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Prove or disprove:

a) $f$ is Reimann integrable in the interval $[1,b]$ $\forall$ $b>1$, $f(x) > 0$ $\forall x$, and $\int_1^\infty f(x)dx $ is convergent. Then, $$\lim_{x\rightarrow\infty}f(x)=0$$ b) Does your conclusion remain the same if $f(x)$ can take any value?

I am struggling with this question. Here is what I know but I can't bring the pieces together. Need a rigorous proof.

For part a, since $\int_1^\infty f(x)dx $ is convergent, $\int_1^\infty f(x)dx = \lim_{x \to \infty}\int_1^x f(u)du $ exists.

Intuitively, I think I should show that since the function is positive and the integral is convergent, it must be that there exist a large enough $N$ such that $\forall n >N$, $f(x_n)-f(x_{n-1})<\epsilon$ for any arbitrary $\epsilon>0$ which should imply that the limit of the function is zero. But I am not sure about this.

For part b, my guess is that the conclusion will not hold as we need the function to be positive to prove the statement in part a. But I don't know how can I show it.

Thanks for your help.

Sher Afghan
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  • Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero. – Eclipse Sun Jan 22 '19 at 22:56
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    I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x \notin \mathbb{N}$ and $f(x) = 1$ if $x \in \mathbb{N}$. Maybe you also want assumption that $f$ is continuous? – prosinac Jan 22 '19 at 22:59
  • @prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/ – Sher Afghan Jan 22 '19 at 23:22
  • @SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer. – RRL Jan 22 '19 at 23:27
  • @RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity. – Sher Afghan Jan 22 '19 at 23:30
  • Also think about $\int_1^\infty \sin(x^2) , dx$. – RRL Jan 22 '19 at 23:47
  • @RRL but $sin(x^2)$ is not positive for all $x>1$. Am I missing something? – Sher Afghan Jan 23 '19 at 00:04
  • This helps you with part (b). Also Riemann integrable function does not have to be continuous everywhere. It can have finite, or countable or generally measure zero discontinuities. But that gives you a trivial counterexample. – RRL Jan 23 '19 at 00:09
  • "N such that ∀n>N, f(xn)−f(xn−1)<ϵ for any arbitrary ϵ>0 which should imply that the limit of the function is zero." That's almost the definition for limit of a sequence (you're missing absolute value bars), but it doesn't work for a continuous function. For instance, suppose $f(x)$ is $0$ on the rationals and $1$ on the irrationals. Then for any $n$, $f(n)-f(n-1)=0$, but $\lim_{x \rightarrow \infty}$ doesn't exist. – Acccumulation Jan 23 '19 at 16:32
  • @Acccumulation Yeah I realized that I was on a completely wrong track. – Sher Afghan Jan 23 '19 at 18:07

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If $f$ is positive (and even continuous) , then we can have $\int_1^\infty f(x) \, dx$ converge and $f(x) \not\to 0$ as $x \to \infty$.

Take $f(x) = x^{-2} + \sum_{k=2}^\infty \phi_k(x)$ where

$$\phi_k(x) = \begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} \leqslant x \leqslant k \\ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\\ 0& \text{otherwise} \end{cases}$$

In this case $\limsup f(x) = 1$ and $\liminf f(x) = 0$ so the limit does not exist, but $$\int_1^\infty f(x) \, dx = 1 + \sum_{k=2}^\infty \frac{1}{k^2} < \infty$$

RRL
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