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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?

Nine Thousand
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    There's a small issue is with your use of variables. I suggest switching the $x$ and $t$ around in your $2$ integral definitions so everything is then consistent. – John Omielan Aug 08 '19 at 04:27

2 Answers2

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No.

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

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

marty cohen
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  • I see, the discontinuity at 0 causes the integrals to differ. However, if the integrals began at 1 instead of 0, then the primitives would in fact be asymptotic, correct? Edit: in fact, there are cases in which they are, see here. – Nine Thousand Aug 08 '19 at 05:03
  • @trytryagain: Given that your integrals start at 0, the example should perhaps have been $f(x) = g(x) + \frac{1}{x+1}$, or something like that, instead. But the singularity at $x=0$ is just a slight annoyance which is really irrelevant here, the point is that the integral $\int_1^\infty \frac{1}{t} , dt$ is divergent. So even if you start integrating at 1, the primitives are not asymptotic. – Hans Lundmark Aug 08 '19 at 07:22
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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 $\int x dx = o(\frac{1}{2}x^2)$.

Varun Vejalla
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