Asymptotic behaviour of a counting function I was studying a proof of Prime Number Theorem from Stein's Complex Analysis:

Theorem: Let $\pi(x)$ be the prime counting function. Then
  $$ \pi(x) \sim \frac{x}{\log x}. $$

The proof makes sense, but is mysterious to me since lots of steps seem arbitrary. I tried to dig into why each step should be carried out, and contemplate the following scheme.
Let $ 0 < p_1 < p_2 < ... $ be any sequence of increasing positive real numbers. Define the counting function to be
$$ \pi(x) := \{p_i < x\}. $$
Throughout this post, we are interested in the asymptotic behaviour of this function. To attack this problem, one defines one of its sibling function
$$ \psi(x) := \Sigma_{p_i < x} \log p_i.$$
My guess is that 

Guess: Given the above notations
  $$\pi(x) \log(x) \sim \psi (x);$$ however, I don't have a proof.

EDIT: As pointed out below, $\pi(x)\log(x) \sim \psi(x)$ means $\pi(x)\log(x)/\psi(x)\to 1$, and there is a counterexample given below too. What I originally meant in my guess is $\pi(x)\log(x) \approx \psi(x)$, meaning the limsup and liminf of the quotient are finite values.
Questions


*

*Is my guess correct?

*If my guess is correct, why does one pass from $\pi$ to its sibling $\psi$? I have heard this might do something with the meromorphicity of something, but I cannot make it straightforward.

*In Stein's book, he further more defined another sibling
$$ \psi_1 := \int \psi. $$ Why this extra step?



Another amazing ingredient for me is the Perron-like formulae. Here are some of them:
Denote $(c)$ be the straight line from $c-i\infty$ to $c+i\infty$ for some fixed positive number $c$. Then
$$ \int_{(c)} \frac{x^s}{s}\,ds $$
is either $1, \frac{1}{2}$, or $0$ depends on the relation between $x$ and $1$. Apparently, this was a weapon for number theorists to attack counting problems with contour integral methods. Another variation is to change the $s$ in the denominator to $s(s+1)$. See more on this wikipedia page.
Question


*I am very curious about this idea's history and any applications about it. If you happen to know more about it, please let me know.


Thank you very much in advance!
 A: In fact, $$\pi(x)\sim\frac{x}{\ln x}\iff\psi(x)\sim x$$
Let $1<y<x$,
$$\pi(x)-\pi(y)=\sum_{y<p\leqslant x}{1}\leqslant\sum_{y<p\leqslant x}{\frac{\ln p}{\ln y}}\leqslant\frac{\psi(x)}{\ln y}.$$ In particular, if $x>e$, take $y=\frac{x}{\ln^2(x)}$ in the above inequality. Moreover, if $x>1$, 
$$ \psi(x)=\sum_{p\leqslant x}{\ln p}\leqslant\sum_{p\leqslant x}{\left\lfloor\frac{\ln x}{\ln p}\right\rfloor\ln p}\leqslant\pi(x)\ln x.$$ In the end
$$ \forall x>e,\,\frac{\psi(x)}{x}\leqslant\frac{\pi(x)\ln x}{x}\leqslant\frac{1}{\ln x}+\frac{\psi(x)\ln x}{x(\ln x-2\ln(\ln x))} $$
It follows directly that $\pi(x)\sim\frac{x}{\ln x}\iff\psi(x)\sim x$.
Concerning your last formula, the Hadamard's factorization theorem states that there exists $(a,b)\in\mathbb{C}^2$ such that
$$ \zeta(s)=\frac{e^{as+b}}{s-1}\prod_{\rho\in\Omega}{\left(1-\frac{s}{\rho}\right)e^{\frac{s}{\rho}}} $$
where $\Omega$ is the set of the roots of $\zeta$. Take the logarithm and differentiate, you have
$$ \frac{\zeta'(s)}{\zeta(s)}=\frac{\zeta'(0)}{\zeta(0)}+\frac{s}{1-s}+\sum_{\rho\in\Omega}{\frac{s}{\rho(s-\rho)}} $$
Let $\overset{\sim}{\Omega}$ the set of the non trivals zeros of $\zeta$. Using your formula, you then have that
$$ \psi(x)=x-\frac{\zeta'(0)}{\zeta(0)}-\frac{1}{2}\ln\left(1-\frac{1}{x^2}\right)-\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{x^{\rho}}{\rho}} $$
for all $x$ that can't be written as $p^k$ with $k\in\mathbb{N}^*$ (you can deduce the general equation by adding $\pm 1$ in one side of the previous equality) and $$ \frac{\psi(x)}{x}=1-\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{x^{\rho-1}}{\rho}}+\underset{x\rightarrow +\infty}{o}(1) $$
However, $$\left|\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{x^{\rho-1}}{\rho}}\right|\leqslant\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{x^{\mathrm{Re}(\rho)-1}}{\rho}}\leqslant x^{\frac{\mathrm{Re}(\rho)-1}{2}}\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{x^{\frac{\mathrm{Re}(\rho)-1}{2}}}{\rho}}$$
Since the non trivial roots of $\zeta$ have a real part in $]0,1[$, you have that $$x^{\frac{\mathrm{Re}(\rho)-1}{2}}=\underset{|\rho|\rightarrow +\infty}{o}\left(\frac{1}{|\rho|^2}\right) $$ and a lemma states that $\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{1}{|\rho|^2}}$ converges. Finally
$$ \lim\limits_{x\rightarrow +\infty}{x^{\frac{\mathrm{Re}(\rho)-1}{2}}\sum_{\rho\in\overset{\sim}{\Omega}}{\frac{x^{\frac{\mathrm{Re}(\rho)-1}{2}}}{\rho}}}=0 $$ and $$\psi(x)\sim x$$ so that $$\pi(x)\sim\frac{x}{\ln x}$$
A: The guess seems to be wrong. Let $p_i=2^i$. Then $\pi(x)$ is roughly $\log_2x$, and $$\psi(x)=\sum_{i<\log_2x}i\log2$$ is roughly $\log2(\log_2x)^2/2=(\log x)^2/(2\log2)$, whereas $\pi(x)\log x$ is essentially $\log_2x\log x=(\log x)^2/\log2$. The guess is off by a factor of $2$ in this case. 
