Do asymptotic functions have asymptotic definite integrals?

Question

If $f(x) \sim g(x)$, and $$ I_f = \int_{0}^{t} f(x)dx $$ and $$ I_g = \int_{0}^{t} g(x)dx $$

then does $I_f \sim I_g$ (as $t$ goes to infinity) hold? If not, in what situations does it not hold? How would one go about proving such a relation? Furthermore, what might good sources for further reading be?

Answer 1

No.

You can have $f(x) \sim g(x)$ and $\int (f(x)-g(x))dx \to \infty$.

An easy example is $f(x) = g(x) + \frac1{x}$.

Answer 2

This should work. However, the error for the integral will be the integral of the error for the function. For example, let's say that you are approximating $f(x)$ using $g(x)$ and the error is $o(x)$. The new error will be $\int x dx = o(\frac{1}{2}x^2)$.