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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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    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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    @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