A few years ago, I assigned something a little weaker, $$\sum_{n\le x}{\Lambda(n)\over n}=\log x+r(x){\rm\ with\ }|r(x)|\le2$$ and it turned out to be the hardest problem on the assignment. Here's the proof I eventually wrote up.
Let $s=[x]$. Note $$\sum_{n\le x}{\Lambda(n)\over n}=\sum_{n\le s}{\Lambda(n)\over n}$$ and $0\le\log x-\log s\lt1$. Now
$$\log s!=\sum_{m\le s}\log m=\sum_{p^r\le s}\log p\sum_{m\le s,p^r\mid m}1=\sum_{p^r\le s}\log p[s/p^r]=\sum_{n\le s}\Lambda(n)[s/n]$$
$$=\sum_{n\le s}\Lambda(n)\left({s\over n}-\left\lbrace{s\over n}\right\rbrace\right)=s\sum_{n\le s}{\Lambda(n)\over n}-\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace$$
Now $$\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace\le\sum_{n\le s}\Lambda(n)\le Cs$$ for some constant $C$ by the Prime Number Theorem (which is actually overkill, as one can prove without PNT that that last sum is bounded by $2s$ for all $s$). Also, comparing $\log s!=\sum\log n$ to $\int_1^s\log t\,dt$ we get $$\log s!=s\log s-s+b(s){\rm\ with\ }|b(s)|\lt\log s+1$$ (cf. Stirling's formula). Putting it together, $$\sum_{n\le x}{\Lambda(n)\over n}=(1/s)\log s!+(1/s)\sum_{n\le s}\Lambda(n)\left\lbrace{s\over n}\right\rbrace=\log s-1+C+b(s)/s=\log x+r(x)$$ with $|r(x)|\le2$ since $C\le2$.
EDIT: The stronger result is proved in the textbook by Bateman and Diamond, Analytic Number Theory, pages 100-102. First they prove that, with the usual definitions, $\psi(x)\sim x$ implies $M(x)=o(x)$, where $M(x)$ is the summatory function for the Möbius $\mu$-function. Then from the result on $M(x)$ they deduce the result you want. It's a bit long to write out as I'd have to explain the notation introduced earlier in the chapter.
MORE EDIT: It's also Proposition 3.4.4 of Jameson's textbook, The Prime Number Theorem, and it's proved on page 91 of Tenenbaum and Mendes France, The Prime Numbers and Their Distribution, in the Student Mathematical Library series of the American Math Society. Again, the proofs require too much previously developed material for me to attempt to write them out here.
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 $c_1$.
Best Answer
The reasoning is flawed because $f\sim g$ most certainly does NOT imply $f’\sim g’$.
For example, take $f(x)\equiv 0$ and $g(x)= \frac1N \sin N^2x$.