Let's assume that $f:[1,\infty) \to [0,\infty)$.

Can someone provide an example of a function where the improper integral $\int\limits_1^{\infty} f(x)dx$ doesn't exist because we can find two sequences $\left(a_n\right)_{n\in\mathbb {N}}$ and $\left(b_n\right)_{n\in\mathbb {N}}$ with $\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}b_n=\infty$ such that $\lim\limits_{n\to\infty}\int\limits_1^{a_n} f(x)dx<\infty$, $\lim\limits_{n\to\infty}\int\limits_1^{b_n} f(x)dx<\infty$ and $\lim\limits_{n\to\infty}\int\limits_1^{a_n} f(x)dx \neq \lim\limits_{n\to\infty}\int\limits_1^{b_n} f(x)dx $?


1 Answer 1


I don't think such a case is possible, indeed $$g : x \rightarrow \int_1^x f(t)dt$$ is a monotonous function and as such admits a limit as $x \to \infty$. This limit can be either finite or $+\infty$ but in either case for any sequence $x_n$ that goes to $+\infty$, we have that $lim_{n\to\infty}g(x_n) \to lim_{x\to\infty}g(x)$.

  • $\begingroup$ Supposed we don't know that $g$ admits a limit (finite or infinite). Why is it not possible that $\int\limits_1^{x_n}f(t)dt$ jumps between two finite values as $x_n \to \infty$ where $x_n$ is not monotone? $\endgroup$
    – Philipp
    Commented Oct 16, 2020 at 18:02
  • $\begingroup$ Basically even if $x_n$ oscillates while going up to $\infty$, it still goes up to $\infty$ meaning that for any $M>0$, there will be a $n_0$ such that $x_n>M$ as long as $n>n_0$. Basically, at some point you can't go back down even if you oscillate. $\endgroup$ Commented Oct 19, 2020 at 11:14

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