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Please help me to find a function with a given criterion.

I need a non-negative function $f(x)$ such that

  1. $\lim\limits_{x \rightarrow \infty }f(x) $ does not exists

  2. $\int\limits_0^{\infty}f(x)\ dx$ exists

I don't know whether such a function exist.

I have tried several function but none of them suit. If no such function exist then please help me to sketch a proof.

David G. Stork
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  • Any other constraints? For instance, does the function need to be continuous? Is this the Riemann integral? – John Griffin Aug 15 '17 at 15:05
  • To understand @zhw. 's example see https://en.wikipedia.org/wiki/Fresnel_integral –  Aug 15 '17 at 15:11
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    @EugenCovaci Unfortunately that example was not nonnegative. I deleted that comment, but probably shouldn't have. The function I mentioned is $f(x)= \sin (x^2).$ – zhw. Aug 15 '17 at 16:45
  • @zhw. Anyway, your example proves a function doesn't have to have a limit to $+ \infty$ for the improper integral to exist. –  Aug 15 '17 at 16:59

4 Answers4

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The typical example is to let $f$ be mostly $0$, but have narrower and narrower, evenly spaced spikes from $0$ to $1$ and back to $0$ again as $x\to \infty$. The narrowness of the spikes lets the integral converge, but stops $f$ from having a limit.

Arthur
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Consider the popcorn function/Thomae's function, which is defined on $[0,1]$. This has Riemann integral $0$, and is discontinuous only on the rationals. Make a copy of this on the higher intervals, e.g., $[1,2]$, $[2,3]$, $\dots$, $[n, n+1]$. Then, clearly, $f$ is non-negative, and $\lim \limits_{x \to \infty} f(x)$ doesn't exist (why?), and $\int \limits_{0}^{\infty} f(x)\,dx = \sum \limits_{n = 0}^{\infty} \int \limits_{n}^{n + 1} f(x)\,dx = \sum \limits_{n = 0}^{\infty} 0 = 0$.

EDIT: By the way, based on your answer history it looks like you proved Thomae's function is continuous on the irrationals, so it looks like you have some experience with this example.

layman
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Here's a real-analytic example:

$$f(x) = \left (\frac{2+\cos x}{3}\right )^{x^3}.$$

The function inside the parentheses equals $1$ at the points $2n\pi, n = 0,1,2,\dots$ Everywhere else it takes values in $[1/3,1).$ The exponent $x^3$ grows rapidly and smashes the thing down close to $0$ at a fast enough rate to lead to a convergent integral. (Why $x^3?$ It has something to do with Laplaces's method.)

I include this just for fun. Triangular spikes are the way to go for intuition.

zhw.
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Take $f:[0, + \infty) \rightarrow [0, + \infty)$ such that $f(x)=x$ if $x \in \mathbb{N}$ and $f(x)=0$ if $x \notin \mathbb{N}$

Also note the $f(x)$ is non-negative.

We have that the limit if $f(x)$ as $x \rightarrow + \infty$ does not exist.

From Lebesgue's theorem, $f$ is Riemman integrable because is discontinuous in a countable set and every countable set has measure zero.

A set $A$ has measure zero if $\forall \epsilon>0$ exist a sequence of open intervals $(a_n,b_n)$ such that $A \subseteq \bigcup_{n=1}^{\infty}(a_n,b_n)$ such that $\sum_{n=1}^{\infty}(b_n-a_n) \leqslant \epsilon$

Now $\int_0^{\infty}f(x)= \sum_{k=1}^{\infty} \int_{k-1}^kf(x)dx$

The O.P as an exercise he/she can prove that $\int_{k-1}^kf(x)dx=0, \forall k \in \mathbb{N}$ and conclude that $\int_0^{\infty}f(x)dx=0$

Marios Gretsas
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