Consider $f(s) = \frac{\zeta'(s)}{s \zeta(s)} - \frac{1}{s-1}$, $f_u(s) = \int_{1}^{u} \frac{\psi(x) - x}{x^{s+1}} dx$. Using these function it's possible to show asymptotic law (we just need to verify that $\lim_{u \to \infty} f_u(1) = f(1)$). It's easy to show that $f_u(s)$ is entire and $f(s)$ is analytic when $Re(s) > 1$. Also we may notice that $f(s)$ has a pole. The lemma given in my proof says : for $\Gamma(R,\mu)$ - is a right-semi circle $|s-1| = R$ and a polygonal chain with points $1+iR, \mu + iR,\mu - iR,1 - iR$ we have that $\displaystyle f(1)-f_u(1) = \frac{1}{2\pi i R} \int_{\Gamma(R,\mu)} (f(s)-f_u(s))u^{s-1} \left(\frac{s-1}{R} + \frac{R}{s-1}\right) ds$. The proof is about using residue theorem, but I'm not sure about it here. Maybe it should be smth like $\displaystyle f(1)-f_u(1) = \frac{1}{2\pi i R} \int_{\Gamma(R,\mu)} (f(s)-f_u(s))u^{s-1} \frac{R}{s-1} ds$?
The same proof is given by Simple Analytic Proof of the Prime Number Theorem