Recall that the Mertens function is defined via: $$M(n):=\sum_{n\ge x\ge 1} \mu(x)$$
Where $\mu$ is the Möbius function.
Littlewood proved that if $M\left(x\right)=o_x(x^{\frac{1}{2}+\varepsilon})$ for every $\varepsilon>0$ then Riemann Hypothesis follows. RH states that all the zeros of the Riemann zeta function $\zeta\left(s\right)$ in the critical strip: $0\leq Re\left(s\right)\leq1$, reside on the line $Re\left(s\right)=\frac{1}{2}$.
I am following the proof from the book Riemann's Zeta Function by Harold M. Edwards, page 260, in the book he states that:
$$\frac{1}{\zeta\left(s\right)}=s\int_{0}^{\infty}\frac{M\left(x\right)}{x^{s+1}}dx=s\int_{1}^{\infty}\frac{M\left(x\right)}{x^{s+1}}dx$$
And then proceed to say that if $M\left(x\right)$ grows less rapidly than $x^{\alpha}$ for some $\alpha>0$ than this integral for $\frac{1}{\zeta\left(s\right)}$ converges for all $s$ in the halfplane $\left\{ Re\left(a-s\right)<0\right\} =\left\{ Re\left(s\right)>a\right\} $ and therefore, by analytic continuation, the function $\frac{1}{\zeta\left(s\right)}$ is analytic in this halfplane. So substitute $\alpha$ with $\frac{1}{2}+\varepsilon$ will achieve that no poles exists in the half-plane $\left\{ Re\left(s\right)>\frac{1}{2}\right\}$.
So my question is this proves for $\left\{ Re\left(s\right)>\frac{1}{2}\right\}$ but what about the other half-plane $\left\{ 0<Re\left(s\right)<\frac{1}{2}\right\}$ ?