I read this proof of the prime number theorem:
PRIME NUMBER THEOREM:
$ψ\left(x\right)-x=O\left(xe^{-c\sqrt{\log \left(x\right)}}\right)$ for some effective $c\in\mathbb{R}_{+}$
PROOF:
$ψ\left(x\right)-x=O\left(∑_{_{|γ|\leq T}}|\frac{x^{\rho}}{\rho}|+\frac{x}{T}\log ^2\left(x\right)\right)$ (where the sum is taken over the non trivial zeros $\rho+iγ$ of the $\zeta$ function with multiplicity what i know as explicit Riemann formula).
Than one has $∑_{_{|γ|\leq T}}|\frac{x^{\rho}}{\rho}|=O\left(x^{1-\frac{c}{\sqrt {\log \left(x\right)}}}\log ^2\left(T\right)\right)$ (I can give more datail if needed but the idea is using the zero free region to estimete $x^{\rho}$ and the Riemann Von Mngoldt formula for $∑\frac{1}{\rho}$).
So by substituting $T=e^{\sqrt {\log \left(x\right)}}$ one get the result.
MY OPINION
With this substitution i get $ψ\left(x\right)-x=O\left(x\log \left(x\right)e^{-c\sqrt{\log \left(x\right)}}\right)$ someone can help me to find the mistake please? Do i miss something?
Best Answer
Yes.
$x\log(x)e^{-c\sqrt{\log(x)}}) =xe^{\log\log(x)-c\sqrt{\log(x)}}) $ and since $\dfrac{\log\log(x)}{\sqrt{\log(x)}} \to 0$,
$\begin{array}\\ \log\log(x)-c\sqrt{\log(x)} &=-c\sqrt{\log(x)}(1-\frac{\log\log(x)}{\sqrt{\log(x)}})\\ &\lt -c_1\sqrt{\log(x)}\\ \end{array} $
for a different $c_1$.