Relating the two errors terms in the prime number theorem

Question

I would like to show that the two errors terms in the prime number theorem $\pi(x)-\frac{x}{\log{x}}$ and $\psi(x)-x$ are quite similar (or differing by something like a factor of $(\log{x})^{\varepsilon}$). This is my attempt: by partial summation, $$ \pi(x)= \frac{\psi(x)}{\log{x}}+\int_{2}^{x}\frac{\psi(t)}{t\log^2(t)}dt + O(\sqrt{x}). $$ This shows that $$ \pi(x)-\frac{x}{\log{x}} = \frac{\psi(x)-x}{\log{x}}+\int_{2}^{x}\frac{\psi(t)}{t\log^2(t)}dt + O(\sqrt{x}). $$ The problem now is that the the order of the integral is bigger than the error term. So I was wondering what would be the way of relating the two errors. Intuitively they should be the same but I’ve been not able to prove this formally

Answer 1

In fact they're not the same! We do expect $\psi(x)-x$ to be very roughly $\sqrt x$ in size (that's the Riemann hypothesis). However, we would only expect something this good for $\pi(x)-\mathop{\rm li}(x)$ where li$(x) = \int^x \frac{dt}{\log t}$ is the logarithmic integral. For $\pi(x)-\frac x{\log x}$ the error term is provably of size $\frac x{(\log x)^2}$.