\\[6pt] &=\zeta(s)\left(1-2^{1-s}\right)\tag{3} \end{align} $$ Formula $(3)$ follows since the terms of an alternating series are the terms of the non-alternating series minus twice the even terms.
Formula $(3)$ increases the domain to $\operatorname{Re}(s)\gt0$. $\eta$ also comes in handy to define $$ \eta(s)\,\Gamma(s)=\int_0^\infty\frac{t^{s-1}}{e^t+1}\,\mathrm{d}t\tag{4} $$ which also converges for $\operatorname{Re}(s)\gt0$. However, we can integrate $(4)$ by parts $k$ times to get $$ \bbox[5px,border:2px solid #C0A000]{\eta(s)\,\Gamma(s)=\frac{(-1)^k}{s(s+1)\cdots(s+k-1)}\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\frac1{e^t+1}\,\mathrm{d}t}\tag{5} $$ $(5)$ agrees with $(4)$ for $\operatorname{Re}(s)\gt0$ and converges for $\operatorname{Re}(s)\gt-k$. Thus, $(5)$ gives an analytic continuation of $\zeta(s)$ for $\operatorname{Re}(s)\gt-k$.
Using this method, $\zeta(0)=-\frac12$ is computed in this answer and $\zeta(-1)=-\frac1{12}$ is computed in this answer.
Proving $\boldsymbol{\zeta(-2n)=0}$
Note that $(5)$ can be rewritten as $$ \eta(s)\,\Gamma(s+k)=(-1)^k\int_0^\infty t^{s+k-1}\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\tag{6} $$ Setting $s=1-k$ in $(6)$ gives $$ \begin{align} \eta(1-k) &=(-1)^k\int_0^\infty\frac{\mathrm{d}^k}{\mathrm{d}t^k}\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\,\mathrm{d}t\\[6pt] &=(-1)^{k-1}\frac{\mathrm{d}^{k-1}}{\mathrm{d}t^{k-1}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{7} \end{align} $$ Set $k=2n+1$ and we get $$ \eta(-2n) =\frac{\mathrm{d}^{2n}}{\mathrm{d}t^{2n}}\left.\left(\frac12-\frac12\frac{e^{t/2}-e^{-t/2}}{e^{t/2}+e^{-t/2}}\right)\right|_{t=0}\tag{8} $$ For $n\ge1$, the right side of $(8)$ is an odd function, so evaluating at $t=0$, and applying $(3)$, yields $$ \bbox[5px,border:2px solid #C0A000]{\zeta(-2n)=0}\tag{9} $$
Analytic Continuation of $\boldsymbol{\zeta}$ Using the Euler-Maclaurin Sum Formula
As described in this answer, this answer, and this answer, for $\operatorname{Re}(s)\gt-2m-1$, $\zeta(s)$ can be represented as $$ \hspace{-12pt}\bbox[5px,border:2px solid #C0A000]{\zeta(s)=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac{1}{k^s}-\left(\frac{1}{1-s}n^{1-s}+\frac12n^{-s}-\sum_{k=1}^m\frac{B_{2k}}{2k}\binom{s+2k-2}{2k-1}n^{-s-2k+1}\right)\right]}\tag{10} $$ where the formula in the parentheses is obtained from the Euler-Maclaurin Sum Formula.
