# Thread: Whether the limit of f(x) is 0 when improper integral is convergent?

1. ## Whether the limit of f(x) is 0 when improper integral is convergent?

Assume f(x) positive and continuous,for improper integrals of the first kind, is there an analog of the n-th term divergence test for series ? It would say:
$\displaystyle \int_{a}^{\infty} f(x)dx\ converges \Rightarrow \lim_{x \to +\infty} f(x) = 0.$
Give a proof or counterexample.

I think $\displaystyle \lim_{x \to +\infty} f(x) \neq 0$,but I can not give a counterexample.

2. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

Yes, that's correct. If $\displaystyle \int_0^\infty{f(x)}\ dx$ converges, and $\displaystyle \lim_{x\rightarrow\infty}f(x)$ exists, then it must be zero. Otherwise, $\displaystyle f(x)$ stays above some positive value $\displaystyle f_0$ when x is large enough, say $\displaystyle x>M$, and the integral is greater than the divergent integral $\displaystyle \int_M^\infty{f_0}\ dx$.

The assumption that $\displaystyle \lim_{x\rightarrow\infty}f(x)$ exists is critical. There are functions whose integral converges but have no limit as x goes to infinity. As a hint, consider the function defined on [0,1], $\displaystyle f(x)=1$ when $\displaystyle x<a_0$ and $\displaystyle f(x)=0$ when $\displaystyle x\ge{a_0}$. You can pick any number $\displaystyle a_0$ between 0 and 1 to be the value of its integral on [0,1].

- Hollywood

3. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

Originally Posted by zhongbeyond
Assume f(x) positive and continuous,for improper integrals of the first kind, is there an analog of the n-th term divergence test for series ? It would say:
$\displaystyle \int_{a}^{\infty} f(x)dx\ converges \Rightarrow \lim_{x \to +\infty} f(x) = 0.$
Give a proof or counterexample.
I think $\displaystyle \lim_{x \to +\infty} f(x) \neq 0$,but I can not give a counterexample.
Originally Posted by hollywood
The assumption that $\displaystyle \lim_{x\rightarrow\infty}f(x)$ exists is critical.
Where in the OP is that fact stated?

I don't think it is.

4. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

No, he didn't say, but it's the additional assumption he needs to prove his result.

Without that additional assumption, the result is false. And we know a little bit about what a counterexample might look like (or what it doesn't look like).

- Hollywood

5. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

sin(x^2) is a counterexample.

6. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

Originally Posted by hollywood
No, he didn't say, but it's the additional assumption he needs to prove his result.

Without that additional assumption, the result is false. And we know a little bit about what a counterexample might look like (or what it doesn't look like).

- Hollywood
Limit of f(x) can be $\displaystyle \infty$.
A graph of such function appears in my head:

Where f(x) have the following three properties:
(1)f(x) is a triangle on $\displaystyle [n,n+\frac{1}{n^{3}}]$;
(2)f(x)'s height is n;
(3)base is $\displaystyle \frac{1}{n^{3}}$.

So,f(x) is positive and continuous and converges.

But I can not find such expression for it.

7. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

Originally Posted by SworD
sin(x^2) is a counterexample.
This function is not positive.

8. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

Originally Posted by Plato
Where in the OP is that fact stated?

I don't think it is.
Limit of f(x) can be $\displaystyle \infty$.
A graph of such function appears in my head:

Where f(x) have the following three properties:
(1)f(x) is a triangle on $\displaystyle [n,n+\frac{1}{n^{3}}]$;
(2)f(x)'s height is n;
(3)base is $\displaystyle \frac{1}{n^{3}}$.

So,f(x) is positive and continuous and converges.

9. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

Yes thanks for the correction.

A simpler example can be a function which is zero everywhere except in [n, n+1/n^3], and inside such intervals it forms a triangle who's peak is at f(x) = 1. Basically, it would be a series of thinner and thinner triangles, each of the same height yet increasingly skinnier enough for convergence.

10. ## Re: Whether the limit of f(x) is 0 when improper integral is convergent?

It sounds like he wants something positive. You both have the right idea, though - take a function whose integral converges and add spikes. The spikes should become narrower so that the integral of the spikes converges, but the height of the spikes can not go to zero, meaning the limit of the function doesn't exist.

- Hollywood