In this paper is written that the prime number theorem in the form $\psi(x) = ( 1 + o(1) ) x$ is elementary equivalent to $$\sum_{n \le x } \frac{\Lambda(n)}{n} = \log x - \gamma + o(1) $$
I started to prove this as follows. By partial summation,
\begin{align*} \sum_{n \le x } \frac{\Lambda(n)}{n} &= \psi(x) \frac{1}{x} - \int_1^x \psi(t) d\left( \frac{1}{t} \right)\\ &= 1 + o(1) + \int_1^x \frac{t + o(1) t}{t^2} dt\\ & = 1 + \log x + \text{something} \end{align*} In order to treat the $o(1)$ inside the integral correctly, it seems natural to split the integral from $0$ to $x_0$ and from $x_0$ up to $x$, where $x_0$ is such that I can bound the $o(1)$ somehow. I cannot make it work however. I also think that I am on the wrong path because I do not see how the $\gamma$ will appear. Can somebody give a hint about how to prove the statement at the top of the page? I think that "elementary equivalent" means no complex analysis.
By Perron's formula we have $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+\infty}-\frac{\zeta'}{\zeta}\left(s+1\right)\frac{x^{s}}{s}ds $$ with $x\notin\mathbb{N} $. We have a double pole at $s=0 $ with residue $\log\left(x\right)-\gamma $ and it is possible to show (with the same tecniques used to show the prime number theorem in the form $\psi\left(x\right)=x+O\left(x\exp\left(-c_{1}\log^{1/2}\left(x\right)\right)\right) $) that the contribute of the other poles is infinitesimal. Then $$\sum_{n\leq x}\frac{\Lambda\left(n\right)}{n}=\log\left(x\right)-\gamma+o\left(1\right). $$