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i did a search for such function but didn't found anything useful/complete ! , like this :

Integrable function $f$ on $\mathbb R$ does not imply that limit $f(x)$ is zero

is there any function that is integrable and $\lim_{x \to \infty}f(x) \neq0 $ and $\infty$ ??

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    $\begingroup$ Do you mean "is there any function that is integrable and [such that] $\lim_{x \to \infty}f(x) \neq0 $ and $\lim_{x\to \infty} f(x) \ne \infty$ ?? $\endgroup$ – Namaste Jan 19 '13 at 23:45
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    $\begingroup$ $f(x)=1$ for $x$ an integer and $0$ otherwise. More interesting examples can be obtained... $\endgroup$ – David Mitra Jan 19 '13 at 23:46
  • $\begingroup$ I'm not sure I understand the question. Like amWhy, I don't know what the last part means. If I may be picky, $\lim\limits_{x\to\infty}f(x)\neq 0$ is not the best way to state the negation of $\lim\limits_{x\to\infty}f(x)=0$ (because one of the possibilities, which will occur here, is that the limit doesn't exist). $\endgroup$ – Jonas Meyer Jan 20 '13 at 2:35
  • $\begingroup$ i'm sorry , the last question is a different one , i just wanted to avoid functions with infinit limit cuz i saw some example and they are complicated to my level ! , and yes even if the function has no limit it still answers the question ! i gotta study all these answers anyway . $\endgroup$ – Lofaif Jan 20 '13 at 4:17
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If the limit $L:=\lim_{x\to\infty} f(x)$ exists and is nonzero, then surely $\int_0^b f(x)\,dx$ grows essentially like $Lb$ as $b\to\infty$ (because for big $b$, $\int_{b}^{b+1} f(x)\,dx\approx \int_b^{b+1}L\,dx$). Note that the question you linked to talks about the $\limsup$, not the $\lim$.

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Take a function whose graph is a sequence of triangles whose bases are the $x$ axis, and the $n$-th triangle has size of $\frac1{n^2}$.

The integral of this function is finite, but there is no limit at $\infty$.

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YOu can do this with a picture. At each integer, draw a bump with area $1/2^n$ under it. This gives the graph of a function $f$ with $$\int_0^\infty f(x)\,dx = 1$$ but you do not have $f(x) \to 0$ as $x\to\infty$. In fact, you can draw these bumps as tall as you would like so you could have $$\limsup_{x\to\infty} f(x) = +\infty.$$

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You can also form a function using the geometric series. So let $f_n = \chi_{[n,n+ \frac{1}{2^n}]}$ and set $f = \sum_{n=1}^{\infty} f_n$. $f$ is 1 infinitely often so it doesn't tend to 0, but it's clearly integrable.

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$f(x) = \sum_{n=1}^\infty 1_{[n,n+n^{-2}]}(x)$

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  • $\begingroup$ @Sanchez thanks, fixed $\endgroup$ – Ilya Jan 20 '13 at 9:20
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If we're talking about Riemann integration, then we can also take $$f(x)=\sin(x^2).$$

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