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Are there examples of functions $f$ such that $\int_0^\infty f\text{d}x$ exists, but $\lim_{x\to\infty}f(x)\neq 0$?

I curious because I know for infinite series, if $a_n\not\to 0$, then $\sum a_n$ diverges. I'm wondering if there is something similar for improper integrals.

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    $f(x)=\begin{cases} 1 &\text{if } x \in \mathbb Z \\ 0 &\text{otherwise}\end{cases}$?2012-09-16
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    The classical [Fresnel Integral](http://en.wikipedia.org/wiki/Fresnel_integral)2012-09-16
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    https://math.stackexchange.com/questions/2527941/if-f-in-l1-bbb-r-dx-then-prove-that-for-almost-every-x-in-bbb-r-lim-lim2018-05-30
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    https://math.stackexchange.com/questions/2401286/does-this-integral-converge-or-diverge-int-bbb-r-left-frac2-cos-x3?noredirect=1&lq=12018-05-30

3 Answers 3

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It depends on what $\int$ means and what else you know about $f$. Here is a continuous example with limits of Riemann integrals:

Let $h(x)=\max\{1-|x|,0\}$ and set $f(x)=\sum_{n=1}^\infty (-1)^n h(nx-n^2)$. This function has up and down "bumps" around integers that become smaller and smaller in area, but have fixed height.

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    +1. Let me suggest to get rid of the signs, which are not necessary for a counterexample, and to present a function such as $f(x)=\sum\limits_nh(n^2x-n^3)$.2012-09-16
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Here is one more example: $$ f(x)=x\sin (x^4) $$ This is infinitely differentiable unbounded function without limit at infinity but with the finite improper Riemann integral over $\mathbb{R}_+$: $$ \int_0^{+\infty}x\sin(x^4)dx=\{t=x^4\}=\frac{1}{4}\int_0^{+\infty}\frac{\sin t}{\sqrt{t}} dt=\frac{1}{4}\sqrt{\frac{\pi}{2}} $$

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If $\lim_{x\to+\infty}f(x)=l>0$, then $\exists M>0:l-\varepsilonM$, and so

$$ \int_M^{+\infty}f(x)dx>\int_M^{+\infty}(l-\varepsilon)dx=+\infty $$

if $\varepsilon$ is sufficiently small.

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    -1, as this $\operatorname{lim}\neq 0$ doesn't mean $\operatorname{lim}=l>0$, as the other examples show.2012-09-16
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    @Vobo: I don't say this, there is an "If" as first word of the answer. Anyway I expect some downvotes.2012-09-16
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    So maybe you have a condition for your limit to be $0$. If $\displaystyle\int_0^{+\infty}f(x)dx$ is convergent and $\displaystyle\lim_{x\to +\infty} f(x)$ exists, then this limit is zero.2012-09-16
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    @enzotib: Ok, your statement is true, but it doesn't answer the question.2012-09-16
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    @Vobo: it at least completes the other answers, none of which points out explicitly that the only possibility is that the limit does not exist.2012-09-16