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Find an example of a function $f:[0,\infty)\rightarrow \mathbb{R}$ integrable on all intervals such that $\lim_{n\rightarrow \infty}\int_0^n f$ converges as a limit of a sequence, but such that $\int_0^\infty f$ does not exist.


I think a kind of function whose integral oscillates between positive and negative region as $n\rightarrow \infty$ may do the job here but I am not sure. Appreciate your help.

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    $\begingroup$ The statement is surely missing an $n$ somewhere; presumably the limit is of the integral over $[0,n]$? Otherwise, you have the limit of a constant. $\endgroup$ – Arturo Magidin Feb 3 at 23:45
  • $\begingroup$ Yes there was a typo, fixed it. $\endgroup$ – Sher Afghan Feb 3 at 23:47
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    $\begingroup$ Your idea is fine. Take $f(x) = \sin(2\pi x)$. $\endgroup$ – Arturo Magidin Feb 3 at 23:51
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Just define $f$ separately on the intervals $(n,n+\frac 1 2)$ and $(n+\frac 1 2,n+1)$, $n=1,2,...,$ so that $\int_n^{n+\frac 1 2} f(x)dx=1$ and $\int_{n+\frac 1 2} ^{n+1} f(x)dx=-1$.

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    $\begingroup$ E.g. $f(x)=2$ for $n\le x<n+1/2$ and $f(x)=-2$ for $n+1/2\le x<n+1.$ $\endgroup$ – DanielWainfleet Feb 4 at 3:03

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