Proving $\boldsymbol{\zeta(-2m)=0}$
If we plug $s=-2m$ into $(10)$, for $m\ge1$, the formula in parentheses exactly matches the sum outside the parentheses by Faulhaber's Formula. In fact, the Euler-Maclaurin Sum Formula is one way to prove Faulhaber's Formula. This means that $$ \bbox[5px,border:2px solid #C0A000]{\zeta(-2m)=0}\tag{11} $$ We can also plug $s=-2m+1$, for $m\ge1$, into $(10)$ and note that Faulhaber's Formula does not include the constant term. Thus, we get $$ \zeta(-2m+1)=-\frac{B_{2m}}{2m}\tag{12} $$
Functional Equation for $\boldsymbol{\zeta}$
The Fourier Transform of $e^{-\pi x^2t}$ is $$ \begin{align} \int_{-\infty}^\infty e^{-\pi x^2t}e^{-2\pi ix\xi}\,\mathrm{d}x &=\int_{-\infty}^\infty e^{-\pi(x-i\xi/t)^2t}e^{-\pi\xi^2/t}\,\mathrm{d}x\\ &=\frac1{\sqrt{t}}e^{-\pi\xi^2/t}\tag{13} \end{align} $$ Applying the Poisson Summation Formula to $(13)$ says that $$ 1+2\sum_{n=1}^\infty e^{-\pi n^2t} =\frac1{\sqrt{t}}+\frac2{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\tag{14} $$ Note that for $s\gt1$, $$ \begin{align} &\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}\\ &=\sum_{n=1}^\infty\frac1{n^s}\int_0^\infty e^{-\pi t}t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=\int_0^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=\int_0^1\left(\frac1{2\sqrt{t}}-\frac12+\frac1{\sqrt{t}}\sum_{n=1}^\infty e^{-\pi n^2/t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t +\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{1-\large s}2}\,\frac{\mathrm{d}t}t +\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^{\frac{\large s}2}\,\frac{\mathrm{d}t}t\\ &=-\frac1{1-s}-\frac1s+\int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)\left(t^{\frac{1-\large s}2}+t^{\frac{\large s}2}\right)\,\frac{\mathrm{d}t}t\tag{15} \end{align} $$ The following integral is increasing in $\alpha$. For $\alpha\ge0$, we have $$ \begin{align} \int_1^\infty\left(\sum_{n=1}^\infty e^{-\pi n^2t}\right)t^\alpha\,\mathrm{d}t &=\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\int_{\pi n^2}^\infty e^{-t}\,t^\alpha\,\mathrm{d}t\\ &\le\sum_{n=1}^\infty\frac1{\pi^{\alpha+1}n^{2\alpha+2}}\left(\int_{\pi n^2}^\infty e^{-t}\,\mathrm{d}t\right)^{1/2}\left(\int_{\pi n^2}^\infty e^{-t}t^{2\alpha}\,\mathrm{d}t\right)^{1/2}\\ &\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac{e^{-\pi n^2/2}}{n^{2+2\alpha}}\\ &\le\frac{\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+1}}\sum_{n=1}^\infty\frac2{\pi n^{4+2\alpha}}\\ &=\frac{2\,\Gamma(2\alpha+1)^{1/2}}{\pi^{\alpha+2}}\zeta(4+2\alpha)\tag{16} \end{align} $$ Thus, the last integral in $(15)$ defines an entire function. Therefore $(15)$ defines $\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}}$ for all $s\in\mathbb{C}$. Since $(15)$ is invariant under $s\leftrightarrow1-s$, we have $$ \bbox[5px,border:2px solid #C0A000]{\zeta(s)\frac{\Gamma(s/2)}{\pi^{s/2}} =\zeta(1-s)\frac{\Gamma((1-s)/2)}{\pi^{(1-s)/2}}}\tag{17} $$
Best Answer
The term in analytic number theory is "zero-free regions". Any proof of the prime number theorem will produce such a region, and the region is equivalent to the error term in bounds for $\pi(x)$ and to the lower bounds in nonvanishing theorems $|\zeta (1+it)|> 0$. At present, the known zero-free regions are asymptotic to the line $Re(s)=1$: at height $h$ all zeros are at least at distance $d(h)$ from the line with $\lim_{|h| \to \infty} d(h) = 0$.
(Added: on the subject of later improvements, if any region not asymptotic to $Re(s)=1$ were demonstrated it would be a giant advance in number theory, comparable to Wiles' proof of the modularity conjecture and Fermat's Last Theorem. In the analogous function field case there is an algebraic technique for boosting Beta < 1 to Beta < (1/2)+epsilon, and the latter is the Riemann hypothesis. You can be sure that if a zero free region were proven that had finite distance from the boundary of the critical strip, we would all have heard about it. Some known bounds are listed at http://www.math.uiuc.edu/~ford/wwwpapers/zeros.pdf )