I am reading the book of the prime number theorem by G.J.O.Jameson on p.192
If the Riemann Hypothesis were true, the contour of integration in the proof of Theorem 5.1.1 could be shifted almost to the line $ \operatorname{Re}(s) = \frac{1}{2} $. A modified version of the argument would then suggest that
$ \psi(x) - x = O(x^c) \quad \text{for every } c > \frac{1}{2}. $
where $ \psi(x) = \sum_{n \leq x} \Lambda(n) $ is the Chebyshev function
theorem 5.1.1 Consider a sequence $ a(n) $ and an associated function $ f(s) $ satisfying the following conditions:
(FT1) The Dirichlet series
$\sum_{n=1}^{\infty} \frac{a(n)}{n^s} $ converges absolutely to $ f(s) $ for $ \operatorname{Re}(s) > 1 $.
(FT2) The function $f(s) $ has the form
$ f(s) = \frac{\alpha}{s-1} + \alpha_0 + (s-1) h(s), $ where $ h(s) $ is differentiable at $s = 1 $.(FT3a) There exist constants $ c_0, M_0 $ such that $ f(s) $ is differentiable in the region $ |t| \geq 2, x \geq 1 - \frac{c_0}{\log |t|} $, and satisfies the bound
$ |f(x + i t)| \leq M_0 (\log |t|)^3. $(FT3b) The function $ f(s) $ is differentiable at all points $x + i t $ (except s = 1 ) for $ |t| \leq 2 $ and $ x \geq 1 - \frac{c_0}{\log 2} $.
then the summatory function
$ A(x) = \sum_{n \leq x} a(n) $ satisfies the integral bound
$ \left| \int_x^\infty \frac{A(y) - \alpha y}{y^2} \, dy \right| \leq K \exp(-c (\log x)^{1/2}) $ for sufficiently large x , where $ K, c $ are constants.
My question is:
How $\psi(x)-x$ is a bound of the form $ O(x^{1/2 + \epsilon}) $ directly from Theorem 5.1.1 under RH?
