Let $t \in \mathbb{R},f:\mathbb{R} \rightarrow \mathbb{R}:t \mapsto f(t)$.

$f$ is required to have:

  1. $\displaystyle\lim_{t \rightarrow \infty} f(t) = L$, where $L>0$ (could be $\infty$, so the limit exists, only it could be $\infty$)

  2. $\displaystyle\int_{0}^{\infty} f(t) dt < \infty$

Is that possible to construct such $f$?


$f$ can be any function continuous or discontinuous.

I found the discussion here which is involving "tent" function but in that example, the limit does not exist.

Note 2: I have edited the title. Thank you for your fedback !!!

Thank you for any answers or comments.

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    $\begingroup$ Your title and your body ask opposite questions. Anyway, the answer to your title question is "yes" and the answer to your body question is "no." $\endgroup$ – Qiaochu Yuan Feb 16 '12 at 1:16
  • $\begingroup$ Thank you for your comment... I am sorry for the confusion. I have corrected the title. Yes, the correct question in the body of this post... $\endgroup$ – netsurfer Feb 16 '12 at 1:38
  • 1
    $\begingroup$ After a while the function is greater than $L/2$ (if $L$ is finite) or greater than $1$ (if the limit is $+\infty$). $\endgroup$ – André Nicolas Feb 16 '12 at 1:41
  • $\begingroup$ As Qiaochu said, there is no such function. $\endgroup$ – David Mitra Feb 16 '12 at 1:42

If the improper integral $\int_0^\infty f(x)\, dx$ exists, then $\int_a^\infty f(x)\, dx$ exists for any number $a>0$.

Since $\lim\limits_{x\rightarrow\infty} f(x)=L>0$ (possible infinite), there are positive numbers $x_0$ and $\alpha$ so that $f(x)>{\alpha}$ for all $x\ge x_0$. Then $$ \int_{x_0}^\infty f(x)\,dx =\lim_{b\rightarrow\infty} \int_{x_0}^b f(x)\,dx\ge \lim_{b\rightarrow\infty} \bigl[ (b-x_0)\cdot\alpha \bigr]= \infty. $$

Thus $\int_0^\infty f(x)\,dx$ diverges to $\infty$.

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    $\begingroup$ Thank you, David... So I conclude that the limit should go to zero or undefined (not exist), in order the improper integral to converge $\endgroup$ – netsurfer Feb 16 '12 at 2:37
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    $\begingroup$ @netsurfer Yes; but "in order for the improper integral to $possibly$ be finite" $\endgroup$ – David Mitra Feb 16 '12 at 2:39
  • $\begingroup$ Yes, $possibly$ be finite. That's the more precise statement. $\endgroup$ – netsurfer Feb 16 '12 at 7:06

If $$\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = L$$

Then for some $M$ large enough

$$\int\limits_M^{M + 1} {f\left( t \right)dt \sim } L $$

So the integral as a whole will not converge. The idea is that the function will eventually be a constant, so integrating it will give the area of a rectangle of height $L$ and unbounded width.


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