I want to prove that \begin{align} \int_0^x O(t^n)dt \neq O(x^{n+1}) \end{align} At first I tried to partially integrate but it did not work.
Thanks,
Axel
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I don't understand exactly what the integral means. $O(t^n)$ is not a function, but a class of functions. Can you explain? – saulspatz Apr 15 '21 at 14:22
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My apologies for the last comment, I overlooked something – Stephen Donovan Apr 15 '21 at 14:31
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@saulspatz does it make more sense if the integrand is a(x)*x^n instead, where a(x) is limited function when close to 0? – axel Apr 15 '21 at 14:51
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If you want the integrand to be $O(n)$ then you would need $|a(x)|<C$ for $x>K$ for some positive constants $C$ and $K$. The behavior near $0$ doesn't affect that, but of course the integral needn't even converge. If $a$ is bounded, then the integral is $O(x^{n+1})$. For example, $\frac1x$ is $O(x^n)$ as $x\to\infty$ for any positive integer $n$. – saulspatz Apr 15 '21 at 15:01