Question: I need to bound Riemann zeta function on the vertical line $s=1/2+\epsilon+it$ where $0\leq t\leq T+\epsilon$, $\epsilon>0$ is arbitrarily small and $T+\epsilon>3$ is not an ordinate of zero of zeta.
My try: This wiki page contains the following integral for $\zeta(s)$ with $s=\sigma+it$ valid for all $s\in \mathbb{C}$:
$$(s-1)\zeta(s) = \frac{\pi}{2}\int_{-\infty}^{+\infty} \frac{(\frac{1}{2}+ix)^{1-s}}{\cosh^2(\pi x)} dx \tag{1}$$
Denote the following $M=\max_{|z|=\frac{3}{2}}{\left|\zeta\left(\frac{1}{2}+iTz\right)\right|} $ where $T>3$.
So take $s=\frac{1}{2}+iTz$ so we have $s=\frac{1}{2}+\frac{3}{2}iT e^{i\theta}=\left(\frac{1-3T\sin \theta}{2}\right)+i\left(\frac{3T\cos \theta}{2}\right)$
Note that $s\neq 1$ since $s=1$ means $z=-\frac{i}{2T}$ so that $|z|=\frac{1}{2T}=\frac{3}{2}$ which gives $T=\frac{1}{3}$ and this contradicts the fact that $T>3$.
By $(1)$ we can write $$|\zeta(s)|\leq \frac{\pi}{2}\int_{-\infty}^{+\infty} \frac{\left|(\frac{1}{2}+ix)^{1-s}\right|}{\cosh^2(\pi x)} dx \tag{2}$$ $$\left|\left(\frac{1}{2}+ix\right)^{1-s}\right|=\left|\left(\frac{1}{2}+ix\right)^{1-\sigma-it}\right|=\left|\left(\frac{1}{2}+ix\right)^{1-\sigma}\right|\left|\left(\frac{1}{2}+ix\right)^{-it}\right| $$ $$\Rightarrow \left|\left(\frac{1}{2}+ix\right)^{1-s}\right|= \left(\frac{1}{4}+x^2\right)^{\frac{1-\sigma}{2}} e^{t\tan^{-1}(2x)}\tag{3}$$ So using $(3)$ in $(2)$ we have $$|\zeta(s)|\leq \frac{\pi}{2}\int_{-\infty}^{+\infty} \frac{\left(\frac{1}{4}+x^2\right)^{\frac{1-\sigma}{2}} e^{t\tan^{-1}(2x)}}{\cosh^2(\pi x)} dx \tag{4}$$ So splitting the integral in RHS of $(4)$ $$\frac{2}{\pi}|\zeta(s)|\leq \int_{0}^{+\infty} \frac{\left(\frac{1}{4}+x^2\right)^{\frac{1-\sigma}{2}} e^{t\tan^{-1}(2x)}}{\cosh^2(\pi x)} dx+\int_{-\infty}^{0} \frac{\left(\frac{1}{4}+x^2\right)^{\frac{1-\sigma}{2}} e^{t\tan^{-1}(2x)}}{\cosh^2(\pi x)} dx \tag{5}$$ In the second integral of $(5)$, subtitute $x=-y$ so we get $$\frac{2}{\pi}|\zeta(s)|\leq \int_{0}^{+\infty} \frac{\left(\frac{1}{4}+x^2\right)^{\frac{1-\sigma}{2}} (e^{t\tan^{-1}(2x)}+e^{-t\tan^{-1}(2x)})}{\cosh^2(\pi x)} dx \tag{6}$$ Hence $$\frac{2}{\pi}|\zeta(s)|\leq \int_{0}^{+\infty} \frac{\left(\frac{1}{4}+x^2\right)^{\frac{1-\sigma}{2}} (\cosh(t\tan^{-1}(2x))}{\cosh^2(\pi x)} dx \tag{7}$$ Now $\sigma=\frac{1-3T\sin \theta}{2}$ and $t=\frac{3T\cos \theta}{2}$
I am struggling to prove the required expression.
It will be better use another integral representation of $\zeta(s)$:
$$ \zeta(s)={s\over s-1}-s\int_1^\infty{x-\lfloor x\rfloor\over x^{s+1}}\mathrm dx $$
Set $s=\frac12+iT$, then we have for $T\ge2$ that $|\zeta(s)|=O(T)$. This indicates that $\log|\zeta(s)|<\log T+O(1)$, and your result follows.