If $f$, $g$ are positive functions, $f(x)\sim g(x)$ for large $x$ means that $\lim_{x\rightarrow \infty}f(x)/g(x)=1$. In the book "The Riemann Zeta function" by H. M. Edwards, in Chapter 4, as the last step in proving that $\psi(x)\sim x$ for large $x$ (where $\psi(x)$ is von Mangoldt's function), he first derives the asymptotics of the antiderivative of $\psi(x)$ as $$\int_0^x \psi(t) dt \sim \frac{x^2}{2},$$ and from that he concludes $\psi(x)\sim x$ with a proof that only uses the fact that $\psi(x)$ is positive and increasing. In other words, the proof in the book proves the following fact:
Suppose $f(x)$ is positive and increasing, and suppose that $\int_0^xf(t)dt \sim x^2/2$ for large $x$. Then $f(x)\sim x$ for large $x$.
I would like to know under what conditions on $g$ the same fact remains true if we replace $x^2/2$ with a differentiable function $g$. In other words, what conditions on $g$ will guarantee that if $\int_0^x f(t)dt \sim g(x)$, then $f(x)\sim g'(x)$, where $f$ is positive and increasing, but not necessarily continuous (van Mangoldt's function is not continuous).